Mistake museum
The exhibits here are not silly slips — they are mistakes that clever people make too. For each one, find the flaw yourself first, then open the diagnosis. 224 exhibits so far.
Diagnosis It is a product, yes, but is still a difference of squares, so it can be split further. It is like writing as and handing in your test.
“Keep going until no factor can be split any further”
From: What Is Factorization?
Diagnosis The last operation is an addition, so the whole thing is still a sum. A factorization must be a product as a whole.
“Look at the last operation. It must be multiplication”
From: What Is Factorization?
Diagnosis The second term is taken out completely, but what is left is not “nothing”: it is . Drop it, and multiplying back gives only : a term is missing.
“When a whole term is taken out, a 1 stays behind”
Diagnosis In the bracket, and still share the factor . Take out the largest common factor: the gcd of the coefficients, and each letter to its lowest power.
“After taking it out, look inside the bracket once more”
Diagnosis Taking out means dividing every term by : , not . Multiply back: , the wrong sign.
“Take out a minus sign, and every sign in the bracket flips”
Diagnosis A square of side contains and , and also two pieces of size . Leave them out and the area is short by .
“A perfect square has three terms, and the middle one is 2ab”
From: Using the Formulas
Diagnosis is still a difference of squares, so you can use the formula again.
“After using a formula, check every factor again”
From: Using the Formulas
Diagnosis Multiply back: , which has two extra middle terms. In the difference of cubes, the second bracket has plus , so that the middle terms cancel.
“Cube formulas: same sign first, opposite sign after”
From: Using the Formulas
Diagnosis The second group is . When you take out , , so the bracket should be , not .
“With a minus sign in front of a bracket, every sign inside flips”
From: Factoring by Grouping
Diagnosis The last operation on the right is still an addition: it is a sum of two terms, not a product.
“Grouping is only a means; the result must be a product”
From: Factoring by Grouping
Diagnosis An extra appeared out of nowhere, so the value changed. The first equals sign is already false: at the left side is and the right side is .
“Whatever you add, you must subtract”
Diagnosis “ cannot be factored” is about and treated as letters. Here happens to be a perfect square, so adding it and subtracting it produces a difference of squares.
“A sum of squares can hide a difference of squares”
Diagnosis The sum really is , but : the constant has the wrong sign. The constant term is negative, so the two numbers must have opposite signs.
“Check the sign of the product before you hunt for the sum”
From: The Cross Method
Diagnosis Multiply back and the middle term is , not . When the leading coefficient is not , where each number sits changes what the cross products add up to.
“It is a cross, so the positions are part of the answer”
From: The Cross Method
Diagnosis The constant is right, but the coefficient of comes out as , not , and the coefficient of is , not . Two of the three crosses are wrong.
“The double cross has three crosses, and you must check every one”
Diagnosis The right-hand side is still a sum. Factoring the quadratic part is only the first cross; the linear terms and the constant have not gone into the brackets yet.
“The first two columns are only the start; the third column finishes the job”
Diagnosis The equation is true, but each bracket still has a common factor , so it is not factored completely.
“Common factor first, formula second. Never the other way round”
From: Putting It All Together
Diagnosis There is no in the problem; it is only a temporary label. After substituting back to , both factors are still quadratics with three terms, so you have to keep checking.
“Substitute back, then check again”
From: Putting It All Together
Diagnosis The divisor is at . The Remainder Theorem says: dividing by leaves the remainder ; and , so .
“Divide by x + 2, substitute −2”
Diagnosis The leading coefficient is , so the candidates also include , with denominator . : the factor was missed.
“Numerator from the constant term, denominator from the leading coefficient”
Diagnosis One equation was never checked: the coefficient of , , should equal , but . There are more equations than unknowns, and every one of them must hold.
“Once you have solved, substitute into the equations you did not use as well”
Diagnosis is another way to split it. Try : and give , .
“When splitting the constant, remember that minus times minus is plus”
Diagnosis Swapping and gives , which is not the same expression (take : the original is , the swapped one is ). It is unchanged only under the rotation .
“Symmetric ⇒ cyclic, but cyclic ⇏ symmetric”
Diagnosis The original has degree , while has degree only . A linear factor is still missing; the two cannot differ by just a constant.
“Count the degrees first, then pin down the remaining factor”
Diagnosis Over the rationals that really is the end. But the problem says over the reals, so is allowed, and splits further.
“First read “over which numbers”, then decide whether you are done”
Diagnosis only says it has no real roots, so it cannot be split over the reals. Over the complex numbers, every quadratic has two roots and splits into two linear factors.
““Cannot be factored” must always come with “over which numbers””
Diagnosis Eisenstein's criterion is a sufficient condition: if the conditions hold, the polynomial is irreducible; if they do not, it tells you nothing. is of course irreducible over the rationals.
“When a sufficient condition fails, the question stays open”
From: Irreducible Polynomials
Diagnosis Having no rational root only means there is no linear factor. A quartic could still be a product of two quadratics: .
“No root ≠ no factor (when the degree is 4 or more)”
From: Irreducible Polynomials
Diagnosis is a slot, and a negative number can go into it. Put in and becomes , which is positive; put in and is . The minus sign only means “take the opposite”.
“A letter carries its own sign; the minus sign just means “the opposite””
From: Look at a and 1
Diagnosis is just . Take lot of away from lots of and you are left with lots of , not the number .
“No coefficient written means the coefficient is 1, not 0”
From: Look at a and 1
Diagnosis Put into the slot and you get : the denominator is , so it is undefined.
“Before you divide, ask “could this be 0?””
From: Look at a and 1
Diagnosis “Moving from the right to the left” really means subtracting from both sides, so what appears on the left is , not . The same goes for moving to the right: you subtract from both sides.
“Moving a term means adding or subtracting on both sides; crossing the equals sign flips its sign”
Diagnosis Multiplying both sides by means multiplying every term by . The and the have no denominator, but they are on the balance too, so they must become and . Substituting back gives it away: the left side is , the right side is .
“When you multiply, everything on the balance gets multiplied”
Diagnosis Before dividing by , nobody asked whether could be . When the equation is , and every number is a solution, not just .
“Before dividing by a letter, split into “it is 0” and “it is not 0””
Diagnosis A coefficient of only tells you that you cannot divide. Whether there is a solution depends on the right side: when the right side is also , the equation becomes , and there are infinitely many solutions.
“0·x = b: if b ≠ 0, no solution; if b = 0, everything is a solution”
Diagnosis Two numbers have absolute value : and . On the number line there are two points units from , one on each side.
“An absolute value is a distance, so look on both sides”
Diagnosis Each result only counts inside its own piece. does not satisfy , and does not satisfy . Substituting back shows it too: when the left side is and the right side is .
“A number found in one piece must be checked against that piece”
Diagnosis Subtracting means subtracting the whole expression: . The being subtracted becomes .
“Subtracting an equation means adding its opposite, so every term changes sign”
Diagnosis “Both sides” includes the constant on the right of the equals sign. Multiply only the left and you get a different line, which no longer passes through the original intersection.
“When you multiply both sides, the right side is a side too”
Diagnosis Together the two new equations still contain all three unknowns , and . The problem has not got any smaller.
“When eliminating, pick one unknown and get rid of it completely”
Diagnosis ④ and ⑤ both come from ① and ② only; equation ③ never appears. Only two of the three conditions are used, so you cannot get a unique answer.
“However many equations there are, make every one of them do some work”
Diagnosis expands to , which has an extra compared with the original. A corner filled in out of nowhere has to be given back, or the expression changes.
“Whatever you add, subtract again”
Diagnosis The is added inside the bracket, but the bracket is multiplied by , so in total an extra was added.
“If the leading coefficient is not 1, take it out first and complete the square inside the bracket”
Diagnosis Usually that is true. But here the left side is a sum of two non-negative numbers: neither can be negative, so neither can make up for the other. A sum of forces both terms to be .
“If a sum of squares is 0, every square is 0”
Diagnosis might be , and you cannot divide by . In fact, substitute : . It is exactly the root that was divided away.
“Never just cancel an expression containing the unknown; move the terms first, then factor”
Diagnosis Two numbers have square : and . Write only one and you block the road .
“When you take square roots, write both the positive and the negative one”
Diagnosis Here , and the formula has , which is . Read off each coefficient together with its sign.
“Write down a, b and c with their signs first, then put them into the formula”
Diagnosis also satisfies , but then the equation becomes , a linear equation with only one root. Saying “two roots” already requires it to be a quadratic equation.
“If the x² coefficient contains a letter, first write “it is not 0””
From: The Discriminant
Diagnosis “Has real roots” includes the case of two equal roots. When the equation is with root , which certainly counts as having real roots.
“Real roots means ≥; distinct real roots means >”
From: The Discriminant
Diagnosis , so . Lose the minus sign and drops from to .
“If a and c have opposite signs, Δ must be positive, a handy check”
From: The Discriminant
Diagnosis Expand and the linear term is , so the sum of the roots is the opposite of the linear coefficient. Here the roots are and , and their sum is clearly .
“The sum flips sign, the product does not”
From: Vieta's Formulas
Diagnosis A sum of squares of two real numbers can never be negative! The trouble is that at we get : the equation has no real roots at all, so do not exist.
“Vieta's formulas are about roots, so first make sure the roots exist”
From: Vieta's Formulas
Diagnosis You read the sum and product straight off the coefficients only for an equation like , with leading coefficient . Here the leading coefficient is , so divide both sides by first.
“Make the leading coefficient 1 first, or remember to divide by a”
From: Vieta's Formulas
Diagnosis and are values of , not of . The question asks for ; is just a helper we brought in ourselves.
“A substitution is a round trip. Whatever you replaced, you must turn back at the end”
From: Higher-Degree Equations
Diagnosis means . The square of a real number is never negative, so this branch has no real solutions. And the square of is , not .
“When you substitute, bring along the range of the new letter too”
From: Higher-Degree Equations
Diagnosis When both sides are , so it is a root too. Dividing by assumes , and that step quietly threw this root away.
“Reduce the degree by factoring, not by cancelling”
From: Higher-Degree Equations
Diagnosis At the denominator is , so neither side of the original equation means anything. is only a root of the polynomial equation , not of the original. A root like this is called an extraneous root.
“The last step of a fractional equation is always the check”
From: Fractional Equations
Diagnosis “Multiply both sides by ” means multiply every term. The has no denominator, but it must be multiplied too: it becomes . Substitute back: the left side is and the right side is . Not equal.
“When you multiply by the common denominator, do not skip a single term”
From: Fractional Equations
Diagnosis At the left side is and the right side is . A square root can never equal a negative number. What actually satisfies is , which becomes exactly the same equation as the original once you square.
“Squaring mixes up a = b and a = −b, so you must substitute back into the original equation”
From: Radical Equations
Diagnosis is not : there is a in the middle. Substitute back: the left side is and the right side is . Not equal.
“You square the whole right side, using the perfect square formula”
From: Radical Equations
Diagnosis is on the circle but not on the line: . The quadratic equation gives two values of for each , and only one of them also satisfies the linear equation.
“Substitute back into the linear equation, so each x gets exactly one y”
Diagnosis Dividing by assumes . But does satisfy (), and then . So is a solution too, and the division threw it away. This is the opposite of an extraneous root: a lost root.
“Never divide by something that could be 0. Turn it into “a product equals 0” instead”
Diagnosis Try : , and . Adding to adds to the left side, and taking from takes away . They do not cancel.
“The step for x is the coefficient of y, the step for y is the coefficient of x, one up and one down”
Diagnosis The question asks for integer solutions, not positive integer solutions. Two negative numbers can also multiply to : , , , .
“First make sure whether you want integer solutions or positive integer solutions”
Diagnosis A perfect-square only guarantees that the square root comes out exactly, so the roots are rational. The formula still divides by : , giving and .
“A square Δ is a necessary condition, not a sufficient one”
Diagnosis The problem only says the roots are integers, not positive integers. are integer factorizations too. Leave them out and you lose and .
“Integer factors can be positive or negative. List them all, then sift”
Diagnosis Average speed is total distance ÷ total time. The return trip is slower, so it takes longer and makes up more of the total time. That pulls the average speed toward .
“First ask “total ÷ total”. Do not rush to add the two speeds and divide by 2”
Diagnosis The two 's in this equation are not the same length of time: the truck has been driving hour longer than the car. You must first say clearly whose start is measured from.
“When you choose the unknown, say exactly from which moment and which point it is measured”
Diagnosis is how long B has been traveling. A left hour later and traveled for only hours. The equation was solved correctly, but the question was answered wrongly.
“The last step is to answer, not to solve”
Diagnosis Multiplying both sides by leaves the solutions unchanged only if . The answer is exactly , which makes the denominators of the original equation , so the original equation means nothing there.
“After clearing denominators or squaring both sides, you must go back and check the roots in the original equation”
From: A Map of Equations
Diagnosis When , the coefficient of is and the expression becomes . Its graph is a horizontal line: that is not a linear function.
“For a linear function y = kx + b, write k ≠ 0 first”
Diagnosis An intercept is a coordinate of the crossing point, so it can be negative: put and you get . For the crossing with the -axis put : gives . Do not flip the sign out of habit.
“To find where a line meets one axis, set the other coordinate to 0”
Diagnosis Vertex form is , with subtracted inside the bracket. , so . Or think directly: when is the squared term ? When .
“The vertex's x-coordinate is the x that makes the squared term 0”
Diagnosis The added inside the bracket gets multiplied by the outside, so you really added . You must subtract afterwards, not .
“Whatever you add inside the bracket, take back “coefficient × it” outside”
Diagnosis also satisfies , but then the equation is , which has only one root. To talk about “two roots”, the equation must be quadratic.
“Before you use the discriminant, check that the leading coefficient is not 0”
Diagnosis The problem says “function”, not “quadratic function”. When it is the line , which also meets the -axis at exactly one point.
“If it does not say “quadratic”, discuss the case where the leading coefficient is 0”
Diagnosis When you multiply both sides of an inequality by a negative number, the inequality sign must flip. The correct result is .
“Multiply (or divide) by a negative and the sign turns round; check with a number at the end”
Diagnosis “” is not a meaningful range. says that the distance from to the origin is greater than , which is two pieces of the number line.
“You cannot take square roots of both sides of an inequality the way you do with an equation”
Diagnosis When the inequality becomes , which also holds for every . The method above silently assumed it was quadratic.
“If the leading coefficient contains a letter, look at the case where it equals 0 separately”
Diagnosis only happens at , and is not in . On this interval the function keeps going up.
“First ask whether the vertex is inside the interval”
Diagnosis The minimum is at the vertex only when the axis falls inside the interval (). For the minimum is ; for it is .
“If the axis can move, split into three cases, left, inside and right”
Diagnosis A perfect-square only guarantees that the roots are rational. The roots are , that is, and : the numerator must also be divisible by .
“A square Δ is only the entry ticket; you still have to check the roots”
Diagnosis The problem did not say “quadratic equation”. When the equation is , whose only root is also an integer, and this has been missed.
“If the leading coefficient contains a parameter, look at it being 0 on its own first”
Diagnosis is not in the allowed range. On the piece the profit keeps rising with the price, so the highest point is at the edge of the range.
“In word problems the variable always has a range; write the range before you look for the max or min”
Diagnosis The horizontal axis is time, not horizontal distance. The ball moves on one vertical line the whole time; the “” on the graph is seconds, not meters.
“Read the axes before you read the graph”
Diagnosis “The sum of two sides is greater than the third” must hold for all three pairings. Of course and . The dangerous one is against : .
“Add the two shortest and compare with the longest”
Diagnosis Splitting into two cases is right, but for each case you must ask “does this triangle exist?” With legs of , : there is no such triangle.
“In a case analysis, go back and check every case”
Diagnosis “Area ratio = base ratio” only holds when the heights are the same. If the two bases are not on the same line, or the triangles do not share the apex, the heights are usually different.
“First make sure what is shared, then compare”
Diagnosis is how the shared side itself is split, and that compares with . The area ratio of two triangles on a common side is the ratio of the distances from the two other vertices to .
“The shared side is not in the ratio; the two other vertices are”
Diagnosis The equal angle is not the angle between the two known sides and ; it is opposite one of them. The same two sides and the same opposite angle can give two different triangles, one wide and one narrow.
“The A in SAS is written in the middle, and it must sit in the middle too”
From: Congruent Triangles
Diagnosis In a congruence statement the order of the letters is the correspondence: , , . corresponds to , not .
“Line up corresponding vertices when you write it, then just read by position”
From: Congruent Triangles
Diagnosis Scaling by multiplies the base by and the height by , so the area is multiplied by .
“Lengths scale by k, areas by k²”
From: Similar Triangles
Diagnosis In the side corresponding to is , not . is correct, but it equals .
“Part to part, or whole to whole; never mix them”
From: Similar Triangles
Diagnosis When all three points are inside the sides, no straight line can pass through all three sides at once. In this case a product of is about Ceva: the three lines , , are concurrent.
“One formula, two theorems; count the points on extensions first”
Diagnosis The third ratio is upside down. The correct version walks once round the triangle, , and every ratio is “vertex to dividing point : dividing point to next vertex”.
“Start at a vertex and walk once round, back to the start”
Diagnosis The point at equal distances from the three vertices lies on the perpendicular bisectors of the sides. It is the center of the circumcircle: the circumcenter. The incenter is where the three angle bisectors meet; it is at equal distances from the three sides, and it is the center of the incircle.
“Circle through the vertices, circumcenter; circle hugging the sides, incenter”
Diagnosis In an obtuse triangle the circumcenter and the orthocenter move outside; in a right triangle they land on a side or a vertex. When is obtuse, .
“Circumcenter and orthocenter can go out; centroid and incenter never do”
Diagnosis The midpoint triangle really does have half the original perimeter, but it is usually not the shortest. The shortest one needs “angle of incidence = angle of reflection” at every vertex, and midpoints cannot do that (unless the original triangle is equilateral).
“Balanced is not the same as shortest; the shortest path is the path of light”
Diagnosis In an obtuse triangle two of the feet lie on extensions of the sides, so the orthic triangle is not even inscribed. In this case the perimeter has no minimum: the closer and get to , the closer the perimeter gets to ( is the altitude), but it never reaches it.
“Before using a result, check its conditions”
Diagnosis The problem does not say whether is a leg or the hypotenuse. If is the hypotenuse, the third side is .
“Ask first: which side is the hypotenuse?”
From: Right Triangles
Diagnosis The converse of the Pythagorean theorem checks whether the squares of the two shorter sides add up to the square of the longest side. Here the longest side is .
“Only the longest side can be the hypotenuse”
From: Right Triangles
Diagnosis could be the apex angle or a base angle. If it is a base angle, the other base angle is also and the apex angle is .
“In an isosceles triangle, “an angle” or “a side” always raises the question “which one?””
From: Isosceles Triangles
Diagnosis The three lines that coincide are the bisector of the apex angle, the median to the base and the altitude to the base. Lines from the base vertex to a leg do not have this property (unless the triangle happens to be equilateral).
“The three lines only coincide on the axis of symmetry”
From: Isosceles Triangles
Diagnosis The proof needs “all three sides equal ” so that can be taken out as a common factor. When the sides are not all equal, the sum is different near a long side and near a short side.
“The constant comes from “all three sides equal”; drop one and it is gone”
From: Equilateral Triangles
Diagnosis A rotation only guarantees . To get you also need the angle of rotation to be : then is isosceles with apex angle , and only that makes it equilateral.
“60° is the key word, so write it down”
From: Equilateral Triangles
Diagnosis Each side of the midpoint quadrilateral is half of one of the diagonals and parallel to it. Its shape is decided by the two diagonals of the original quadrilateral, not directly by its sides.
“For the midpoint quadrilateral, look at the diagonals, not the sides”
Diagnosis At the dented vertex, the interior angle is the one greater than (on the inside of the quadrilateral). The diagonal that lies inside still splits the quadrilateral into two triangles.
“An interior angle is measured on the inside, and it can be more than 180°”
Diagnosis One pair of opposite sides is parallel and the other pair is equal. An isosceles trapezoid satisfies both conditions, and it is not a parallelogram.
“Parallel and equal must be about the same pair of sides”
From: Parallelograms
Diagnosis The diagonals of a parallelogram bisect each other, but they are usually not equal. Equal diagonals are a property of rectangles.
“Bisecting each other ≠ equal”
From: Parallelograms
Diagnosis The criterion says “a parallelogram with equal diagonals is a rectangle”. Drop “parallelogram” and an isosceles trapezoid, or even a very irregular quadrilateral, can have equal diagonals.
“The criteria for rectangles and rhombuses all stand on the shoulders of “parallelogram””
From: Rectangles and Rhombuses
Diagnosis needs perpendicular diagonals (rhombus, square, kite). In a general parallelogram the diagonals do not meet at a right angle, and the area is smaller than this.
“A formula comes with its conditions: perpendicular first, then ½·d₁·d₂”
From: Rectangles and Rhombuses
Diagnosis Four equal sides only guarantee a rhombus. Push a square frame out of shape: the sides stay the same length, but the angles change.
“A square must pass two tests: equal sides and a right angle”
From: Squares
Diagnosis “Bisect each other” is still missing. Two segments of the same length that are perpendicular but do not cross at their midpoints have endpoints that do not form a square.
“Bisect, equal, perpendicular; you cannot leave out a single one”
From: Squares
Diagnosis The midline (joining the midpoints of the legs) is half the sum of the bases. Half the difference of the bases is a different segment: the one joining the midpoints of the diagonals.
“Midpoints of the legs → half the sum; midpoints of the diagonals → half the difference”
From: Trapezoids
Diagnosis By the textbook definition, a trapezoid has exactly one pair of parallel opposite sides; the other pair (the legs) is not parallel. A parallelogram has two such pairs, so it does not count.
“The definition of a trapezoid contains the word “only””
From: Trapezoids
Diagnosis Equal opposite angles are a property of parallelograms. In a cyclic quadrilateral, opposite angles are supplementary: the two arcs that and stand on make up the whole circle.
“Cyclic quadrilateral: opposite angles are supplementary, and an exterior angle equals the opposite interior angle”
Diagnosis The result for a tangential quadrilateral is : the sums of the two pairs of opposite sides are equal, not the opposite sides themselves.
“Tangential quadrilateral: a + c = b + d”
Diagnosis When the arc is more than half the circle, the segment contains the center, so it is the sector plus a triangle. Subtracting the triangle throws away a piece that belongs to the segment.
“First ask whether the center is inside the segment or outside it”
Diagnosis gives the arc length, which is a length. An area must have in it.
“Arc length is a fraction of the circumference; sector area is the same fraction of the circle's area”
Diagnosis Any two diameters bisect each other at the center, yet they can meet at any angle. The converse of the perpendicular-diameter theorem needs the bisected chord to be not a diameter.
“The converse hides a condition in brackets — the chord is not a diameter”
From: Symmetry of the Circle
Diagnosis The problem never says the two chords are on the same side of the center. One can be above it and one below, with the center between them.
“When a circle problem comes without a figure, first ask “could it be drawn another way?””
From: Symmetry of the Circle
Diagnosis Chord splits the circle into two arcs. With the vertex on the major arc the inscribed angle is ; with the vertex on the minor arc it is . The problem says “the inscribed angle on the chord” without saying which side the vertex is on, so you need both.
“An arc gives one inscribed angle; a chord gives two”
From: Angles in a Circle
Diagnosis “Half the arc it stands on” only holds when the vertex is on the circle. is inside, so is bigger than : you also have to add half of the opposite arc .
“First check whether the vertex is on, inside or outside the circle, then pick the formula”
From: Angles in a Circle
Diagnosis There are two kinds of tangency: the small circle can touch the big one from outside (externally tangent) or from inside (internally tangent). The problem only says “tangent”, so you need both.
“Tangency comes in an inside and an outside version, and so does “no common point””
Diagnosis When the small circle hides entirely inside the big one, the circles also have no common point.
“Two different stretches of the number line both mean “no common point””
Diagnosis Infinitely many lines pass through , and only one of them is the tangent. The test has two conditions: the line passes through the outer end of a radius, and it is perpendicular to that radius. Both are needed.
“Common point known — join the radius, prove perpendicular. Unknown — drop the perpendicular, prove it is a radius”
From: Tangents to a Circle
Diagnosis The tangent–chord angle equals the inscribed angle on the arc it encloses. If happens to lie on that enclosed arc, then stands on the other arc and equals .
“The tangent–chord angle equals an inscribed angle only when both stand on the same arc”
From: Tangents to a Circle
Diagnosis If and are on opposite sides of line , then on one circle these two angles would be supplementary ( and ), not equal. A angle on each side means the four points are not concyclic.
“Same side — look for equal angles; opposite sides — look for supplementary angles”
From: Concyclic Points
Diagnosis says : that is the condition for a trapezoid or parallelogram. Concyclic points need two opposite angles to be supplementary.
“Opposite angles supplementary means concyclic; adjacent angles supplementary means parallel”
From: Concyclic Points
Diagnosis Both segments in the theorem are measured from : and . is the chord length, not measured from .
“Every segment in the power-of-a-point theorems has one end at P”
From: The Power of a Point
Diagnosis The two segments you multiply must lie on the same line: the two pieces of one chord multiplied together equal the two pieces of the other chord multiplied together.
“Multiply the two pieces on the same line”
From: The Power of a Point
Diagnosis Points on the perpendicular bisector are equally far from the two centers, that is, . The radical axis needs . The two agree only when the radii are equal.
“The radical axis compares powers, not distances to the centers”
Diagnosis The power is defined as , in that order. When is inside the circle it is negative. Its absolute value is the product of the two pieces of any chord through .
“Power = distance squared minus radius squared, negative inside the circle”
Diagnosis Ptolemy's theorem multiplies opposite sides: with , and with . Two adjacent sides share a vertex, so they are not a pair.
“Each product uses all four vertices exactly once”
Diagnosis The theorem is about a cyclic quadrilateral : the four points lie on the circle in the order , , , , so and are the diagonals. If the order around the circle is , , , , then is a side and the equation fails.
“First work out which segments are the diagonals”
Diagnosis The point equally far from the three vertices is the circumcenter (where the perpendicular bisectors of the sides meet). The incenter is where the three angle bisectors meet, and it is equally far from the three sides.
“The circumcenter is equidistant from the vertices; the incenter is equidistant from the sides”
Diagnosis is the angle at the circumcenter: (a central angle is twice the inscribed angle, when is acute). The angle at the incenter has to be worked out from the angle bisectors.
“At the circumcenter it is 2∠A; at the incenter it is 90° + ½∠A”
Diagnosis The central angle is right, but can be on the major arc or on the minor arc. The inscribed angles in the two positions are supplementary.
“In circle problems, count the cases first: which arc is the point on? Which side of the center is the chord? Internally or externally tangent?”
Diagnosis . The hides in the and the hides in the : the was split up between the two factors, and neither one got a whole .
“A composite number can be split up; a prime cannot”
Diagnosis The definition of divisibility is “there is an integer with ”. , and is an integer.
“Go back to the definition: can you find that q?”
Diagnosis A prime must have exactly two different positive divisors, and has only one. More importantly: if counted as a prime, then , and factorization would no longer be unique.
“1 is neither prime nor composite”
From: Primes and Composites
Diagnosis , and the primes up to that are . The was never tried, and .
“Go all the way up to the square root, and skip none”
From: Primes and Composites
Diagnosis . It is only guaranteed not to be divisible by . Nobody promised it is not divisible by a bigger prime.
“Claim only what your proof actually uses”
From: Primes and Composites
Diagnosis and share the factor : they “overlap”. only has to hold the least common multiple of and , which is , not .
“Check whether they are coprime before combining two divisibilities”
Diagnosis A remainder of only says “it divides exactly, you can stop”. The answer is the divisor that divides exactly, which is the last nonzero remainder.
“The answer is the last nonzero remainder”
Diagnosis Choosing a divisor means making two independent decisions: the exponent of is one of , and the exponent of is one of . Every combination gives a different divisor, so you multiply.
“Decisions made in steps multiply”
Diagnosis and . A common multiple only needs the higher exponent of each prime: . The product counts both bars, so it has an extra factor of .
“The lcm takes the higher bar, not both”
Diagnosis is true, but is not a multiple of . and share the factor : in , part of the is “eaten” by the .
“Add, subtract and multiply freely; before cancelling, check coprimality”
Diagnosis A remainder must be between and . means going back marks from , which lands on mark . And and differ by , not a multiple of , so they are not even in the same column.
“Negative numbers go backwards round the clock”
Diagnosis Congruent numbers can be swapped in the base. The exponent is not a number being added or multiplied, and there is no such rule for it. In fact , while .
“Reduce the base mod m and the exponent mod the period”
Diagnosis They are all even. Divided by , they leave only the remainders , each one twice; the classes with remainder have no representative at all.
“Counting is not enough; check for collisions”
Diagnosis . is always a multiple of , so it can only land on the three places , with arrows crowded onto each.
“If the multiplier is coprime to the modulus, the line-up stays in order”
Diagnosis is not prime, so the theorem says nothing about it. In fact , so .
“Check the condition before using the theorem: is p prime?”
Diagnosis , so . The theorem needs : start from , multiply by , and once you fall into you never get out.
“a must not be a multiple of p”
Diagnosis . It just happens that , so . The theorem says “prime ⇒ remainder 1”, not “remainder 1 ⇒ prime”.
“The theorem only works in one direction”
Diagnosis Parity (or coloring) is only a security check: whatever fails it is impossible, but whatever passes it is not necessarily possible. For example, the removed cells might cut off a single cell in a corner. Black and white can still be equal, but the board obviously cannot be tiled.
“Parity can rule things out, never in”
From: Odd and Even Numbers
Diagnosis is even, so and always have the same parity.
“Sum and difference are both odd or both even”
From: Odd and Even Numbers
Diagnosis “A square leaves or when divided by ” is a condition every square must satisfy, not one that is enough. Plenty of numbers leave remainder : are not squares.
“A remainder table rules out; it does not certify”
From: Perfect Squares
Diagnosis , but neither nor is a square. The exponent of the prime is : even in total, but odd in each factor.
“Coprime first, then each one is a square”
From: Perfect Squares
Diagnosis Everything this gives really is a solution, but it misses two thirds of them: is also a solution (), and it is not on the list. Neighboring solutions differ by only , not .
“Take the smallest possible step”
Diagnosis Put in : . Increasing makes the left side bigger, so has to decrease to make up for it.
“One goes up, the other goes down”
Diagnosis The question asks for integer solutions, and are valid splits too. They give and .
“For integer solutions, divisors come with both signs”
Diagnosis If one modulus finds no problem, that only shows this modulus is not good enough. Switch to mod : the left side is , the right side leaves , and there is an immediate contradiction.
“One modulus letting it through does not mean all of them do”
Diagnosis The range is a necessary condition: every solution is inside the box, but most points in the box are not solutions. Of the cells, only are solutions.
“Fencing in the range is only the first half”
Diagnosis is a Pythagorean triple, but is not a sum of two positive squares (, , all fail), so the formula does not give it directly.
“Reduce to a primitive triple first, then use the formula”
From: The Pythagorean Equation
Diagnosis is plainly a solution. divided by is still , so nothing “gets smaller”: the descent stops here and gives no contradiction.
“Only 0 can keep getting “smaller” forever”
From: The Pythagorean Equation
Diagnosis Choosing a top and choosing pants are things you do one after the other: every top can go with each of the pairs of pants. You only add when the cases are “either this or that”.
“Steps multiply, cases add. First ask yourself: “and then” or “or”?”
From: Counting
Diagnosis When A shakes hands with B, that handshake is counted once among “A's ” and again among “B's ”. Every handshake has been counted times.
“When you finish counting, ask: how many times did I count each thing?”
From: Counting
Diagnosis are multiples of both and , so they were counted once in each case. The addition principle needs cases that do not overlap.
“Before you add, check whether the cases have anything in common”
From: Counting
Diagnosis The pigeonhole principle only promises that there is a box with at least . Putting all balls in one box is also allowed: then one box has and four boxes are empty.
“The pigeonhole principle only says “there is” and “at least””
From: The Pigeonhole Principle
Diagnosis is certainly enough, but the question asks for the smallest number. The unluckiest case is socks of each color ( socks, and still no of a color). Then just more is sure to complete a set.
“Imagine the unluckiest case first, then add just 1”
From: The Pigeonhole Principle
Diagnosis Two points in the same strip can be one at the top and one at the bottom, more than apart. A box is only useful if “being in the same box” automatically gives the conclusion you want.
“A good box: if two things share it, the conclusion holds”
From: The Pigeonhole Principle
Diagnosis There are thousands upon thousands of ways to place the dominoes, and you tried only a tiny fraction. “I didn't find one” is not the same as “there isn't one”. Maybe try number 21 would have worked?
“Examples can show “possible”, never “impossible””
From: Coloring
Diagnosis The number of squares dividing evenly is only a necessary condition. Color the board in colors along the diagonals: every bar covers exactly one square of each color. But the four colors have squares, which are not equal.
“The right area only gets you past the first check”
From: Coloring
Diagnosis Coloring can only prove that something is impossible. Equal counts only mean “this coloring found no contradiction”. Color again with the diagonals going the other way: the counts are , and there is the contradiction.
“A coloring can veto, but it can never approve”
From: Coloring
Diagnosis A matching invariant only means “this invariant does not object”. Try another invariant: the product of the four corners. Every flip changes exactly or corners, so it never changes either. Right now it is , and the goal has .
“An invariant is a veto, not a permit”
From: Operations and Games
Diagnosis You take , leaving ; your opponent takes , leaving . From then on, whenever you take , your opponent takes , so each round removes : , and the last stone goes to your opponent.
“Games are not won by taking the most, but by leaving your opponent in a losing position”
From: Operations and Games
Diagnosis Your opponent's first move takes the th coin, right in the middle. Its mirror image is itself, so you cannot copy it. Now there are coins on each side, and it is your opponent who gets to copy you.
“Before using symmetry, check whether something sits on the mirror line”
From: Operations and Games
Diagnosis “At most ” is true, but nobody can actually place : once a row has kings, the next row has almost no room. An upper bound that cannot be reached is not the answer.
“Every upper bound needs a construction to meet it”
Diagnosis Being unable to add one only shows that this particular placement is stuck. A different placement, at the crossings of rows 1, 3, 5 and columns 1, 3, 5, holds .
“A dead end does not mean you have gone the farthest”
Diagnosis One guard can watch neighboring squares in the same row at once. “Every square must be watched” does not mean “every square needs its own guard”.
“When you count “at least how many”, first make sure no one can do two jobs”
Diagnosis expands to , which is more than the original . The corner you filled in was borrowed; if you never subtract it again, the value of the expression changes.
“Add a term, and you must take it away again”
Diagnosis The added inside the brackets is doubled by the outside: , so the extra amount is really , not .
“What you borrow inside the brackets, pay back inside the brackets”
Diagnosis is not just any number: it is , so . would mean , and no real can do that. Treating as if it were and taking square roots creates two fake solutions out of nowhere.
“When you substitute, carry the new variable's range across too”
From: Substitution line
Diagnosis Only half of the substituting back was done. Once the lens comes off, each factor must be checked again in the -world: and can both still be factored by the cross method.
“After substituting back, ask once more “can it factor further?””
From: Substitution line
Diagnosis The sum of the squares of two real numbers can never be negative: the answer itself is shouting “something is wrong”. When , , so the equation has no real roots at all, and “the sum of the roots” and “the product of the roots” are promises nobody can keep.
“First make sure the roots exist, then talk about their sum and product”
Diagnosis satisfies , but then the equation is , which has only one root. The discriminant is a tool for quadratic equations: before using it, check that the coefficient of is not .
“First check whether a is 0, then compute Δ”
Diagnosis With as the main variable, is a “coefficient”. But it contains a letter, so it might be exactly : when the equation is , with no solution; when it is , and every is a solution. Dividing by loses both of these cases.
“When a coefficient contains a letter, first ask whether it could be 0”
Diagnosis “Without loss of generality ” needs the expression to stay the same when any two letters are swapped. This expression is only cyclic: gives , while gives , so swapping and changes the value. There are possible orderings, and cycling can only merge them into classes, not .
“Symmetric expressions allow a full ordering; cyclic ones only let you choose the largest”
From: Symmetry line
Diagnosis What you proved is “less than ”, but the problem asks for “less than ”. is bigger than , so this bound is too loose and proves nothing. The trouble is the first term: enlarging to throws away in one go.
“If a bound is too loose, bound fewer terms”
From: Bounding line
Diagnosis The right-hand inequality is equivalent to . When , , so it is not squeezed at all, and is in fact a perfect square.
“Use an inequality only from where it starts to hold”
From: Bounding line
Diagnosis In the area formula the “height” must be perpendicular to the chosen “base”. is a leg: it is perpendicular to the other leg, , not to the hypotenuse, .
“Whichever base you choose, use the height perpendicular to it”
From: The Area Method line
Diagnosis With and as bases, the two heights are the distances from to and from to . They are equal only when lies on the bisector of .
“A ratio of areas equals the ratio of the bases only when the heights are the same”
From: The Area Method line
Diagnosis Two segments can be added only if , , are on one straight line. The rotation only guarantees ; it does not guarantee that the pieces join up straight. In a square they happen to, but in a general quadrilateral they may not.
“Before joining pieces, check they are collinear”
Diagnosis Rotation does not change lengths. Since , cannot land on , so the rotated triangle does not fit against the original figure, and you get nothing out of it.
“Equal segments sharing a vertex are what make a rotation work”
Diagnosis In the corresponding sides are and . is only a piece of ; it is not a side of either triangle.
“Corresponding sides must be actual sides of the two triangles”
Diagnosis Area is a two-dimensional quantity, like “length × width”. If the base is scaled by , the height is scaled by too, so the area is scaled by .
“Lengths scale by the ratio, areas by its square”
Diagnosis Four points are concyclic when opposite angles are supplementary, not equal. , so these four points are not on one circle, and the inscribed angle theorem does not apply.
“First prove the points are concyclic, then use the circle”
Diagnosis They are equal only when and are on the same side of the chord (on the same arc). If they are on opposite sides, the two angles stand on two different arcs that together make the whole circle, so they are supplementary.
“Same arc, equal angles; opposite arcs, supplementary angles”
Diagnosis is the length of the segment , so it must be positive. is a root of the equation, but not a solution of this geometry problem, and quietly dropping the minus sign makes no sense at all.
“Take every root of the equation back to the figure and check it”
Diagnosis The vertex is at , but the problem only allows . If the vertex is outside the allowed range, its value is never reached.
“First write down which values x can take, then look for the maximum or minimum”
Diagnosis is even too. An even sum only tells you that and have the same parity: both even or both odd. Forget the “both odd” half and your argument is missing a whole case.
“Even = even + even, and also = odd + odd”
From: Parity line
Diagnosis Matching parity is only a necessary condition. Actually try it: the squares are just , and no two of them add up to . In fact, look at remainders when dividing by : the sum of two odd squares leaves , while leaves .
“Parity can rule things out, but it can never rule them in”
From: Parity line
Diagnosis , yet neither nor is a square. They share the prime factor : has one , has three, and together that makes four, an even number.
“Check coprime first, then split the square”
Diagnosis A prime written as a product of two integers can be , but also . Here gives , and is prime too.
“Integer factor pairs come in positive and negative versions”
Diagnosis When a modulus gives no contradiction, all it tells you is that this modulus does not help. Matching remainders are a long way from “really equal”. Switch to mod : the left side can only leave , and the right side leaves , a contradiction at once.
“A modulus can prove “guilty”, never “innocent””
Diagnosis When , , which also leaves on division by , but does not leave on division by . The you cancelled shares a factor with the modulus , and cancelling like that loses solutions.
“Only cancel a number that is coprime to the modulus; otherwise divide the modulus too”
Diagnosis The right-hand inequality simplifies to . For it is simply false: gives and gives , both squares.
“Estimation handles the big numbers; check the small ones by hand”
From: Estimation line
Diagnosis There are plenty of squares less than : , … A bound on one side rules nothing out. When , is a square.
“You need both an upper and a lower bound, and they must be consecutive”
From: Estimation line
Diagnosis Two numbers with the same remainder have a difference that is a multiple of , but not necessarily a sum. For example, all leave remainder , and any two of them add up to something with remainder . The boxes were right; the conclusion was read wrongly.
“Before building the boxes, ask what two in the same box would mean”
Diagnosis That is not wrong, but it says too little. If every box had at most , there would be at most objects, which is not enough room for . So you can guarantee .
“Suppose no box is full enough, and see whether everything still fits”
Diagnosis Coloring can only say “impossible”, never “possible”. Equal color counts are only a necessary condition: if the counts are wrong it certainly cannot be done, but if they are right it still might not be. In fact, color it with 3-color diagonals running the other way and that cell gives the game away: this board cannot be covered.
“Coloring only ever says no; to say yes, show a tiling”
Diagnosis The checkerboard coloring is made to measure for 1×2 dominoes: each domino covers exactly one black and one white. A 1×3 strip may cover two black and one white, or two white and one black, so for this tile the coloring is not fixed and of course shows nothing.
“Look at the shape of the tile first, then decide how to color”
Diagnosis When , , and you cannot divide by . The equation is then , and every number is a solution, not just .
“Before dividing by a letter, ask whether it can be zero”
Diagnosis Each case has to pass two tests. The base must be positive, so ; the two equal sides must add up to more than the base, , so . With equal sides there is no triangle at all (for example ).
“After working out each case, go back and check that case’s conditions”
Diagnosis However many cases you check, it is still only finitely many. In fact it stays prime all the way up to , but for , , which is not prime. Without the step “from to ”, a pattern can break at any moment.
“Checking is guessing; the recursion is the proof”
Diagnosis A recursion only tells you “how each term comes from the two before it”; it does not tell you where to start. Here and (both dominoes upright, or both lying flat), not .
“If the recursion uses several earlier terms, count that many starting values”
Diagnosis There is no smallest positive fraction: however small is, is a positive fraction too, and smaller. Reasoning about something that does not exist proves nothing.
“Make sure the extreme one exists before you take it”
Diagnosis Several numbers can tie for largest. The neighbors are only not larger than it, and they may well be equal to it. In fact, the conclusion of the circle problem is exactly that “they are all equal”.
“The largest need not be unique”
Diagnosis Every handshake involves two people, so in “how many times did each person shake hands” it is counted twice. counts handshake-ends, not handshakes. The real number would be , which is not a whole number!
“Before you count twice, ask how many times each thing is counted”
Diagnosis Only one position was checked. Something true in a special position may well be false somewhere else. For example, “the altitude and the median coincide” is true in an isosceles triangle but false in a general one.
“Special values find the answer; they do not prove it”
Diagnosis These values make both sides at once, and holds for every , so you learn nothing.
“Be special, but not so special that both sides become 0”
Diagnosis Count by hand: you get regions, not . Five examples were right and the sixth is wrong. Examples can help you guess; they cannot prove anything for you.
“However many examples you have, they are not a proof”
From: Guess, Then Prove line
Diagnosis When , , which is composite. You can even see it without calculating: when , all three terms are multiples of .
“A conjecture is either proved or knocked down by a counterexample”
From: Guess, Then Prove line