Symmetry

In one sentence

a^{2} + b^{2} a \leftrightarrow b b^{2} + a^{2} (unchanged)

If swapping two letters leaves the expression exactly as it was, the two letters are completely equal partners. Equal partners bring three benefits:

Whatever you prove for one letter automatically holds for the other: half the work.
You can put them in order (“without loss of generality $a \leq b$ ”) without losing any case.
A symmetric expression in two letters can always be written using only $a + b$ and $ab$ .

Predict first

In $a^{2} (b - c) + b^{2} (c - a) + c^{2} (a - b)$ , set $a = b$ . What do you get?

0

2 b^{2} (b - c)

It depends on the values of

b, c

Choose an answer first; then you can see the result.

Stop 1 · Factoring cyclic expressions (Vol. 1)

Example: factor a²(b − c) + b²(c − a) + c²(a − b)

Step 1 / 5

Try a special value
Set $a = b$ and the expression becomes $0$ . So it has the factor $(a - b)$ .

$a = b \Rightarrow 0 \Rightarrow (a - b)$
Cycle the letters: nothing changes
Count degrees
Find the constant
The result

Stop 2 · Vieta's formulas (Vol. 2)

The two roots of $x^{2} - 4 x + 1 = 0$ are $2 \pm 3$ , and which one is “the first root” and which is “the second” is a purely human choice. So any question with a definite answer must be asking about a quantity that does not change when the roots are swapped, and such a quantity can be worked out from the sum $s = 4$ and the product $p = 1$ alone.

Example: a and b are the roots of x² − 4x + 1 = 0. Find a² + b², a³ + b³ and a²b + ab²

Step 1 / 4

First: is it symmetric?
Swap $a$ and $b$ : $a^{2} + b^{2}$ does not change. So it can be written using only $s = a + b$ and $p = ab$ .

$a^{2} + b^{2} = s^{2} - 2 p = 16 - 2 = 14$
Sum of cubes
Take out a common factor
A counterexample

Stop 3 · Symmetric systems of equations (Vol. 2)

If swapping $x$ and $y$ leaves both equations of a system unchanged, do not rush to eliminate. Switch to $s = x + y$ and $p = x y$ first, and the degree will come down.

Example: solve the system x² + y² = 13, x + y + xy = 11

Step 1 / 6

Spot the symmetry
Swap $x$ and $y$ : neither equation changes. Let $s = x + y$ and $p = x y$ .

$x^{2} + y^{2} = s^{2} - 2 p$
Rewrite
Eliminate p
s = 5
s = −7
Solutions come in pairs

Stop 4 · Isosceles triangles (Vol. 4)

In a figure, “unchanged by a swap” means reflection symmetry. Fold an isosceles triangle along its axis of symmetry and $B$ and $C$ swap places, and so do $A B$ and $A C$ .

Example: in isosceles triangle ABC with AB = AC, M and N are the midpoints of AC and AB. Prove that BM = CN

Step 1 / 5

Find the axis of symmetry
Since $A B = A C$ , the perpendicular bisector of the base $B C$ passes through $A$ . It is the axis of symmetry of the whole triangle.
Reflect
Conclusion
Not just midpoints
If it is not isosceles

Practice

Exercise 1Basic

$a$ and $b$ are the two roots of $x^{2} - 3 x - 2 = 0$ . Find $a^{2} + b^{2}$ .

Exercise 2Advanced

How many real solutions $(x, y)$ does the system $x^{2} + y^{2} = 10$ , $x y = 3$ have?

Exercise 3Challenge

Given that $a^{3} (b - c) + b^{3} (c - a) + c^{3} (a - b) = k (a - b) (b - c) (c - a) (a + b + c)$ holds for all $a, b, c$ , find the constant $k$ .

In one sentence

Stop 1 · Factoring cyclic expressions (Vol. 1)

Example: factor a²(b − c) + b²(c − a) + c²(a − b)

Stop 2 · Vieta's formulas (Vol. 2)

Example: a and b are the roots of x² − 4x + 1 = 0. Find a² + b², a³ + b³ and a²b + ab²

Stop 3 · Symmetric systems of equations (Vol. 2)

Example: solve the system x² + y² = 13, x + y + xy = 11

Stop 4 · Isosceles triangles (Vol. 4)

Example: in isosceles triangle ABC with AB = AC, M and N are the midpoints of AC and AB. Prove that BM = CN

Practice

Mistake Museum · Exhibits from this stop

Assuming “a ≥ b ≥ c” for a cyclic expression too

Stops on this line