Kvant Math Problem 1097

Consider small examples of isosceles triangles whose vertices have integer coordinates.

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Problem

The coordinates of the vertices of an isosceles triangle are integers. Prove that the square of its base is an even number.

V. V. Proizvolov

Exploration

Consider small examples of isosceles triangles whose vertices have integer coordinates. The simplest such triangles are right triangles aligned with the axes. For instance, the points $(0,0)$, $(2,0)$, $(1,1)$ form an isosceles triangle with base $(0,0)-(2,0)$ of length $2$, and $2^2 = 4$ is even. Another example is $(0,0)$, $(4,0)$, $(2,3)$; the base squared is $16$, also even. Attempting to find a counterexample where the base squared is odd fails for small integers, suggesting a parity constraint is in play.

Examining the parity of coordinates, note that if the base endpoints are both even or both odd in both $x$ and $y$, the squared distance is even. If one is even and one is odd, the squared difference in coordinates is odd. The key difficulty is showing that for an isosceles triangle with integer vertices, the parity of coordinates cannot produce a squared base that is odd.

Another angle is vector analysis. Let $\mathbf{AB}$ be the base vector and $\mathbf{AC}$ and $\mathbf{BC}$ the equal sides. The squared distances involve sums of squares of integers. The problem reduces to showing that if two squared distances are equal, the remaining squared distance must be even. The challenge is formalizing this parity argument rigorously.

Problem Understanding

The problem asks to prove that in any isosceles triangle whose vertices all have integer coordinates, the square of the length of the base is an even integer. This is a Type B problem, as the statement is universal and requires proof. The core difficulty is controlling the parity of squared distances when the triangle has integer coordinates. The insight from small examples and parity considerations suggests that the squared length of the base is necessarily divisible by $2$.

Proof Architecture

Lemma 1: In any isosceles triangle with integer coordinates, the midpoint of the base has coordinates that are either both integers or both half-integers. Sketch: The midpoint formula $(\frac{x_1+x_2}{2}, \frac{y_1+y_2}{2})$ shows this directly.

Lemma 2: The vertex opposite the base must be equidistant from the base endpoints. Sketch: By definition of isosceles triangles, the squared distances to the base endpoints are equal; expand these distances as sums of squares of integers.

Lemma 3: If a line segment connects two integer-coordinate points, and a point equidistant from both endpoints lies on the perpendicular bisector, then the squared length of the segment is even. Sketch: Express the coordinates algebraically; parity arguments show that the sum of squared differences in coordinates must be even.

The hardest step is Lemma 3, as careless parity handling can lead to a false conclusion. The main argument combines the first two lemmas to apply the parity constraint.

Solution

Let $A=(x_1,y_1)$, $B=(x_2,y_2)$, and $C=(x_3,y_3)$ be the vertices of an isosceles triangle with integer coordinates. Suppose $AB$ is the base. Then $AC=BC$, which implies

$$(x_3-x_1)^2 + (y_3-y_1)^2 = (x_3-x_2)^2 + (y_3-y_2)^2.$$

Expanding both sides, we obtain

$$x_3^2 - 2x_3 x_1 + x_1^2 + y_3^2 - 2y_3 y_1 + y_1^2 = x_3^2 - 2x_3 x_2 + x_2^2 + y_3^2 - 2y_3 y_2 + y_2^2.$$

Subtracting $x_3^2 + y_3^2$ from both sides, we get

$$-2 x_3 x_1 + x_1^2 - 2 y_3 y_1 + y_1^2 = -2 x_3 x_2 + x_2^2 - 2 y_3 y_2 + y_2^2.$$

Rewriting, we have

$$2 x_3 (x_2 - x_1) + 2 y_3 (y_2 - y_1) = x_2^2 - x_1^2 + y_2^2 - y_1^2.$$

Factor both sides:

$$2 x_3 (x_2 - x_1) + 2 y_3 (y_2 - y_1) = (x_2 - x_1)(x_2 + x_1) + (y_2 - y_1)(y_2 + y_1).$$

Divide both sides by $1$ (no factor 2 yet). Now observe that the left-hand side is divisible by $2$, since $2 x_3 (x_2 - x_1) + 2 y_3 (y_2 - y_1)$ is clearly even. Therefore, the right-hand side must also be even:

$$(x_2 - x_1)(x_2 + x_1) + (y_2 - y_1)(y_2 + y_1) \equiv 0 \pmod{2}.$$

Now, $(x_2 - x_1)^2 + (y_2 - y_1)^2 = (x_2 - x_1)(x_2 - x_1) + (y_2 - y_1)(y_2 - y_1)$ is congruent modulo $2$ to the sum above, because for any integers $a,b$, $a^2 \equiv a \cdot a \equiv a \cdot a \pmod{2}$. More precisely, note that $(x_2 - x_1)^2 + (y_2 - y_1)^2 = (x_2 - x_1)(x_2 - x_1) + (y_2 - y_1)(y_2 - y_1)$, which is an integer congruent modulo $2$ to $(x_2 - x_1)(x_2 + x_1) + (y_2 - y_1)(y_2 + y_1)$ because $(x_2 - x_1)(x_2 + x_1) - (x_2 - x_1)^2 = (x_2 - x_1)(x_1 + x_2 - x_2 + x_1) = 2(x_2 - x_1)x_1$ is divisible by $2$, and similarly for the $y$ term.

Thus, the squared length of the base $AB$,

$$AB^2 = (x_2 - x_1)^2 + (y_2 - y_1)^2,$$

is even. This completes the proof.

∎

Verification of Key Steps

The crucial step is verifying the parity congruence:

$$(x_2 - x_1)(x_2 + x_1) + (y_2 - y_1)(y_2 + y_1) \equiv (x_2 - x_1)^2 + (y_2 - y_1)^2 \pmod{2}.$$

Re-derive independently: $(x_2 - x_1)(x_2 + x_1) - (x_2 - x_1)^2 = (x_2 - x_1)(2 x_1)$, which is divisible by $2$. Similarly for $y$. Therefore the left-hand side being even implies $AB^2$ is even. Testing with concrete examples, $(0,0),(1,0),(0,1)$ fails to form isosceles; $(0,0),(2,0),(1,1)$ yields $AB^2 = 4$, which is even, confirming consistency.

Alternative Approaches

A vector approach using the midpoint of the base simplifies the argument. Let $M$ be the midpoint of $AB$; then the third vertex $C$ lies on the line perpendicular to $AB$ through $M$. Since $M$ has coordinates with either both integers or both half-integers, the distance squared from $A$ to $B$ must be divisible by $2$ to maintain integer coordinates for $C$. This approach emphasizes geometry but reduces to the same parity argument. The algebraic expansion is preferred for rigor because it explicitly handles all integer possibilities without assumptions about coordinate alignment.