2010 USAMO Problems/Problem 4

Problem

Let $ABC$ be a triangle with $\angle A = 90^{\circ}$ . Points $D$ and $E$ lie on sides $AC$ and $AB$ , respectively, such that $\angle ABD = \angle DBC$ and $\angle ACE = \angle ECB$ . Segments $BD$ and $CE$ meet at $I$ . Determine whether or not it is possible for segments $AB, AC, BI, ID, CI, IE$ to all have integer lengths.

Solution

We know that angle $BIC = 135^{\circ}$ , as the other two angles in triangle $BIC$ add to $45^{\circ}$ . Assume that only $AB, AC, BI$ , and $CI$ are integers. Using the Law of Cosines on triangle BIC,

$[asy] import olympiad; // Scale unitsize(1inch); // Shape real h = 1.75; real w = 2.5; // Points void ldot(pair p, string l, pair dir=p) { dot(p); label(l, p, unit(dir)); } pair A = origin; ldot(A, "$A$", plain.SW); pair B = w * plain.E; ldot(B, "$B$", plain.SE); pair C = h * plain.N; ldot(C, "$C$", plain.NW); pair D = extension(B, bisectorpoint(C, B, A), A, C); ldot(D, "$D$", D-B); pair E = extension(C, bisectorpoint(A, C, B), A, B); ldot(E, "$E$", E-C); pair I = extension(B, D, C, E); ldot(I, "$I$", A-I); // Segments draw(A--B); draw(B--C); draw(C--A); draw(C--E); draw(B--D); // Angles import markers; draw(rightanglemark(B, A, C, 4)); markangle(Label("$\scriptstyle{\frac{\theta}{2}}$"), radius=40, I, B, E); markangle(Label("$\scriptstyle{\frac{\theta}{2}}$"), radius=40, C, B, I); markangle(Label("$\scriptstyle{\frac{\pi}{4} - \frac{\theta}{2}}$"), radius=40, I, C, B); markangle(Label("$\scriptstyle{\frac{\pi}{4} - \frac{\theta}{2}}$"), radius=40, D, C, I); markangle(Label("$\scriptstyle{\frac{3\pi}{4}}$"), radius=10, B, I, C); [/asy]$

$BC^2 = BI^2 + CI^2 - 2BI\cdot CI \cdot \cos 135^{\circ}$ . Observing that $BC^2 = AB^2 + AC^2$ is an integer and that $\cos 135^{\circ} = -\frac{\sqrt{2}}{2},$ we have

$BC^2 - BI^2 - CI^2 = BI\cdot CI\cdot \sqrt{2}$

and therefore,

$\sqrt{2} = \frac{BC^2 - BI^2 - CI^2}{BI\cdot CI}$

The LHS ( $\sqrt{2}$ ) is irrational, while the RHS is the quotient of the division of two integers and thus is rational. Clearly, there is a contradiction. Therefore, it is impossible for $AB, AC, BI$ , and $CI$ to all be integers, which invalidates the original claim that all six lengths are integers, and we are done.

Solution 2

The answer is no.

Suppose otherwise. It is easy to see (through simple angle chasing) that $\angle DIC=45^{\circ}$ . Also, since $I$ is the incenter, we have $\angle IAC = 45^{\circ}$ . Using the Law of Cosines, we have $CD^2=IC^2+ID^2-\sqrt{2}(IC)(ID),$ so that $CD$ is irrational. But $\triangle IAC \sim \triangle DIC$ , thus $IC^2=CD\cdot AC$ , implying that $CD$ is rational, contradiction. $\blacksquare$

Solution 3

The result can be also proved without direct appeal to trigonometry, via just the angle bisector theorem and the structure of Pythagorean triples. (This is a lot more work).

A triangle in which all the required lengths are integers exists if and only if there exists a triangle in which $AB$ and $AC$ are relatively-prime integers and the lengths of the segments $BI, ID, CI, IE$ are all rational (we divide all the lengths by the $\gcd(AB, AC)$ or conversely multiply all the lengths by the least common multiple of the denominators of the rational lengths).

Suppose there exists a triangle in which the lengths $AB$ and $AC$ are relatively-prime integers and the lengths $IB, ID, CI, IE$ are all rational.

Since $CE$ is the bisector of $\angle ACB$ , by the angle bisector theorem, the ratio $IB : ID = CB : CD$ , and since $BD$ is the bisector of $\angle ABC$ , $CB : CD = (AB + BC) : AC$ . Therefore, $IB : ID = (AB + BC) : AC$ . Now $IB : ID$ is by assumption rational, so $(AB + BC) : AC$ is rational, but $AB$ and $AC$ are assumed integers so $BC : AC$ must also be rational. Since $BC$ is the hypotenuse of a right-triangle, its length is the square root of an integer, and thus either an integer or irrational, so $BC$ must be an integer.

With $AB$ and $AC$ relatively-prime, we conclude that the side lengths of $\triangle ABC$ must be a Pythagorean triple: $(2pq, p^2 - q^2, p^2 + q^2)$ , with $p > q$ relatively-prime positive integers and $p+q$ odd.

Without loss of generality, $AC = 2pq, AB = p^2 - q^2, BC = p^2+q^2$ . By the angle bisector theorem,

$\begin{align*} AE &= \dfrac{AB \cdot AC}{AC + CB} = \dfrac{2pq(p^2-q^2)}{p^2 + q^2 + 2pq} = \dfrac{2pq(p-q)}{p+q} \end{align*}$

Since $\triangle CAE$ is a right-triangle, we have:

$\begin{align*} CE^2 &= AC^2 + AE^2 = 4p^2q^2 + \left(\dfrac{2pq(p-q)}{p+q}\right)^2 = 4p^2q^2\left[1 + \left(\dfrac{p-q}{p+q}\right)^2\right] \\ &= \frac{4p^2q^2}{(p+q)^2}\left[(p+q)^2 + (p-q)^2\right] = \frac{4p^2q^2}{(p+q)^2}(2p^2 + 2q^2) \end{align*}$

and so $CE$ is rational if and only if $2p^2 + 2q^2$ is a perfect square.

Also by the angle bisector theorem,

$\begin{align*} AD &= \dfrac{AB \cdot AC}{AB + BC} = \dfrac{2pq(p^2-q^2)}{p^2 + q^2 + p^2 - q^2} = \dfrac{q(p^2-q^2)}{p} \end{align*}$

and therefore, since $\triangle DAB$ is a right-triangle, we have:

$\begin{align*} BD^2 &= AB^2 + AD^2 = (p^2-q^2)^2 + \left(\dfrac{q(p^2-q^2)}{p}\right)^2 \\ &= (p^2-q^2)^2\left[1 + \frac{q^2}{p^2}\right] = \frac{(p^2-q^2)^2}{p^2}(p^2 + q^2) \end{align*}$

and so $BD$ is rational if and only if $p^2 + q^2$ is a perfect square.

Combining the conditions on $CE$ and $BD$ , we see that $2p^2+2q^2$ and $p^2+q^2$ must both be perfect squares. If it were so, their ratio, which is $2$ , would be the square of a rational number, but $\sqrt{2}$ is irrational, and so the assumed triangle cannot exist.

Video Solution

https://www.youtube.com/watch?v=Dh9H_DNDMAg

2010 USAMO (Problems • Resources)
Preceded by Problem 3	Followed by Problem 5
1 • 2 • 3 • 4 • 5 • 6
All USAMO Problems and Solutions

2010 USAJMO (Problems • Resources)
Preceded by Problem 5	Followed by Last Problem
1 • 2 • 3 • 4 • 5 • 6
All USAJMO Problems and Solutions

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