Difference between revisions of "2010 USAMO Problems/Problem 4"

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==Solution 2==
 
==Solution 2==
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The answer is no.
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Suppose otherwise. It is easy to see (through simple angle chasing) that <math>\angle DIC=45^{\circ}</math>. Also, since <math>I</math> is the incenter, we have <math>\angle IAC = 45^{\circ}</math>. Using the Law of Cosines, we have <cmath>CD^2=IC^2+ID^2-\sqrt{2}(IC)(ID),</cmath> so that <math>CD</math> is irrational. But <math>\triangle IAC \sim \triangle DIC</math>, thus <math>IC^2=CD\cdot AC</math>, implying that <math>CD</math> is rational, contradiction. <math>\blacksquare</math>
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==Solution 3==
 
The result can be also proved without direct appeal to trigonometry,
 
The result can be also proved without direct appeal to trigonometry,
 
via just the angle bisector theorem and the structure of Pythagorean
 
via just the angle bisector theorem and the structure of Pythagorean
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their ratio, which is <math>2</math>, would be the square of a rational number,
 
their ratio, which is <math>2</math>, would be the square of a rational number,
 
but <math>\sqrt{2}</math> is irrational, and so the assumed triangle cannot exist.
 
but <math>\sqrt{2}</math> is irrational, and so the assumed triangle cannot exist.
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==Video Solution==
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https://www.youtube.com/watch?v=Dh9H_DNDMAg
  
 
==See also==
 
==See also==
 
{{USAMO newbox|year=2010|num-b=3|num-a=5}}
 
{{USAMO newbox|year=2010|num-b=3|num-a=5}}
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{{USAJMO newbox|year=2010|num-b=5|after=Last Problem}}
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[[Category: Olympiad Geometry Problems]]
 
{{MAA Notice}}
 
{{MAA Notice}}

Latest revision as of 20:59, 2 October 2021

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

See also

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

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