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

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==Problem==
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Let <math>ABC</math> be a triangle with <math>\angle A = 90^{\circ}</math>. Points <math>D</math>
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and <math>E</math> lie on sides <math>AC</math> and <math>AB</math>, respectively, such that <math>\angle
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ABD = \angle DBC</math> and <math>\angle ACE = \angle ECB</math>. Segments <math>BD</math> and
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<math>CE</math> meet at <math>I</math>.  Determine whether or not it is possible for
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segments <math>AB, AC, BI, ID, CI, IE</math> to all have integer lengths.
 +
 +
==Video Solution in 3 minutes!!!==
 +
https://www.youtube.com/watch?v=Z9fjesTOs1Q
 +
 
==Solution==
 
==Solution==
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We know that angle <math>BIC = 135^{\circ}</math>, as the other two angles in triangle <math>BIC</math> add to <math>45^{\circ}</math>. Assume that only <math>AB, AC, BI</math>, and <math>CI</math> are integers. Using the [[Law of Cosines]] on triangle BIC,
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<center>
 +
<asy>
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import olympiad;
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 +
// Scale
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unitsize(1inch);
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 +
// Shape
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real h = 1.75;
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real w = 2.5;
 +
 +
// Points
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void ldot(pair p, string l, pair dir=p) { dot(p); label(l, p, unit(dir)); }
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pair A = origin; ldot(A, "$A$", plain.SW);
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pair B = w * plain.E; ldot(B, "$B$", plain.SE);
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pair C = h * plain.N; ldot(C, "$C$", plain.NW);
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pair D = extension(B, bisectorpoint(C, B, A), A, C); ldot(D, "$D$", D-B);
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pair E = extension(C, bisectorpoint(A, C, B), A, B); ldot(E, "$E$", E-C);
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pair I = extension(B, D, C, E); ldot(I, "$I$", A-I);
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// Segments
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draw(A--B); draw(B--C); draw(C--A);
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draw(C--E); draw(B--D);
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// Angles
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import markers;
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draw(rightanglemark(B, A, C, 4));
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markangle(Label("$\scriptstyle{\frac{\theta}{2}}$"), radius=40, I, B, E);
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markangle(Label("$\scriptstyle{\frac{\theta}{2}}$"), radius=40, C, B, I);
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markangle(Label("$\scriptstyle{\frac{\pi}{4} - \frac{\theta}{2}}$"), radius=40, I, C, B);
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markangle(Label("$\scriptstyle{\frac{\pi}{4} - \frac{\theta}{2}}$"), radius=40, D, C, I);
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markangle(Label("$\scriptstyle{\frac{3\pi}{4}}$"), radius=10, B, I, C);
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</asy>
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</center>
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<math>BC^2 = BI^2 + CI^2 - 2BI\cdot CI \cdot \cos 135^{\circ}</math>. Observing that <math>BC^2 = AB^2 + AC^2</math> is an integer and that <math>\cos 135^{\circ} = -\frac{\sqrt{2}}{2},</math> we have
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<center>
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<cmath>
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BC^2 - BI^2 - CI^2 = BI\cdot CI\cdot \sqrt{2}
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</cmath>
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</center>
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and therefore,
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<center>
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<cmath>
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\sqrt{2} = \frac{BC^2 - BI^2 - CI^2}{BI\cdot CI}
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</cmath>
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</center>
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The LHS (<math>\sqrt{2}</math>) 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 <math>AB, AC, BI</math>, and <math>CI</math> to all be integers, which invalidates the original claim that all six lengths are integers, and we are done.
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==Solution 2==
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The answer is no.
 +
 +
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==
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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
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only if there exists a triangle in which <math>AB</math> and <math>AC</math> are
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relatively-prime integers and the lengths of the segments <math>BI, ID, CI, IE</math> are all rational
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(we divide all the lengths by the <math>\gcd(AB, AC)</math> or
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conversely multiply all the lengths by the least common multiple
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of the denominators of the rational lengths).
 +
 +
Suppose there exists a triangle in which the lengths <math>AB</math> and <math>AC</math> are
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relatively-prime integers and the lengths <math>IB, ID, CI, IE</math> are all rational.
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 +
Since <math>CE</math> is the bisector of <math>\angle ACB</math>, by the angle bisector
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theorem, the ratio <math>IB : ID = CB : CD</math>, and since <math>BD</math> is the
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bisector of <math>\angle ABC</math>, <math>CB : CD = (AB + BC) : AC</math>. Therefore,
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<math>IB : ID = (AB + BC) : AC</math>. Now <math>IB : ID</math> is by assumption rational,
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so <math>(AB + BC) : AC</math> is rational, but <math>AB</math> and <math>AC</math> are assumed integers
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so <math>BC : AC</math> must also be rational. Since <math>BC</math> is the hypotenuse of
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a right-triangle, its length is the square root of an integer,
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and thus either an integer or irrational, so <math>BC</math> must be an integer.
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With <math>AB</math> and <math>AC</math> relatively-prime, we conclude that the side
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lengths of <math>\triangle ABC</math> must be a Pythagorean triple: <math>(2pq, p^2
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- q^2, p^2 + q^2)</math>, with <math>p > q</math> relatively-prime positive integers
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and <math>p+q</math> odd.
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Without loss of generality, <math>AC = 2pq, AB = p^2 - q^2, BC = p^2+q^2</math>.
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By the angle bisector theorem,
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<center>
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<cmath>
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\begin{align*}
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AE &= \dfrac{AB \cdot AC}{AC + CB} = \dfrac{2pq(p^2-q^2)}{p^2 + q^2 + 2pq}
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  = \dfrac{2pq(p-q)}{p+q}
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\end{align*}
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</cmath>
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</center>
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Since <math>\triangle CAE</math> is a right-triangle, we have:
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<center>
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<cmath>
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\begin{align*}
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  CE^2 &= AC^2 + AE^2
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      = 4p^2q^2 + \left(\dfrac{2pq(p-q)}{p+q}\right)^2
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      = 4p^2q^2\left[1 + \left(\dfrac{p-q}{p+q}\right)^2\right] \
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      &= \frac{4p^2q^2}{(p+q)^2}\left[(p+q)^2 + (p-q)^2\right]
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      = \frac{4p^2q^2}{(p+q)^2}(2p^2 + 2q^2)
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\end{align*}
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</cmath>
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</center>
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and so <math>CE</math> is rational if and only if <math>2p^2 + 2q^2</math> is a perfect square.
 +
 +
Also by the angle bisector theorem,
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<center>
 +
<cmath>
 +
\begin{align*}
 +
AD &= \dfrac{AB \cdot AC}{AB + BC} = \dfrac{2pq(p^2-q^2)}{p^2 + q^2 + p^2 - q^2}
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  = \dfrac{q(p^2-q^2)}{p}
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\end{align*}
 +
</cmath>
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</center>
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and therefore, since <math>\triangle DAB</math> is a right-triangle, we have:
 +
<center>
 +
<cmath>
 +
\begin{align*}
 +
  BD^2 &= AB^2 + AD^2
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      = (p^2-q^2)^2 + \left(\dfrac{q(p^2-q^2)}{p}\right)^2 \
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      &= (p^2-q^2)^2\left[1 + \frac{q^2}{p^2}\right]
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      = \frac{(p^2-q^2)^2}{p^2}(p^2 + q^2)
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\end{align*}
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</cmath>
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</center>
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and so <math>BD</math> is rational if and only if <math>p^2 + q^2</math> is a perfect square.
 +
 +
Combining the conditions on <math>CE</math> and <math>BD</math>, we see that
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<math>2p^2+2q^2</math> and <math>p^2+q^2</math> must both be perfect squares. If it were so,
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their ratio, which is <math>2</math>, would be the square of a rational number,
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but <math>\sqrt{2}</math> is irrational, and so the assumed triangle cannot exist.
 +
 +
==Solution 4==
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We proceed by contradiction.
  
We know that angle <math>BIC = \frac{3\pi}{4}</math>, as the other two angles in triangle <math>BIC</math> add to <math>\frac{\pi}{4}</math>. Assume that only <math>AB, BC, BI</math>, and <math>CI</math> are integers. Using the [[Law of Cosines]] on triangle BIC,
+
FTSOC, let <math>AB, BI, ID</math> have integer lengths. Then <math>BD = BI + ID \in \mathbb{Z}</math> as well. By trigonometry, <cmath>BD = \frac{AB}{\cos{(\frac{\angle ABC}{2})}}.</cmath>
 +
Rearranging we find <math>\cos{(\frac{\angle ABC}{2})} = \frac{AB}{BD} \in \mathbb{Q}</math>. As <math>0 < \angle ABC < \frac{\pi}{2}</math> so <math>0 < \frac{\angle ABC}{2} < \frac{\pi}{4}</math> but there are no possible angles in the interval that result in a rational cosine by Niven's theorem. So we have contradiction.
  
<math>BC^2 = BI^2 + CI^2 - 2BI*CI*cos \frac{3\pi}{4}</math>. Observing that <math>BC^2 = AB^2 + AC^2</math> and that <math>cos \frac{3\pi}{4} = -\frac{\sqrt{2}}{2}</math>, we have
+
~Aaryabhatta1.
  
<math>AB^2 + AC^2 - BI^2 - CI^2 = BI*CI*\sqrt{2}</math>
+
==Video Solution==
 +
https://www.youtube.com/watch?v=Dh9H_DNDMAg
  
<math>\sqrt{2} = \frac{AB^2 + AC^2 - BI^2 - CI^2}{BI*CI}</math>
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==See also==
 +
{{USAMO newbox|year=2010|num-b=3|num-a=5}}
 +
{{USAJMO newbox|year=2010|num-b=5|after=Last Problem}}
  
Since the right side of the equation is a rational number, the left side (i.e. <math>\sqrt{2}</math>) must also be rational. Obviously since <math>\sqrt{2}</math> is irrational, this claim is false and we have a contradiction. Therefore, it is impossible for <math>AB, BC, BI</math>, and <math>CI</math> to all be integers, which invalidates the original claim that all six lengths are integers, and we are done.
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[[Category: Olympiad Geometry Problems]]
 +
{{MAA Notice}}

Latest revision as of 18:47, 31 January 2024

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.

Video Solution in 3 minutes!!!

https://www.youtube.com/watch?v=Z9fjesTOs1Q

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.

Solution 4

We proceed by contradiction.

FTSOC, let $AB, BI, ID$ have integer lengths. Then $BD = BI + ID \in \mathbb{Z}$ as well. By trigonometry, \[BD = \frac{AB}{\cos{(\frac{\angle ABC}{2})}}.\] Rearranging we find $\cos{(\frac{\angle ABC}{2})} = \frac{AB}{BD} \in \mathbb{Q}$. As $0 < \angle ABC < \frac{\pi}{2}$ so $0 < \frac{\angle ABC}{2} < \frac{\pi}{4}$ but there are no possible angles in the interval that result in a rational cosine by Niven's theorem. So we have contradiction.

~Aaryabhatta1.

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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