Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

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

m (→‎Solution)
(→‎Solution)
Line 7: Line 7:
  
 
==Solution==
 
==Solution==
 +
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, BC, BI</math>, and <math>CI</math> are integers. Using the [[Law of Cosines]] on triangle BIC,
 +
<center>
 +
<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);
  
−
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, BC, BI</math>, and <math>CI</math> are integers. Using the [[Law of Cosines]] on triangle BIC,
+
// 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>
 +
</center>
  
 
<math>BC^2 = BI^2 + CI^2 - 2BI*CI*cos 135^{\circ}</math>. Observing that <math>BC^2 = AB^2 + AC^2</math> and that <math>cos 135^{\circ} = -\frac{\sqrt{2}}{2}</math>, we have
 
<math>BC^2 = BI^2 + CI^2 - 2BI*CI*cos 135^{\circ}</math>. Observing that <math>BC^2 = AB^2 + AC^2</math> and that <math>cos 135^{\circ} = -\frac{\sqrt{2}}{2}</math>, we have
Line 17: Line 50:
  
 
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.
 
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.
 +
 +
==Solution 2==
 +
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 <math>AB</math> and <math>AC</math> are
 +
relatively-prime integers and the lengths of the segments <math>BI, ID, CI, IE</math> are all rational
 +
(we divide all the lengths by the <math>\gcd(AB, AC)</math> 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 <math>AB</math> and <math>AC</math> are
 +
relatively-prime integers and the lengths <math>IB, ID, CI, IE</math> are all rational.
 +
 +
Since <math>CE</math> is the bisector of <math>\angle ACB</math>, by the angle bisector
 +
theorem, the ratio <math>IB : ID = CB : CD</math>, and since <math>BD</math> is the
 +
bisector of <math>\angle ABC</math>, <math>CB : CD = (AB + BC) : AC</math>. Therefore,
 +
<math>IB : ID = (AB + BC) : AC</math>. Now <math>IB : ID</math> is by assumption rational,
 +
so <math>(AB + BC) : AC</math> is rational, but <math>AB</math> and <math>AC</math> are assumed integers
 +
so <math>BC : AC</math> must also be rational. Since <math>BC</math> is the hypotenuse of
 +
a right-triangle, its length is the square root of an integer,
 +
and thus either an integer or irrational, so <math>BC</math> must be an integer.
 +
 +
With <math>AB</math> and <math>AC</math> relatively-prime, we conclude that the side
 +
lengths of <math>\triangle ABC</math> must be a Pythagorean triple: <math>(2pq, p^2
 +
- q^2, p^2 + q^2)</math>, with <math>p > q</math> relatively-prime positive integers
 +
and <math>p+q</math> odd.
 +
 +
Without loss of generality, <math>AC = 2pq, AB = p^2 - q^2, BC = p^2+q^2</math>.
 +
By the angle bisector theorem,
 +
<center>
 +
<cmath>
 +
\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*}
 +
</cmath>
 +
</center>
 +
 +
Since <math>\triangle CAE</math> is a right-triangle, we have:
 +
<center>
 +
<cmath>
 +
\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*}
 +
</cmath>
 +
</center>
 +
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,
 +
<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}
 +
  = \dfrac{q(p^2-q^2)}{p}
 +
\end{align*}
 +
</cmath>
 +
</center>
 +
and therefore, since <math>\triangle DAB</math> is a right-triangle, we have:
 +
<center>
 +
<cmath>
 +
\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*}
 +
</cmath>
 +
</center>
 +
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
 +
<math>2p^2+2q^2</math> and <math>p^2+q^2</math> must both be perfect squares. If it were so,
 +
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.

Revision as of 19:37, 12 May 2010

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 $AB, BC, BI$, 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]

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

$AB^2 + AC^2 - BI^2 - CI^2 = BI*CI*\sqrt{2}$

$\sqrt{2} = \frac{AB^2 + AC^2 - BI^2 - CI^2}{BI*CI}$

Since the right side of the equation is a rational number, the left side (i.e. $\sqrt{2}$) must also be rational. Obviously since $\sqrt{2}$ is irrational, this claim is false and we have a contradiction. Therefore, it is impossible for $AB, BC, BI$, and $CI$ to all be integers, which invalidates the original claim that all six lengths are integers, and we are done.

Solution 2

The 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.

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2010_USAMO_Problems/Problem_4&oldid=34498"