Difference between revisions of "2010 USAMO Problems/Problem 4"
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==Problem== | ==Problem== | ||
− | Let <math>ABC</math> be a triangle with <math>\angle A = 90^{\ | + | Let <math>ABC</math> be a triangle with <math>\angle A = 90^{\circ}</math>. Points <math>D</math> |
and <math>E</math> lie on sides <math>AC</math> and <math>AB</math>, respectively, such that <math>\angle | and <math>E</math> lie on sides <math>AC</math> and <math>AB</math>, respectively, such that <math>\angle | ||
ABD = \angle DBC</math> and <math>\angle ACE = \angle ECB</math>. Segments <math>BD</math> and | ABD = \angle DBC</math> and <math>\angle ACE = \angle ECB</math>. Segments <math>BD</math> and | ||
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, AC, 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); | ||
− | + | // 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\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 | ||
+ | <center> | ||
+ | <cmath> | ||
+ | BC^2 - BI^2 - CI^2 = BI\cdot CI\cdot \sqrt{2} | ||
+ | </cmath> | ||
+ | </center> | ||
+ | and therefore, | ||
+ | <center> | ||
+ | <cmath> | ||
+ | \sqrt{2} = \frac{BC^2 - BI^2 - CI^2}{BI\cdot CI} | ||
+ | </cmath> | ||
+ | </center> | ||
+ | |||
+ | 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. | ||
+ | |||
+ | ==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. | ||
+ | |||
+ | ==Video Solution== | ||
+ | https://www.youtube.com/watch?v=Dh9H_DNDMAg | ||
+ | |||
+ | ==See also== | ||
+ | {{USAMO newbox|year=2010|num-b=3|num-a=5}} | ||
+ | {{USAJMO newbox|year=2010|num-b=5|after=Last Problem}} | ||
+ | |||
+ | [[Category: Olympiad Geometry Problems]] | ||
+ | {{MAA Notice}} |
Latest revision as of 13:32, 17 June 2020
Problem
Let be a triangle with . Points and lie on sides and , respectively, such that and . Segments and meet at . Determine whether or not it is possible for segments to all have integer lengths.
Solution
We know that angle , as the other two angles in triangle add to . Assume that only , and are integers. Using the Law of Cosines on triangle BIC,
. Observing that is an integer and that we have
and therefore,
The LHS () 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 , and 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 and are relatively-prime integers and the lengths of the segments are all rational (we divide all the lengths by the 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 and are relatively-prime integers and the lengths are all rational.
Since is the bisector of , by the angle bisector theorem, the ratio , and since is the bisector of , . Therefore, . Now is by assumption rational, so is rational, but and are assumed integers so must also be rational. Since is the hypotenuse of a right-triangle, its length is the square root of an integer, and thus either an integer or irrational, so must be an integer.
With and relatively-prime, we conclude that the side lengths of must be a Pythagorean triple: , with relatively-prime positive integers and odd.
Without loss of generality, . By the angle bisector theorem,
Since is a right-triangle, we have:
and so is rational if and only if is a perfect square.
Also by the angle bisector theorem,
and therefore, since is a right-triangle, we have:
and so is rational if and only if is a perfect square.
Combining the conditions on and , we see that and must both be perfect squares. If it were so, their ratio, which is , would be the square of a rational number, but is irrational, and so the assumed triangle cannot exist.
Video Solution
https://www.youtube.com/watch?v=Dh9H_DNDMAg
See also
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.