Difference between revisions of "2014 UMO Problems/Problem 3"

(Solution)
m (Solution)
 
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size(200);
 
size(200);
 
path T=((0,0)--(4,0)--(4,.3)--(3.7,.3)--(3.7,0)--(4,0)--(4,3)--(0,0));
 
path T=((0,0)--(4,0)--(4,.3)--(3.7,.3)--(3.7,0)--(4,0)--(4,3)--(0,0));
D(T,black);
 
 
D(shift(10,0)*rotate(53)*T,black);
 
D(shift(10,0)*rotate(53)*T,black);
 
D(shift(15,5)*rotate(233)*T,black);
 
D(shift(15,5)*rotate(233)*T,black);

Latest revision as of 15:27, 12 February 2020

Problem

Completely describe the set of all right triangles with positive integer-valued legs such that when four copies of the triangle are arranged in square formation shown below, the incenters of the four triangles lie on the extensions of the sides of the smaller square. (Note: the incenter of a triangle is the center of the circle inscribed in that triangle.)

[asy] size(200); path T=((0,0)--(4,0)--(4,.3)--(3.7,.3)--(3.7,0)--(4,0)--(4,3)--(0,0)); D(T,black); D(shift(10,0)*rotate(53)*T,black); D(shift(15,5)*rotate(233)*T,black); D(shift(15,0)*rotate(143)*T,black); D(shift(10,5)*rotate(323)*T,black); [/asy]


Solution

[asy] size(200); path T=((0,0)--(4,0)--(4,.3)--(3.7,.3)--(3.7,0)--(4,0)--(4,3)--(0,0)); D(shift(10,0)*rotate(53)*T,black); D(shift(15,5)*rotate(233)*T,black); D(shift(15,0)*rotate(143)*T,black); D(shift(10,5)*rotate(323)*T,black); pair A, B, C, X, Y, Z, I; A = shift(10,0)*rotate(53)*(0,0); B = shift(10,0)*rotate(53)*(4, 0); C = shift(10,0)*rotate(53)*(4, 3); I = incenter(A, B, C); X = foot(I, B, C); Y = foot(I, A, C); Z = foot(I, A, B); D(incircle(A, B, C), black); D(I -- X, black); D(I -- Y, black); D(I -- Z, black); label("X", X, NE); label("Y", Y, W); label("Z", Z, S); label("A", A, S); label("B", B, NE); label("C", C, NW); label("I", I, N); D(A--Y, green); D(A--Z, green); D(B--Z, red); D(B--X, red); D(C--Y, blue); D(C--X, blue); [/asy] Let $I$ be the incenter of a triangle. Drop $I$ onto the three sides of the triangle, and let the points be $X, Y, Z$ Finally, let $a = AB, b = BC$ so that $a > b$ and let $s = BZ$. Note that $Z$ is also a corner of the square, so $s = a - b$. But then, $AC = CX + AZ = a + b - 2(a-b) = 3b-a$. From the Pythogorean theorem, we also know that $AC^2 = a^2+b^2$. Therefore, \[a^2 + b^2 = (3b-a)^2 = 9b^2-6ab + a^2\] \[\Longleftrightarrow\] \[8b^2 = 6ab\] \[\Longleftrightarrow\] \[b = \frac34 a\] So, the only solutions are of the form $(3k, 4k, 5k).$

See Also

2014 UMO (ProblemsAnswer KeyResources)
Preceded by
Problem 2
Followed by
Problem 4
1 2 3 4 5 6
All UMO Problems and Solutions
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