Difference between revisions of "2014 UMO Problems/Problem 3"
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== Solution == | == 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(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); | ||
+ | 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)); | ||
+ | D(I -- X); | ||
+ | D(I -- Y); | ||
+ | D(I -- Z); | ||
+ | label("X", X, NE,blue); | ||
+ | label("Y", Y, W, blue); | ||
+ | label("Z", Z, S, blue); | ||
+ | label("A", A, S); | ||
+ | label("B", B, NE); | ||
+ | label("C", C, NW); | ||
+ | label("I", I, N, blue); | ||
+ | draw(A--Y, green); | ||
+ | draw(A--Z, yellow); | ||
+ | draw(B--Z, yellow); | ||
+ | draw(B--X, red); | ||
+ | draw(C--Y, green); | ||
+ | draw(C--x, red); | ||
+ | </asy> | ||
+ | Let <math>I</math> be the incenter of a triangle. Drop <math>I</math> onto the three sides of the triangle, and let the points be <math>X, Y, Z</math> | ||
+ | Finally, let <math>a = AB, b = BC</math> so that <math>a > b</math> and let <math>s = BZ</math>. | ||
+ | Note that <math>Z</math> is also a corner of the square, so <math>s = a - b</math>. But then, <math>AC = CX + AZ = a + b - 2(a-b) = 3b-a</math>. From the Pythogorean theorem, we also know that <math>AC^2 = a^2+b^2</math>. Therefore, | ||
+ | <cmath>a^2 + b^2 = (3b-a)^2 = 9b^2-6ab + a^2</cmath> | ||
+ | <cmath>\Longleftrightarrow</cmath> | ||
+ | <cmath>8b^2 = 6ab</cmath> | ||
+ | <cmath>\Longleftrightarrow</cmath> | ||
+ | <cmath>b = \frac34 a</cmath> | ||
+ | So, the only solutions are of the form <math>(3k, 4k, 5k).</math> | ||
== See Also == | == See Also == |
Revision as of 12:09, 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.)
Solution
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); 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)); D(I -- X); D(I -- Y); D(I -- Z); label("X", X, NE,blue); label("Y", Y, W, blue); label("Z", Z, S, blue); label("A", A, S); label("B", B, NE); label("C", C, NW); label("I", I, N, blue); draw(A--Y, green); draw(A--Z, yellow); draw(B--Z, yellow); draw(B--X, red); draw(C--Y, green); draw(C--x, red); (Error compiling LaTeX. 86f063c5cfe7360f5832fe37c074b2df3ee501ba.asy: 34.9: no matching variable 'x')
Let be the incenter of a triangle. Drop onto the three sides of the triangle, and let the points be Finally, let so that and let . Note that is also a corner of the square, so . But then, . From the Pythogorean theorem, we also know that . Therefore, So, the only solutions are of the form
See Also
2014 UMO (Problems • Answer Key • Resources) | ||
Preceded by Problem 2 |
Followed by Problem 4 | |
1 • 2 • 3 • 4 • 5 • 6 | ||
All UMO Problems and Solutions |