2006 AIME II Problem 15

Attempt Problem 15 of the 2006 AIME II below, then check your answer against the professionally curated solution from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 2006 AIME II solutions, or check the answer key.

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15.

Given that x,x, y,y, and zz are real numbers that satisfy x=y2116+z2116,x = \sqrt{y^2 - \frac{1}{16}} + \sqrt{z^2 - \frac{1}{16}}, y=z2125+x2125,y = \sqrt{z^2 - \frac{1}{25}} + \sqrt{x^2 - \frac{1}{25}}, z=x2136+y2136,z = \sqrt{x^2 - \frac{1}{36}} + \sqrt{y^2 - \frac{1}{36}}, and that x+y+z=mn,x + y + z = \frac{m}{\sqrt{n}}, where mm and nn are positive integers, and nn is not divisible by the square of any prime, find m+n.m + n.

Answer: 9
Concepts:system of equationsaltitudeHeron’s Formula
Difficulty rating: 3370
Solution:

Each radical y2116\sqrt{y^2 - \frac{1}{16}} is the leg of a right triangle with hypotenuse yy and other leg 14.\frac{1}{4}. So the first equation says: in a triangle XYZXYZ with x=YZ,x = YZ, y=ZX,y = ZX, z=XY,z = XY, the altitude from XX has length 14,\frac{1}{4}, and its foot splits YZYZ into the two radical lengths. The other equations say the altitudes to sides yy and zz are 15\frac{1}{5} and 16.\frac{1}{6}.

If KK is the area of this triangle, then K=12x14K = \frac{1}{2} \cdot x \cdot \frac{1}{4} gives x=8K,x = 8K, and likewise y=10Ky = 10K and z=12K.z = 12K. These are proportional to 8,10,12,8, 10, 12, and 82+102>122,8^2 + 10^2 \gt 12^2, so the triangle is acute and the altitude feet do land inside the sides. Heron's formula with s=15Ks = 15K gives K2=15K7K5K3K=1575K4, \begin{aligned} K^2 &= 15K \cdot 7K \cdot 5K \cdot 3K \\ &= 1575K^4, \end{aligned} so K2=11575K^2 = \frac{1}{1575} and K=1157.K = \frac{1}{15\sqrt{7}}.

Then x+y+z=30K=27,x + y + z = 30K = \frac{2}{\sqrt{7}}, so m+n=2+7=9.m + n = 2 + 7 = 9.

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