Question

Given that $x,$ $y,$ and $z$ are real numbers that satisfy $x = \sqrt{y^2 - \frac{1}{16}} + \sqrt{z^2 - \frac{1}{16}},$ $y = \sqrt{z^2 - \frac{1}{25}} + \sqrt{x^2 - \frac{1}{25}},$ $z = \sqrt{x^2 - \frac{1}{36}} + \sqrt{y^2 - \frac{1}{36}},$ and that $x + y + z = \frac{m}{\sqrt{n}},$ where $m$ and $n$ are positive integers, and $n$ is not divisible by the square of any prime, find $m + n.$

Answer

Solution:

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

If $K$ is the area of this triangle, then $K = \frac{1}{2} \cdot x \cdot \frac{1}{4}$ gives $x = 8K,$ and likewise $y = 10K$ and $z = 12K.$ These are proportional to $8, 10, 12,$ and $8^2 + 10^2 \gt 12^2,$ so the triangle is acute and the altitude feet do land inside the sides. Heron's formula with $s = 15K$ gives $K^2 = 15K \cdot 7K \cdot 5K \cdot 3K = 1575K^4,$ so $K^2 = \frac{1}{1575}$ and $K = \frac{1}{15\sqrt{7}}.$

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

2006 AIME II Problem 15

Solution:

Problem 15 in Other Years