# 2006 iTest Problems/Problem 29

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## Problem

The altitudes in triangle $ABC$ have lengths 10, 12, and 15. The area of $ABC$ can be expressed as $\tfrac{m\sqrt n}p$, where $m$ and $p$ are relatively prime positive integers and $n$ is a positive integer not divisible by the square of any prime. Find $m + n + p$.

$[asy] import olympiad; size(200); defaultpen(linewidth(0.7)+fontsize(11pt)); pair A = (9,8.5), B = origin, C = (15,0); draw(A--B--C--cycle); pair D = foot(A,B,C), E = foot(B,C,A), F = foot(C,A,B); draw(A--D^^B--E^^C--F); label("A",A,N); label("B",B,SW); label("C",C,SE); [/asy]$

## Solution (credit to jeffisepic)

Let $A$ be the area of the triangle. That makes the three side lengths $\tfrac{A}{5}, \tfrac{A}{6}, \tfrac{2A}{15}$. Since we know the lengths of all three sides, we can use Heron's Formula to solve for $A$. \begin{align*} A &= \sqrt{\frac{A}{4} \cdot \frac{A}{20} \cdot \frac{A}{12} \cdot \frac{7A}{60}} \\ A &= \frac{A^2 \sqrt{7}}{240} \\ 0 &= A(\frac{A\sqrt{7}}{240} - 1) \end{align*} Since the area can not equal 0, $A = 1 \cdot \tfrac{240}{\sqrt{7}} = \tfrac{240\sqrt{7}}{7}$. Thus, $m + n + p = \boxed{254}$.

## See Also

 2006 iTest (Problems) Preceded by:Problem 28 Followed by:Problem 30 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 • 26 • 27 • 28 • 29 • 30 • 31 • 32 • 33 • 34 • 35 • 36 • 37 • 38 • 39 • 40 • U1 • U2 • U3 • U4 • U5 • U6 • U7 • U8 • U9 • U10
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