# 2008 iTest Problems/Problem 74

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## Problem

Points $C$ and $D$ lie on opposite sides of line $\overline{AB}$. Let $M$ and $N$ be the centroids of $\triangle ABC$ and $\triangle ABD$ respectively. If $AB=841$, $BC=840$, $AC=41$, $AD=609$, and $BD=580$, find the sum of the numerator and denominator of the value of $MN$ when expressed as a fraction in lowest terms.

## Solution

Notice that $41^2 + 840^2 = 841^2$ and $580^2 + 609^2 = 841^2$, so $\triangle ABC$ and $\triangle ABD$ are right triangles. Since $\angle ACB = \angle ADB = 90^\circ$, quadrilateral $ACBD$ is cyclic. $[asy] pair A=(0,0),C=(1681/841,34440/841),B=(841,0),D=(370881/841,-353220/841); draw(A--C--B--D--A); dot(A); label("A",A,W); dot(C); label("C",C,NW); dot(B); label("B",B,E); dot(D); label("D",D,S); draw(circle((841/2,0),841/2)); dot((841/2,0)); draw(C--(841/2,0)--D); pair M=(1417924/5046,11480/841),n=(1282/3,-140); dot(M); dot(n); label("M",M,N); label("N",n,E); draw(C--D,dotted); draw(M--n,dotted); [/asy]$

Let $O$ be the midpoint of $AB$. Since $M$ and $N$ are centroids, $CM : MO = DN : DO = 2 : 1$. Thus, by SAS Similarity, $\triangle COD \sim \triangle MON$, so $MN = \tfrac13 CD$.

By Ptolemy's Theorem, $AC \cdot BD + BC \cdot AD = CD \cdot AB$, so \begin{align*} 41 \cdot 580 + 840 \cdot 609 &= 841 \cdot CD \\ 535340 &= 841 \cdot CD \\ CD &= \frac{18460}{29} \end{align*}

That means $MN = \tfrac13 \cdot \tfrac{18460}{29} = \tfrac{18460}{87}$, and the sum of the numerator and denominator in lowest terms is $18460 + 87 = \boxed{18547}$.

## See Also

 2008 iTest (Problems) Preceded by:Problem 73 Followed by:Problem 75 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 • 41 • 42 • 43 • 44 • 45 • 46 • 47 • 48 • 49 • 50 • 51 • 52 • 53 • 54 • 55 • 56 • 57 • 58 • 59 • 60 • 61 • 62 • 63 • 64 • 65 • 66 • 67 • 68 • 69 • 70 • 71 • 72 • 73 • 74 • 75 • 76 • 77 • 78 • 79 • 80 • 81 • 82 • 83 • 84 • 85 • 86 • 87 • 88 • 89 • 90 • 91 • 92 • 93 • 94 • 95 • 96 • 97 • 98 • 99 • 100
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