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2008 Mock ARML 2 Problems/Problem 4

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Problem

Equilateral triangle $ABC$ has a side length of $7$. A ball begins at vertex $A$, rolls through the interior of the triangle, bounces off side $BC$, and settles at point $P$. Given that $BP = 3$ and $CP = 5$, find the total distance that the ball travels.

Solution

[asy] pointpen = black; pathpen = black + linewidth(0.7); pair A=(0,0),B=(7,0),C=7*expi(pi/3);  pair Q=IP(CR(B,3),CR(C,5)), P=OP(CR(B,3),CR(C,5)),D=IP(A--Q,B--C); D(MP("A",A)--MP("B",B)--MP("C",C,N)--cycle); D(A--D--D(MP("P",P)));D(D--D(MP("P'",Q,NE))--C--Q--B,linetype("5 5")); MP("7",(A+B)/2); MP("3",(B+Q)/2,SE); MP("5",(C+Q)/2,NE);  [/asy]

Reflect $P$ across $\overline{BC}$ to point $P'$; since the ball travels in a straight path, it follows that the distance the ball traveled is $AP'$. By symmetry, $BP' = 3, CP' = 5$. By the Law of Cosines on $\triangle BCP'$,

$7^2 = 3^2 + 5^2 - 2 \cdot 3 \cdot 5 \cos \theta \Longrightarrow \theta = 120^{\circ}.$

Since $\angle BAC + \angle BP'C = 60 + 120 = 180$, it follows that quadrilateral $ABP'C$ is a cyclic quadrilateral. By Ptolemy's Theorem,

$3\cdot 7 + 5\cdot 7 = 7 \cdot AP' \Longrightarrow \boxed{AP' = 8}.$

See also

2008 Mock ARML 2 (Problems, Source)
Preceded by
Problem 3
Followed by
Problem 5
1 2 3 4 5 6 7 8
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