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1998 AIME Problems/Problem 6

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Problem

Let A\displaystyle BCD be a parallelogram. Extend \overline{DA} through A to a point P, and let \overline{PC} meet \overline{AB} at Q and \overline{DB} at R. Given that PQ = 735 and QR = 112, find RC.

Solution

There are several similar triangles. \triangle PAQ \displaystyle \sim \triangle PDC, so we can write the proportion:

\frac{AQ}{CD} = \frac{PQ}{PC} = \frac{735}{112 + 735 + RC} = \frac{735}{847 + RC}

Also, \triangle BRQ \displaystyle \sim DRC, so:

\displaystyle \frac{QR}{RC} = \frac{QB}{CD} = \frac{112}{RC} = \frac{CD - AQ}{CD} = 1 - \frac{AQ}{CD} \displaystyle

\frac{AQ}{CD} = 1 - \frac{112}{RC} = \frac{RC - 112}{RC}

Substituting,

\frac{AQ}{CD} = \frac{735}{847 + RC} = \frac{RC - 112}{RC}

\displaystyle735RC = (RC + 847)(RC - 112)
0 = RC^2 - 112\cdot847

Thus, RC = \sqrt{112*847} = 308 \displaystyle.

See also

1998 AIME (ProblemsResources)
Preceded by
Problem 5
Followed by
Problem 7
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15
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