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The diagonals of a square meet at O. The bisector of angle OAB meets BO and BC at N and P respectively. The length of NO is 24. How long is PC?

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Weekly Problem 9 - 2006

weekly problem 9-2006

Points in Pairs

Stage: 4 Challenge Level: Challenge Level:1

We had a couple of correct solutions (500/48 or 10.4 to 3sf) using the Cosine Rule, and one solution which used the much simpler method of similar triangles:

I noticed that, since $OP'=100/OP$ and $OQ'=100/OQ$, the ratio $\frac{OP}{OQ}$ is $\frac{OQ'}{OP'}$ and so the triangles $OPQ$ and $OQ'P'$ are similar.

So, in this question $OP'=100/8$ and, since the ratio between lengths of sides is the same by similar triangles, $$ \frac{P'Q'}{OP'}=\frac{PQ}{OQ}.$$ Therefore, $$P'Q'=OP'\times\frac{PQ}{OQ}=\frac{100}{8}\times\frac{5}{6}=\frac{500}{48}.$$.