Integer Lengths

Geometry Level 4

Given triangle $ABC$ , let $P$ be a point on $AB$ and $Q$ be a point on $AC$ .

$PQ \cap BC=X$ and $\angle PXB=\angle C - \angle B$

$PC \times QB=1000$ and $PB \times QC=371$ .

Given that $BC>PQ>1$ and $BC$ and $PQ$ are integers, find the length of $BC$ .

1 solution

Sam Bealing
Mar 28, 2016

$\angle QCX=180-\angle C \Rightarrow \angle XQC=180-(\angle C- \angle B)-(180-\angle C)=\angle B$ so $\angle CQP=180-\angle B$ so $CQPB$ is cyclic.

By Ptolemy's theorem , we have $BQ \times PC=CQ \times BP+ BC \times PQ$ so $PQ \times BC=1000-371=629=17 \times 37$ .

As $BC>PQ>1$ and $BC$ and $PQ$ are integers, we have $BC=37$ .

It looks like you also need to assume that PQ is an integer.

Calvin Lin Staff - 5 years, 2 months ago

Yes you do, I have edited the problem.

Sam Bealing - 5 years, 2 months ago

Yay! Thanks! :)

Calvin Lin Staff - 5 years, 2 months ago

@Calvin Lin – THE FIGURES DECIEVES TO USE MENELAUS THEOREM...........

Abhisek Mohanty - 5 years, 2 months ago

@Abhisek Mohanty – Please refrain from typing in all caps, as that is considered very rude on the internet. Thanks for your assistance!

You are free to use any theorem / fact that you choose to use. Part of problem solving is to figure out the relevant tool to help you understand.

Calvin Lin Staff - 5 years, 2 months ago

Exact same solution. Beautiful problem.

Sal Gard - 5 years, 2 months ago

Solved the same way.

Niranjan Khanderia - 5 years, 1 month ago