CEMC Fermat, 2014

Geometry Difficulty 4.7 AIME Find the answer Canada

$\triangle PQR$ has $PQ=QR=RP=30$ . Points $S$ and $T$ are on $PQ$ and $PR$ , respectively, so that $ST$ is parallel to $QR$ . Points $V$ and $U$ are on $QR$ so that $TU$ is parallel to $PQ$ and $SV$ is parallel to $PR$ .

If $VS+ST+TU=35$ , the length of $VU$ is

Pick one

Solution

Since $\triangle PQR$ has $PQ=QR=RP$ , then $\triangle PQR$ is equilateral and all of its angles equal $60^\circ$ .

Since $ST$ is parallel to $QR$ , $SV$ is parallel to $PR$ , and $TU$ is parallel to $PQ$ , then all of the angles in $\triangle PST$ , $\triangle SQV$ and $\triangle TUR$ equal $60^\circ$ . In other words, each of these triangles is also equilateral.

Let $SQ=x$ .

Since $\triangle SQV$ is equilateral, then $QV=VS=SQ=x$ .

Since $PQ=30$ , then $PS=30-x$ .

Since $\triangle PST$ is equilateral, then $ST=TP=PS=30-x$ .

Since $PR=30$ , then $TR=30-(30-x)=x$ .

Since $\triangle TUR$ is equilateral, then $TU=UR=TR=x$ .

Since $VS+ST+TU=35$ , then $x+(30-x)+x=35$ or $30+x=35$ and so $x=5$ .

Therefore, $VU=QR-QV-UR=30-x-x=30-5-5=20$ .

Want a route through all this instead of an archive? The track puts 2,604 problems in a working order, from Junior Challenge level to the IMO shortlist.

Source: CEMC, University of Waterloo, licensed CC-BY-NC-4.0. Statement reproduced verbatim; metadata (topic, difficulty) added by this project.