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CEMC Pascal, 2016

Geometry Difficulty 2.5 Junior Find the answer Canada

In the diagram, $PQ$ is perpendicular to $QR$ , $QR$ is perpendicular to $RS$ , and $RS$ is perpendicular to $ST$ .

If $PQ=4$ , $QR=8$ , $RS=8$ , and $ST=3$ , then the distance from $P$ to $T$ is

Pick one

Solution

Extend $PQ$ and $ST$ to meet at $U$ .

Since $QUSR$ has three right angles, then it must have four right angles and so is a rectangle.

Thus, $\triangle PUT$ is right-angled at $U$ .

By the Pythagorean Theorem, $PT^2 = PU^2 + UT^2$ .

Now $PU = PQ + QU$ and $QU = RS$ so $PU = 4 + 8 = 12$ .

Also, $UT = US - ST$ and $US = QR$ so $UT = 8 - 3 = 5$ .

Therefore, $PT^2 = 12^2 + 5^2 = 144 + 25 = 169$ .

Since $\text{PT gt; 0}$ , then $PT = \sqrt{169}=13$ .

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