Maths Olympiad Prep

Library / /186 of 213

, 2018

Geometry Difficulty 3.7 AMC 10/12 Find the answer Canada

Line segments PQPQ and RSRS are parallel. Points T,U,VT, U, V are placed so that ∠QTV=30∘\angle QTV=30^{\circ}, ∠SUV=40∘\angle SUV=40^{\circ}, and ∠TVU=x∘\angle TVU=x^{\circ}, as shown.

A number or a short expression. Fractions can be typed as 3/2, and spacing doesn't matter.

Solution

Consider the following diagram.

[[IMAGE0]]

Since S,T,US,T,U lie on a straight line, ∠STU\angle STU measures 180∘180^{\circ}.
Therefore, ∠RTU=180∘−∠STR=180∘−120∘=60∘\angle RTU=180^{\circ}-\angle STR=180^{\circ}-120^{\circ}=60^{\circ}.
Similarly, Q,U,RQ,U,R lie on a straight line, and so ∠QUR\angle QUR

measures 180∘180^{\circ}.

Therefore, ∠TUR=180∘−∠TUQ=180∘−95∘=85∘\angle TUR=180^{\circ}-\angle TUQ=180^{\circ}-95^{\circ}=85^{\circ}.

The sum of the angles in △TUR\triangle TUR is 180∘180^{\circ}.

Thus, ∠TRU=180∘−∠RTU−∠TUR=180∘−60∘−85∘=35∘\angle TRU=180^{\circ}-\angle RTU-\angle TUR=180^{\circ}-60^{\circ}-85^{\circ}=35^{\circ}.

Since △PQR\triangle PQR is isosceles with PQ=PRPQ=PR, then ∠PQR=∠PRQ=35∘\angle PQR=\angle PRQ=35^{\circ}.

Finally, the sum of the angles in △PQR\triangle PQR is 180∘180^{\circ}, and so x∘=180∘−∠PQR−∠PRQx^{\circ}=180^{\circ}-\angle PQR-\angle PRQ or x∘=180∘−35∘−35∘x^{\circ}=180^{\circ}-35^{\circ}-35^{\circ} and so x=110x=110.

Want a route through all this instead of an archive? The track puts 2,444 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.