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CEMC Fermat, 2016

Geometry Difficulty 4.7 AIME Find the answer Canada

In the diagram, PQRSPQRS represents a rectangular piece of paper. The paper is folded along a line VWVW so that ∠VWQ=125∘\angle VWQ = 125^\circ. When the folded paper is flattened, points RR and QQ have moved to points R′R' and Q′Q', respectively, and R′VR'V crosses PWPW at YY.

Figure 0

The measure of ∠PYV\angle PYV is

Pick one

Solution

Since points YY, WW and QQ form a straight line segment, then ∠YWV=180∘−∠VWQ\angle YWV = 180^\circ - \angle VWQ and so ∠YWV=180∘−125∘=55∘\angle YWV = 180^\circ - 125^\circ = 55^\circ.

Since Q′Q' is the final position of QQ after folding, then ∠Q′WV\angle Q'WV is the final position of ∠QWV\angle QWV after folding, and so ∠Q′WV=∠QWV\angle Q'WV = \angle QWV.

Thus, ∠Q′WV=∠QWV=125∘\angle Q'WV = \angle QWV = 125^\circ and so ∠Q′WY=∠Q′WV−∠YWV=125∘−55∘=70∘\angle Q'WY = \angle Q'WV - \angle YWV = 125^\circ - 55^\circ = 70^\circ.

Figure 1

Since Q′WQ'W and R′YR'Y are parallel sides of the piece of paper, then ∠R′YW+∠Q′WY=180∘\angle R'YW + \angle Q'WY = 180^\circ, and so ∠R′YW=180∘−∠Q′WY=180∘−70∘=110∘\angle R'YW = 180^\circ - \angle Q'WY = 180^\circ - 70^\circ = 110^\circ.

Finally, ∠PYV\angle PYV is opposite ∠R′YW\angle R'YW so ∠PYV=∠R′YW=110∘\angle PYV = \angle R'YW = 110^\circ.

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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.