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Problem 370

Geometry Difficulty 2.5 Multiple choice CEMC Fermat · Canada · 2015

In the diagram, PQRSPQRS is a square and MM is the midpoint of PSPS.

The ratio of the area of QMS\triangle{QMS} to the area of square PQRSPQRS is

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Official solution

Solution 1

Since PQRSPQRS is a square, then its diagonal SQSQ cuts it into two equal areas.

Therefore, the ratio of the area of PQS\triangle PQS to the area of square PQRSPQRS is 1:21:2.

PQS\triangle PQS can be viewed as having base PSPS and height PQPQ.

MQS\triangle MQS can be viewed as having base MSMS and height PQPQ. (This is because PQPQ is perpendicular to the line containing MSMS.)

Since MS=12PSMS = \frac{1}{2}PS, then the area of MQS\triangle MQS is one-half of the area of PQS\triangle PQS.

Since the ratio of the area of PQS\triangle PQS to the area of square PQRSPQRS is 1:21:2, then the ratio of the area of QMS\triangle QMS to the area of square PQRSPQRS is 1:41:4.

Solution 2

Suppose that the side length of square PQRSPQRS is 2a2a.

Then the area of square PQRSPQRS is (2a)2=4a2(2a)^2 = 4a^2.

Since MM is the midpoint of side PSPS, then PM=MS=aPM=MS=a.

Then QMS\triangle QMS can be seen as having base MSMS and height PQPQ. (This is because PQPQ is perpendicular to the line containing MSMS.)

Since MS=aMS=a and PQ=2aPQ=2a, then the area of QMS\triangle QMS is 12(MS)(PQ)=12a(2a)=a2\frac{1}{2}(MS)(PQ) = \frac{1}{2}a(2a)=a^2.

Therefore, the ratio of the area of QMS\triangle QMS to the area of square PQRSPQRS is a2:4a2a^2:4a^2 which equals 1:41:4.

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