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Japan Junior Mathematical Olympiad First Round

Geometry Difficulty 3.6 AMC 10/12 Prove it Japan

In the diagram given below, the triangles $\triangle OAB$ , $\triangle OBC$ , $\triangle OCD$ are isosceles right triangles with $\angle OAB$ , $\angle OBC$ , $\angle OCD$ being the right angle, respectively. Find the area of the triangle $\triangle OAB$ if the area of the triangle $\triangle OCD$ is 12.

Solution

Since the length of the hypotenuse of an isosceles right triangle is $\sqrt{2}$ times the length of each of the other sides, by letting $OA = x$ we have $OB = \sqrt{2}x$ , $OC = \sqrt{2} \cdot \sqrt{2}x = 2x$ . Then the ratio of the area of the triangle $\triangle OAB$ to that of the triangle $\triangle OCD$ is given by $\frac{x^2}{2} : \frac{(2x)^2}{2} = 1 : 4$ . Thus we conclude that the area of the triangle $\triangle OAB$ is $\frac{12}{4} = 3$ .

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