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Weekly Problem 34 - 2008

Stage: 3 Short Challenge Level: Challenge Level:2 Challenge Level:2

$$\frac{11}{12} \; \text{square units}$$

solution

We want to find out how far the points $A$ and $B$ (the points on both the triangle and sqaure) are from $Y$. The triangle $OQT$ is $3$ units across by $2$ units down. The triangle $OPA$ is similar to $OQT$, and is half its size (since it is one unit down rather than two). So the point $A$ is $\frac{3}{2}$ units from $O$, so $\frac{3}{2}-1=\frac{1}{2}$ unit from $X$, so $1-\frac{1}{2}=\frac{1}{2}$ unit from $Y$

Similarly, the triangle $BST$ is $\frac{1}{3}$ the size of $OQT$, so the point $B$ is $\frac{2}{3}$ unit from $Z$, so $1-\frac{2}{3}=\frac{1}{3}$ unit from $Y$. So the triangle $AYB$ has area $$\frac{1}{2}\times \frac{1}{2}\times {1}{3} \; \text{square units}= \frac{1}{12}\; \text{square units}$$ so the area of the overlap is $$1-\frac{1}{12}\; \text{square units}=\frac{11}{12}\; \text{square units}$$

This problem is taken from the UKMT Mathematical Challenges.

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