Problem 18 · 2012 Math Kangaroo
Stretch
Geometry & Measurement
area
A square ABCD has side-length 2. E is the midpoint of AB and F the midpoint of AD. G is a point on the line CF with 3CG = 2GF. How big is the area of the triangle BEG?
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Answer: B — \(\tfrac{4}{5}\)
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Hint 1 of 2
Drop coordinates on the square so \(E\), \(F\) and \(C\) are easy points.
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Hint 2 of 2
Find \(G\) by splitting \(CF\) in the ratio \(CG:GF = 2:3\); then base \(BE\) lies on the x-axis so the area only needs \(G\)'s height.
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Approach: coordinates with base BE on the x-axis
- Let \(A=(0,0)\), \(B=(2,0)\), \(C=(2,2)\), \(D=(0,2)\); then \(E=(1,0)\) and \(F=(0,1)\).
- \(G\) divides \(CF\) with \(CG:GF = 2:3\), so \(G = C + \tfrac{2}{5}(F-C) = \left(\tfrac{6}{5}, \tfrac{8}{5}\right)\).
- Base \(BE\) lies on the x-axis with length 1, and \(G\)'s height above it is \(\tfrac{8}{5}\), so the area is \(\tfrac12 \cdot 1 \cdot \tfrac{8}{5} = \tfrac{4}{5}\).
- So the area of triangle \(BEG\) is \(\tfrac{4}{5}\), choice B.
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