area chasing, sides // to sidelengths (2016 AMQ Concours p4)

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area chasing, sides // to sidelengths (2016 AMQ Concours p4)

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#1 h Mar 6, 2021, 10:06 PM

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Let be a triangle $\vartriangle ABC$ and parallel segments $PQ, RS$ and $TU$ to $AB, BC$ and $CA$ , respectively. Suppose that these three segments, as shown in the figure below, meet at points $X, Y$ and $Z$ . If each segment $PQ, RS$ and $TU$ subdivides the triangle $\vartriangle ABC$ into two parts of equal area, and if the area of the inner triangle $\vartriangle XY Z$ is $1$ , then what is the area of $\vartriangle ABC$ ?

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This post has been edited 1 time. Last edited by parmenides51, Mar 6, 2021, 10:07 PM

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#2 h Mar 6, 2021, 10:07 PM

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SKeole

#3 h Mar 6, 2021, 11:24 PM

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$BQ:QU:UC=\sqrt{2}:2:\sqrt{2}$ by similarity. Let $BQ=\sqrt{2}x$ , $QU=2x$ , and $UC=\sqrt{2}x$ , meaning $BC=2\sqrt{2}x+2x$ . $\frac{RS}{BC}=\frac{\sqrt{2}}{2}$ by similarity, so $RS=2x+\sqrt{2}x$ (We could also notice that $RS=QC$ to find this more quickly). $RY=ZS=BQ=UC=\sqrt{2}x$ , so we can calculate that $YZ=2x-\sqrt{2}x$ . Because XYZ ~ ABC, we can find the area by finding the similarity ratio. The ratio is $\frac{BC}{YZ}=\frac{2+2\sqrt{2}}{2-\sqrt{2}}=4+3\sqrt{2}$ . The area is $1\cdot($ ratio $)^2=34+24\sqrt{2}$

This post has been edited 1 time. Last edited by SKeole, Mar 6, 2021, 11:24 PM
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