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Difference between revisions of "2022 MMATHS Individual Round Problems/Problem 2"

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==Solution 1==
 
==Solution 1==
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The area of rectangle <math>BEHG</math> is <math>6 \cdot 4 = 24</math>. The area of triangle <math>BEF = \frac {8 \cdot 6}{2} = 24</math>. The area of rectangle <math>BDIF</math> is <math>3 \cdot = 24</math>. The area of triangle <math>DBG</math> is <math>\frac {4 \cdot 3}{2} = 6</math>. Add them all together to get <math>\boxed {78}</math>.
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~Arcticturn

Latest revision as of 21:13, 18 December 2022

Problem

Triangle $ABC$ has $AB = 3, BC = 4,$ and $CA = 5$. Points $D, E, F, G, H,$ and $I$ are the reflections of $A$ over $B$, $B$ over $A$, $B$ over $C$, $C$ over $B$, $C$ over $A$, and $A$ over $C$, respectively. Find the area of hexagon $EFIDGH$.

Solution 1

The area of rectangle $BEHG$ is $6 \cdot 4 = 24$. The area of triangle $BEF = \frac {8 \cdot 6}{2} = 24$. The area of rectangle $BDIF$ is $3 \cdot = 24$. The area of triangle $DBG$ is $\frac {4 \cdot 3}{2} = 6$. Add them all together to get $\boxed {78}$.

~Arcticturn

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