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- Jan 26, 2012

- 4,055

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- Thread starter
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- #1

- Jan 26, 2012

- 4,055

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Remember to read the POTW submission guidelines to find out how to submit your answers!

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- #2

- Jan 26, 2012

- 4,055

1) MarkFL

2) anemone

Solution (from MarkFL):

We can see that the absolute difference is the area of triangle $ACC'$, which we can also see is an isosceles triangle.

We may use the Law of Sines on triangle $ABC'$ to state:

\(\displaystyle \frac{\sin\left(30^{\circ} \right)}{2\sqrt{2}}=\frac{\sin(\theta)}{4}\implies\sin(\theta)=\frac{1}{\sqrt{2}}\implies \theta=\frac{\pi}{4}\)

Thus, we find:

\(\displaystyle \beta=\pi-2\theta=\frac{\pi}{2}\)

And so the area of triangle $ACC'$ is

\(\displaystyle A=\frac{1}{2}\left(2\sqrt{2} \right)^2=4\)

Hence the absolute difference in area of the two triangles is $4$ square units.

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