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$ABCD$ is a square with a side of length 4. P is on AB, S is on CD and Q is on PS such that:

  • $AP = CS$

  • The triangles $PBR$ and $SDQ$ are both equilateral triangles. See the image below.

enter image description here

Calculate the combined area of the 2 triangles.

What would be the easiest way the solve this? I first tried to name SC x, and then make an equality involving x (something equalling 16, the area of the square, and then solving for that x) but that didn't work out.

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Well, it satisfies the conditions of the problem to take $Q=R$ to be the center of the square, in which case the blue triangles have half the area of the square ($8$). So if the problem is well-defined, that must be the answer. – mjqxxxx Feb 15 '13 at 22:07
@mjqxxxx, $Q$ cannot be $R$. – Yimin Feb 15 '13 at 22:08
@mjqxxxx I made a mistake, see the edit – JohnPhteven Feb 15 '13 at 22:08
You might use the relation: $DA = \frac{\sqrt{3}}{2}(PQ+QD)$ , and $\frac{1}{2}PQ + AP =$ distance from $Q$ to line $DA$, which is $\frac{1}{2}QD$ – Yimin Feb 15 '13 at 22:11
up vote 3 down vote accepted

Let's do it in the style that you did not complete. While it is not the easiest way, it is completely mechanical.

We write down faithfully the first calculation that we did. Then we describe a much more streamlined way of doing the same thing.

Let $a=AP=CS$. We find the equation of the "left" side of the bottom triangle. This passes through $(a,0)$ and has slope $\tan(60^\circ)=\sqrt{3}$. So the equation is $$\frac{y-0}{x-a}=\sqrt{3},$$ or equivalently $y=\sqrt{3}x-a\sqrt{3}$.

This line is the same as the "right" side of the upper triangle.

We find the equation of that right side. It has slope $\sqrt{3}$, and passes through the point $(4,4-a)$. So the equation of the right side of the upper triangle is $$\frac{y-4}{x-4+a}=\sqrt{3},$$ or equivalently $y=\sqrt{3}x +4-4\sqrt{3}+a\sqrt{3}$.

Compare the two equations we have obtained. Their constant terms must be the same. This gives us a linear equation for $a$.

Another way: We want the line through $(a,0)$ and $(4-a,4)$ to have slope $\sqrt{3}$. That yields the equation $$\frac{4}{4-2a}=\sqrt{3}.$$

It is slightly more convenient to let the side of the triangles be $b$. then we are talking about the line from $(4-b,0)$ to $(b,4)$. We get the equation $$\frac{4}{2b-4}=\sqrt{3}.$$ Now that we have $b$, we can find the areas of the triangles by using a standard formula.

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You are amazing. How do you know the answers to these types of questions so fast? I understand that a lot of math is just very hard work, but geometry is more intuition, and you just do this like it's nothing. Any short tips? – JohnPhteven Feb 15 '13 at 22:28
It is just a matter of having done mathematics for quite a while. It is the typing that takes a while: when I started, we handwrote, and the typing was done by competent people. – André Nicolas Feb 15 '13 at 22:34

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