Let $ABCD$ be a square with sidelength 1 and $AKL$ is an equilateral triangle where $K$ lies on BC and L lies on CD. What is the area for AKL?

Question

I tried to use the Pythagorean Theorem to find one side of $AKL$, but there isn't enough information. What am I missing?

Answer 1

Let $a$ be the sidelength of the triangle $AKL$. Then $K$ and $L$ are the intersection points between the circle $\omega$ centered in $A$ with radius $a$ and $BC$ and $CD$ respectively. Since both the circle $\omega$ and the square $ABCD$ are symmetric with respect to the line $AC$ then $|BK| = |DL|$.

Let $b = |BK|$. Then by applying the Pythagorean Theorem on $ABK$ we get $$a^2 = 1 + b^2$$

and by applying it on $KCL$ we get

$$a^2 = 2(1-b)^2 = 2 - 4b + 2b^2$$

Subtracting the two equations and solving for $b$ we get

$$b = 2 \pm \sqrt{3}.$$

Note that $0 < b < 1$ so we have to discard the option $2+\sqrt{3}$. Then

$$b = 2 - \sqrt{3}.$$

Now

$$|AKL| = |ABCD| - |ABK| - |KCL| - |ALD| = 1 - \dfrac{b}{2} - \dfrac{(1-b)^2}{2} - \dfrac{b}{2} = \dfrac{1-b^2}{2}$$ $$ = \boxed{2\sqrt{3} - 3}$$

Answer 2

Put it on a cartesian plane $A = (0,0); B= (1,0); C=(1,1); D=(0,1)$ and $K = (1,k)$ and $L=(j, 1)$.

Then ... it all comes out in the wash.

$AK = \sqrt{k^2 + 1} = KL = \sqrt{(1-k)^2 + (1-j)^2} = AL = \sqrt{1 + j^2} ; 1>k > 0; 1>j> 0$

So $k^2 + 1 = k^2 + j^2 - 2(k+j) + 2 = 1+ j^2$

So $j = k$ and $k^2 - 4k + 1 = 0$ so $k = 2 \pm \frac{\sqrt{12}}{2}$ and (as $k < 1$) $k = 2 - \sqrt{3}$

So $AK = KL = AL = \sqrt{8 - 4\sqrt{3}}$

So area of triangle is $\frac 12 b*h = \frac 12 * AK * \frac {\sqrt{3}}2*AK = \frac {\sqrt{3}}4*AK^2 = \frac {\sqrt 3}4*(8 - 4\sqrt{3}) = 2\sqrt{3} - 3$.