The area of intersection of an isosceles triangle with another triangle

Question

I tried graphing the equations that form the two isosceles triangles and integrating the bounded area and got 7.456 as my answer after rounding. The answer key has the answer listed as 7.2 However, these type of problems rarely involve calc and I'm sure there is a much simpler way of solving this problem geometrically.

Answer 1

See the picture below. The thick black square is $6 \times 6$ and the two blue triangles are the ones described by the problem.

As we can see, the geometry induces a secondary square lattice. The width $x$ of one such square, highlighted in green, is easily determined as satisfying the relationship $x^2 + (x/3)^2 = 2^2$, or $x = 3 \sqrt{2/5}.$ Therefore, the area of the dark blue shaded region common to both blue triangles is simply $2x^2 = 36/5$.

(An alternative computation is to observe that the large black square has area $36$, which is equivalent to ten times the area of a single green square; thus a single green square has area $18/5$, and two such squares, which is equivalent in area to the dark blue shaded region, is $36/5$.)

Answer 2

I was good at contest math in my youth, but sucked at the geometry problems. I always ended up using Cartesian coordinates and reducing it to an algebra problem.

Here, setting $A=(0,2)$, $B=(0,-2)$, $C=(6,0)$ and thus $P=(0,0)$, and try to work from there. $AD$ perpendicular and equal in length to $AB$ would mean that $D=(\pm 4,2)$, for example. The condition that $BD$ intersects $AC$ means $D=(4,2)$.

Answer 3

It's clear that the altitude of $\triangle ABK$ is 3. Also, $\triangle ABJ\sim\triangle ICJ$, and so the altitudes are in the ratio 1:4, since $\overline{GI}=12$ (by another pair of similar triangles). Hence the altitude of $\triangle ABJ$ is 6/5.

So the area of the overlap is $4\times(3-6/5)=36/5$.

In fact, we can do this for arbitrary isosceles triangles, with base $2a$ and altitude $b$.

The altitude of $\triangle ABG$ is $b\frac{\overline{AB}}{\overline{AB}+\overline{CH}}=\frac{2ab}{a+b}$ by the similar triangles $\triangle ABG\sim\triangle HCG$

By the similar triangles $\triangle APE\sim\triangle LHE$, we have $\overline{HL}=\frac{b}{a}(b-a)$. So, like in the previous case, the altitude of $\triangle ABF$ works out nicely to be $\frac{2a^2b}{a^2+b^2}$.

So the answer is $2\times\text{Area}(\triangle AFG)=2\times\text{Area}(\triangle ABG)-2\times\text{Area}(\triangle ABF)=2a\left(\frac{2ab}{a+b}-\frac{2a^2b}{a^2+b^2}\right)$. You can check that $a=2,b=6$ gives the answer $36/5$.