Bisecting the area and perimeter

Question

In triangle $ABC$, $AB=16$, $AC=15$, and $BC=13$. Point $D$ is on $AB$, and point $E$ is on $AC$ so that $DE$ bisects both the area and perimeter of triangle $ABC$. (In other words, both $DA+AE$ and $DB+BC+CE$ are equal to half the perimeter.) Find $DE^2$.

Answer 1

Define $d:=AD$ and $e:=AE$. Then the perimeter bisection becomes

$$d+e=(16+15+13)/2=22 \\ e=22-d$$

Using Heron's formula, the area of $\triangle ABC$ is

$$\frac14\sqrt{(16+15+13)(-16+15+13)(16-15+13)(16+15-13)}=6\sqrt{231}$$

so you want $\triangle DAE$ to have an area of $3\sqrt{231}=\sqrt{2079}$. But in order to compute that area using Heron again, you need the distance $f:=DE$. Let's for the moment assume we had that distance given.

\begin{align*} \frac14\sqrt{(d+e+f)(-d+e+f)(d-e+f)(d+e-f)}&=\sqrt{2079}\\ (22+f)(22-2d+f)(2d-22+f)(22-f)&=4^2\cdot2079\\ \bigl(f^2-22^2\bigr)\bigl(f^2-(22-2d)^2\bigr)&=-33264\\ 4d^2f^2 - f^4 - 88df^2 - 1936d^2 + 968f^2 + 42592d - 267520 &= 0 \tag1 \end{align*}

You can use the law of cosines to describe $f$, or rather $f^2$:

\begin{align*} \cos A = \frac{d^2+e^2-f^2}{2de} &= \frac{b^2+c^2-a^2}{2bc} = \frac{15^2+16^2-13^2}{2\cdot15\cdot16}=\frac{13}{20} \\ 20(d^2+(22-d)^2-f^2)&=13(2d(22-d)) \\ 66d^2 - 20f^2 - 1452d + 9680 &= 0 \\ f^2 &= (66d^2 - 1452d + 9680)/20 \tag2 \end{align*}

Now if you plug $(2)$ into $(1)$ you get

\begin{align*} \frac{231}{100}d^4 - \frac{2541}{25}d^3 + \frac{27951}{25}d^2 - 33264 &= 0 \\ d^4 - 44d^3 + 484d^2 - 14400 &= 0 \tag3 \\ \{11-\sqrt{241}\approx-4.5,10,12,11+\sqrt{241}\approx26.5\} &\ni d \end{align*}

The first solution is negative. The last one is greater than $22$ and would therefore lead to negative $e$. So you have two possible choices for $d$ and therefore two possible choices for $D$ and $E$. These two choices are symmetric with respect to the angle bisector at $A$, so the quantity $f^2$ you are asked for remains the same. Plugging either of these solutions into $(2)$ will give you that $f^2$.

Of course, solving the quartic equation $(3)$ is a bit uncomfortable (unless you let a computer do it for you), so you might want to simplify that step. You could have thought about the symmetry before seeing the solutions: since $\triangle DAE$ is determined by its perimeter, its area and the angle at $A$, this description is fully symmetric. So you could use that symmetry in your equation as well. Write $d$ as $d=11+x$ so that whenever $x$ is a solution, $-x$ is a solution as well. Then $(3)$ becomes

$$x^4 - 242x^2 + 241 = 0\tag4$$

which is simply a quadratic equation in $x^2$.

Answer 2

$EK$ is parallel to $CD$ , $K$ on $AB$

$CKD$ and $CED$ have a common base and equal height so they have the same area. It follows that area $CKB$ is half the area of $ABC$. Therefore $K$ is the midpoint of $AB$.

$\frac{\text{AK}}{\text{AD}}=\frac{\text{AE}}{\text{AC}}$

$\frac{8}{AD}=\frac{AE}{15}$

$\text{AD}*\text{AE}=\text{AK}*\text{AC}=120$

$AD+AE=22$ (see the previous solution by MvG)

This gives the two solutions ${AD,AE}={10,12}$ or ${AD,AE}={12,10}$

From here use the Law of Cosines to get $\cos A = 13/20$

and

$\text{ED}^2=10^2+12^2-2*10*12 *\frac{13}{20}=88$