Given triangle $ABC$ with $CE=6 cm $ , $BD=9 cm $ , $\angle A = 60$ , CE , BD are medians. Find the area of $ ABC $

Question

Given triangle $ABC$ with $CE=6 cm $ , $BD=9 cm $ , $\angle A = 60$ , CE , BD are medians. Find the area of $ ABC $

This problem is too difficult for me , the teacher said it is a challenge problem.

Any help will be appreciated , thank you.

Answer 1

Let $AB=c$, $AC=b$ and $BC=a$.

Thus, $a^2=b^2+c^2-bc$, $\frac{1}{2}\sqrt{2a^2+2c^2-b^2}=9$ and $\frac{1}{2}\sqrt{2a^2+2b^2-c^2}=6$, which gives $$\frac{2(b^2+c^2-bc)+2c^2-b^2}{2(b^2+c^2-bc)+2b^2-c^2}=\frac{9}{4}$$

Answer 2

Well let's use the following names:$AC=2x$, $AB=2y$, and $\angle BAC=\alpha$.

Now write the cosine law for both the triangles $ADB$ and $AEC$ both centred at the angle $\alpha$: $$ \begin{cases} ABD: x^2 + 4y^2-4xy\cos\alpha = |DB|^2 \\ AEC: 4x^2 + y^2 - 4xy\cos\alpha= |EC|^2 \end{cases} $$ Now replacing your numerical values gives:

$$ (*) \begin{cases} x^2 + 4y^2- 2xy = 81 \quad(1) \\ 4x^2+y^2-2xy=36 \quad(2) \end{cases} $$

Now there probably should be some slick way to get $xy$ out of this. Since I am not that bright, going to wolfram alpha gives: $$ xy = \frac{1}{7}\left(39 + 5 \sqrt{249} \right) $$

And your area, A, according to a simple trigonometry is: $$ A= 2xy\sin\alpha = \frac{2}{7}\left(39 + 5 \sqrt{249} \right) \frac{\sqrt3}{2} $$

---

Update: According to @Lozenges, System $(*)$ can be solved as follows: Denote $xy=p$, and $x+y=s$. Then, add $(1)$ and $(2)$ to get: $$ 5s^2 - 14 p - 117 = 0. \quad (3)$$ Subtract $(2)$ from $(1)$ to get: $$ s^4 - 4ps^2 -225= 0. \quad (4)$$ Eliminating $s$ from $(3)$ and $(4)$ gives: $$ 7p^2-78p-672=0, $$ whose solution is: $$ p = \frac{1}{7}\left(39 + 5 \sqrt{249} \right). $$

Answer 3

I tried to find an elegant geometric solution but couldn't. Here is the gung-ho algebraic version. First by cosine rule about $A$ in $\Delta ADB$, and $\Delta AEC$, we get $$b^2 + c^2/4 -bc/2 = 36$$ and $$b^2/4 + c^2 -bc/2 = 81$$ From these we get by subtracting $$c^2=b^2 + 60 $$ Substituting this in the first equation we get $$5b^2/4 -21 = b\sqrt{b^2+60} $$ Squaring both sides $$25b^4/16 + 221 -105b^2/2 = b^4 + 60b^2 $$ $$b^4 - 200b^2 + 3536/9 = 0$$ You can solve this to get $b^2$. Since $c = \sqrt{b^2+60}$, and therfore the area is $$1/2 \cdot bc \cdot \sin(60)=\sqrt3/4 \cdot b \cdot \sqrt{b^2+60} $$ $$\Delta^2 = 3/16 \cdot b^2 \cdot (b^2 + 60) $$

Answer 4

Let's express coordinates of the vertices of $\triangle ABC$ in some scaled units ($\mathrm{su}$) such that $\triangle ABC$ is inscribed in the unit circle (with the radius $R=1\,\mathrm{su}$).

Since $\angle BAC=\tfrac\pi3$, let's start with an equilateral $\triangle A_0BC$, inscribed in a unit circle with the center $O$.

In order to get the location of the vertex $A$, we need to rotate $A_0$ by the angle $\phi\in(0,\tfrac{2\pi}3)$ such that $|BD|:|CE|=9:6=3:2$, or $|BD|^2:|CE|^2=9:4$.

Thus the coordinates of vertices in ($\mathrm{su}$) are: \begin{align} A&=(\cos\phi,\sqrt{1-\cos^2\phi}),\, B=(-\tfrac12,-\tfrac{\sqrt3}2),\, C=(-\tfrac12,\tfrac{\sqrt3}2) ,\\ D&=(\tfrac12\cos\phi-\tfrac14,\tfrac12\sin\phi+\tfrac14\sqrt3)) ,\\ E&=(\tfrac12\cos\phi-\tfrac14,\tfrac12\sin\phi-\tfrac14\sqrt3)) . \end{align}

Condition

\begin{align} 4\cdot|B-D|^2&=9\cdot|C-E|^2 \end{align}

simplifies to a quadratic equation in $\cos\phi$:

\begin{align} 4588\cos^2\phi+400\cos\phi-2963 , \end{align} which provides one valid root \begin{align} \cos\phi&=\tfrac{117}{2294}\sqrt{249}-\tfrac{50}{1147} . \end{align}

The area of the triangle $ABC$

\begin{align} S_{\triangle ABC}&=|CF|\cdot|FG| , \end{align} expressed in scaled units (su) \begin{align} S'_{\triangle ABC}&= (\tfrac{\sqrt3}2 \cdot(\tfrac12+\cos\phi))\,\mathrm{su}^2 = (\tfrac{1047}{4588}\sqrt3+\tfrac{351}{4588}\sqrt{83}) \,\mathrm{su}^2. \end{align}

We can convert this value to $\mathrm{cm}^2$, using one of known medians, for example, $CE$:

\begin{align} |CE|^2&=36\,\mathrm{cm}^2 =\left(2+\tfrac14\cos\phi-\tfrac34\sqrt3\sqrt{1-\cos^2\phi}\right) \,\mathrm{su}^2 = (\tfrac{1404}{1147}+\tfrac9{1147}\sqrt{83}\sqrt3) \,\mathrm{su}^2 ,\\ 1\,\mathrm{su}^2&= (\tfrac{208}{7}-\tfrac4{21}\sqrt{83}\sqrt{3}) \,\mathrm{cm}^2 ,\\ S_{\triangle ABC}&= (\tfrac{1047}{4588}\sqrt3+\tfrac{351}{4588}\sqrt{83} )\,\mathrm{su}^2 \cdot (\tfrac{208}{7}-\tfrac4{21}\sqrt{83}\sqrt{3}) \,\frac{\mathrm{cm}^2}{\mathrm{su}^2} = \tfrac{39}7\sqrt3+\tfrac{15}7\sqrt{83} \approx 29.172355 . \end{align}