In an equilateral $\triangle ABC$ : $ DB^2 + DC^2 + BC^2 = 100 $

Question

I have a question that:

There is a point $D$ inside the equilateral triangle ABC. If $$ DB^2 + DC^2 + BC^2 = 100 $$ and the area of $DBC$ is $ 5 \sqrt{3} $, find $AD^2$.

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This is what I tried: $DB = x ,\: DC = y,\: BC = a$. Then $$ x^2 + y^2 + a^2 = 100 \tag{1}$$ $$ \sqrt{ 2(x^2 y^2 + a^2 x^2 + a^2 y^2 )- (a^4 + x^4 + y^4 ) } = 20 \sqrt{3} \tag{2} $$

I want to find $a$, $x$, and $y$, but I have only two equations. What can I do?

Answer 1

Let $|AB|=|BC|=|CA|=a$, $|DB|=m$, $|DC|=n$, $|AD|=q$, $|OB|=\tfrac12a$, $|OH|=x$, $|DH|=y$

We have two constraints \begin{align} m^2+n^2+a^2&=100 \tag{1}\label{1} ,\\ S_{DBC}&=5\sqrt3 \tag{2}\label{2} . \end{align}

From \eqref{1}, \begin{align} (\tfrac12a+x)^2+y^2+ (\tfrac12a-x)^2+y^2+a^2=100 ,\\ x^2+y^2 = 50-\tfrac34a^2 \tag{3}\label{3} , \end{align}

hence one condition is that the point $D$ must be on a circle with the center in the middle of $BC$.

Next, the area of $\triangle DBC$ is constrained, \begin{align} S_{ABC}&=\tfrac12 a y=3\sqrt5 , \end{align}

and we can express the side of the $a$ inn terms of $y$,

\begin{align} a&=\tfrac{10\sqrt3}y . \end{align}

Given that and assuming $x$ positive, we have

\begin{align} x&=\frac1y\cdot\sqrt{50 y^2-225-y^4}= \frac1y\cdot\sqrt{(45-y^2)(y^2-5)} \\ &= \frac1y\cdot\sqrt{(y-\sqrt5)(3\sqrt5-y)(y+\sqrt5)(3\sqrt5+y)} ,\\ \text{thus }\quad y&\in(\sqrt5,3\sqrt5) . \end{align}

Next,

\begin{align} |AD|^2&=q^2=|AE|^2+|AD|^2 \\ &= (\tfrac{\sqrt3}2\,a-y)^2+x^2 \tag{4}\label{4} . \end{align}

After the substitution of expressions for $a$ and $x$ in terms of $y$, everything cancels nicely, and we have the answer

\begin{align} |AD|^2&=20 . \end{align}

Answer 2

Denote $x= AD$, $y= BD$, $z=CD$, $a=BC(=AB=AC)$ and $\alpha=\angle BDC$. So we have: $$(*)\ y^2+z^2+a^2=100,$$ by the area of $\triangle DBC$ condition we have: $$(**)\ yz\sin\alpha=10\sqrt 3,$$ and by cosine rule on $\triangle DBC$ we have: $$(***)\ a^2=y^2+z^2-2yz\cos\alpha.$$

Consider the rotation $\mathcal R_{C,60^\circ}$ mapping $B\mapsto A$, $A\mapsto M$ and $D\mapsto N$. Since $CD=CN$ and $\angle DCN=60^\circ$, $\triangle DCN$ is equilateral, and thus $DN=DC=z$ and $\angle DNC=60^\circ$. Since we mapped $BD$ do $AN$, we have $AN=BD=y$. Also we mapped $\triangle BDC$ to $\triangle ANC$, so $\angle ANC=\angle BDC=\alpha$, wherefrom $\angle AND=\angle ANC-\angle DNC=\alpha-60^\circ$. So, in $\triangle AND$ we have $AD=x$, $AN=y$, $DN=z$ and $\angle AND=\alpha-60^\circ$, so by cosine rule: $$(****)\ x^2=y^2+z^2-2yz\cos(\alpha-60^\circ)= y^2+z^2-yz(\cos\alpha+\sqrt 3\sin\alpha).$$ Multiply $(****)$ by $2$ and substract $(***)$. We get: $$2x^2-a^2= y^2+z^2-2\sqrt 3yz\sin\alpha.$$ By using $(*)$ and $(**)$ we get: $$2x^2= a^2+y^2+z^2-2\sqrt 3\cdot 10\sqrt 3=100-60=40,$$ so $AD^2=x^2=20$.