For a triangle $ABC$, $B=90^\circ , AC=6$, equations of medians through $A$ is $y=2x+4$ and through $C$ is $y=x+3$. What is the area of triangle $ABC$?

I'm really bad at geometry, and to make matters worse, the equations of the medians are given. I have no idea where to start. A hint would be appreciated.

  • $\begingroup$ Notice that instead of $90^o$ you can write $90^\circ$. I changed it. ${}\qquad{}$ $\endgroup$ Dec 1, 2014 at 19:36
  • $\begingroup$ That's nice. Thanks @MichaelHardy. $\endgroup$
    – Rohinb97
    Dec 1, 2014 at 19:41
  • $\begingroup$ The answer is 4, but my solution is not a polished one. $\endgroup$ Dec 3, 2014 at 11:30
  • $\begingroup$ @RicardoCruz, don't hesitate to put your answer. Your answer is correct. $\endgroup$
    – Rohinb97
    Dec 4, 2014 at 18:49

1 Answer 1


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Let denote the sides $|BC|=a$, $|AC|=b$, $|AB|=c$ and medians $|AA_m|=m_a$, $|BB_m|=m_b$, $|CC_m|=m_c$ of $\triangle ABC$. Also, let $x=|GA_m|=\tfrac13|AA_m|$, $y=|GC_m|=\tfrac13|CC_m|$.

Since $\angle ABC=90^\circ$,

\begin{align} |BB_m|=m_b=\tfrac12|AC|=\tfrac12b=3 . \end{align}

It is known that for such right triangle

\begin{align} a^2+c^2&=b^2 \tag{1}\label{1} ,\\ m_a^2+m_c^2&=5m_b^2 ,\\ x^2+y^2&=\tfrac59\,m_b^2 =\tfrac5{36}\,b^2 \tag{2}\label{2} . \end{align}

\begin{align} \triangle GCA_m:\quad |CA_m|^2&= |GA_m|^2+|GC|^2 -2|GA_m|\cdot|GC|\cos\theta ,\\ \tfrac14a^2&= x^2+4y^2-4xy\cos\theta ,\\ \tfrac14a^2&= \tfrac59b^2-3x^2-4xy\cos\theta \tag{3}\label{3} . \end{align}

Similarly, \begin{align} \triangle GC_mA:\quad |AC_m|^2&= |GC_m|^2+|GA|^2 -2|GC_m|\cdot|GA|\cos\theta ,\\ \tfrac14c^2&= 4x^2+y^2-4xy\cos\theta ,\\ \tfrac14c^2&= \tfrac59b^2-3y^2-4xy\cos\theta \tag{4}\label{4} . \end{align}

Combination of \eqref{3}+\eqref{4} with \eqref{1},\eqref{2} results in \begin{align} y&=\frac{b^2}{18\,x\cos\theta} \tag{5}\label{5} \end{align}

Substitution of \eqref{5} into \eqref{2} provides a quadratic equation in $x^2$ with the pair of roots, which represents two combinations of $x^2,y^2$.

\begin{align} x^2&=\tfrac5{72}\,b^2\left(1 \pm \sqrt{1-\frac 1{(\tfrac54\,\cos\theta)^2}}\right) ,\\ y^2&=\tfrac5{72}\,b^2\left(1 \mp \sqrt{1-\frac 1{(\tfrac54\,\cos\theta)^2}}\right) . \end{align}

The area of a general $\triangle ABC$ in terms of its medians is \begin{align} S&=\frac13\sqrt{4m_a^2m_c^2-(m_a^2+m_c^2-m_b^2)^2} , \end{align}

and in case of this right triangle we have

\begin{align} S&=\frac13\sqrt{4m_a^2m_c^2-(4m_b^2)^2} . \end{align}

Using $m_a^2=9x^2$, $m_c^2=9y^2$, we arrive at a nice expression for the area of the right triangle in terms of its hypotenuse $b$ and the angle $\theta$ between its medians $AA_m$ and $CC_m$ \begin{align} S&=\tfrac13\,b^2\tan\theta . \end{align}

The length of hypotenuse is given, and the angle $\theta=\angle A_mGC=\angle AGC_m$ and its tangent and cosine can be derived from the equations $y=2x+4$, $y=x+3$ of the lines through medians,

\begin{align} \tan\theta&=\tan(\arctan(2)-\arctan(1)) =\frac{\tan(\arctan(2))-\tan(\arctan(1))}{1+\tan(\arctan(2))\tan(\arctan(1))} =\frac13 ,\\ \cos\theta&=\frac{3\sqrt{10}}{10} . \end{align}

Thus the sought area is \begin{align} S&=\tfrac13 \cdot 6^2 \cdot \tfrac13=4 . \end{align}


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