Let $\triangle ABC$ be an equilateral triangle and let $P$ be a point on its circumcircle. Let lines $PA$ and $BC$ intersect at $D$ , let lines $PB$ and $CA$ intersect at $E$, and let lines $PC$ and $AB$ intersect at $F$. Prove that the area of $\triangle DEF $ is twice the area of $\triangle ABC$.

I had used my method and it required that $\triangle DEF $ have to be isosceles triangle which means $ P $ should be the mid point of $\widehat {BC}$ or $\widehat{AC}$ or $\widehat {BC}$. But the problem does not said so. So, how to solve it correctly?

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    $\begingroup$ Maybe you can use induction? After proving that the isosceles $\triangle DEF$ has twice the area? $\endgroup$ – John Glenn Mar 21 '18 at 8:08
  • $\begingroup$ @JohnGlenn can I really use it .? There’s no other method? Is it available to use in high school problem ? Specially in geometry. $\endgroup$ – November ft Blue Mar 21 '18 at 8:41
  • $\begingroup$ @John how can we use induction here? Thanks $\endgroup$ – King Tut Mar 21 '18 at 12:46
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    $\begingroup$ very very interesting problem! I had much fun solving it with no trigonometry or even circumference equation. Thanks for posting it. $\endgroup$ – dfnu Mar 5 at 15:10
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    $\begingroup$ @Matteo I am not so sure , my teacher asked me to do , but I believe it was a French Olympiad exercise! $\endgroup$ – November ft Blue Mar 6 at 12:39

Proof (for the case when $P$ lies on arc $BC$):

Let $AB=BC=CA=1$ and $\angle PAC=\theta$.

$$\frac{PD}{\sin\angle PBD}=\frac{PB}{\sin \angle PDB} \implies \frac{PD}{\sin\theta}=\frac{PB}{\sin (120^\circ-\theta)} $$

$$\frac{PB}{\sin\angle PAB}=\frac{AB}{\sin \angle APB}\implies \frac{PB}{\sin(60^\circ-\theta)}=\frac{AB}{\sin 60^\circ}$$

So, $\displaystyle PD=\frac{\sin\theta\sin(60^\circ-\theta)}{\sin60^\circ\sin(120^\circ-\theta)}=\frac{\sin\theta\sin(60^\circ-\theta)}{\sin60^\circ\sin(60^\circ+\theta)}$.

$$\frac{PE}{\sin\angle PAC}=\frac{AE}{\sin \angle APE} \implies \frac{PE}{\sin\theta}=\frac{AE}{\sin 120^\circ} $$

$$ \frac{AE}{\sin\angle ABE}=\frac{AB}{\sin\angle AEB} \implies \frac{AE}{\sin (60^\circ+\theta)}=\frac{AB}{\sin(60^\circ-\theta)}$$

So, $\displaystyle PE=\frac{\sin\theta\sin(60^\circ+\theta)}{\sin120^\circ\sin(60^\circ-\theta)}=\frac{\sin\theta\sin(60^\circ+\theta)}{\sin60^\circ\sin(60^\circ-\theta)}$.

$$\frac{PF}{\sin\angle PBF}=\frac{PB}{\sin \angle PFB} \implies \frac{PF}{\sin(120^\circ-\theta)}=\frac{PB}{\sin \theta}$$

So, $\displaystyle PF=\frac{\sin(120^\circ-\theta)\sin(60^\circ-\theta)}{\sin60^\circ\sin\theta}=\frac{\sin(60^\circ+\theta)\sin(60^\circ-\theta)}{\sin60^\circ\sin\theta}$.

The area of $\triangle DEF$ is

\begin{align*} &\;\frac{1}{2}(PD\cdot PE+ PE\cdot PF+PF\cdot PD)\sin 120^\circ\\ =&\;\frac{\sqrt{3}}{4}\left[\frac{\sin^2\theta+\sin^2(60^\circ+\theta)++\sin^2(60^\circ-\theta)}{\sin^260^\circ}\right]\\ =&\;\frac{1}{\sqrt{3}}\left[\sin^2\theta+(\sin60^\circ\cos\theta+\cos60^\circ\sin\theta)^2+(\sin60^\circ\cos\theta-\cos60^\circ\sin\theta)^2\right]\\ =&\;\frac{1}{\sqrt{3}}\left[\sin^2\theta+2\sin^260^\circ\cos^2\theta+\cos^260^\circ\sin^2\theta\right]\\ =&\;\frac{1}{\sqrt{3}}\left[\frac{3}{2}\sin^2\theta+\frac{3}{2}\cos^2\theta\right]\\ =&\;\frac{\sqrt{3}}{2} \end{align*}

The area of triangle $\triangle ABC$ is



It is very interesting to solve this problem by only using our greek friends Pythagoras and Menelaus, together with some simple algebra.

enter image description here

Initial Observations

Without loss of generality assume that $P$ lies on the half-plane determined by $BC$ and not containing $A$. Note that there is no need to use the circle, once we fix $\angle BPC = 120°$. In the figure above I added point $G$ given by the intersection of line $FD$ with $AC$. Suppose also $\overline{AB}=1$.

Once $\overline{BF} = x$ is chosen, the entire Figure is defined. We aim therefore at showing that, independently of $x$, $$[DEF] = 2[ABC].$$

Characterization of $\triangle BFC$

Below the triangle in question has been isolated, to better help you in the demontrations.

enter image description here

Recall that $\overline{BC} = 1$ and $\overline{BF} = x$. Use Pythagorean Theorem on $\triangle CKF$ in oder to determine $$\overline{CF} = \sqrt{x^2+x+1}.$$ From the similarity $$\triangle BPC \sim \triangle BCF,$$ show that $$\overline{CF}\cdot\overline{CP} = 1,$$ so that, in the end, you have $$ \begin{cases} \overline{CF} = \sqrt{x^2+x+1},\\ \overline{CP} = \frac{1}{\sqrt{x^2+x+1}},\\ \overline{FP} = \frac{x(x+1)}{\sqrt{x^2+x+1}}. \end{cases} $$

Four Applications of Menelaus's Theorem (MT)

MT on $\triangle ACF$ with the line $BE$, gives $$\frac{\overline{CE}}{\overline{AE}} = \frac{1}{x+1}.$$ Together with $\overline{AE}-\overline{CE} = 1$ this leads to $$\overline{CE} = \frac{1}{x}$$ and $$\overline{AE} = \frac{x+1}{x}.$$

Similarly, MT on $\triangle BFC$ and line $AP$, with the known fact that $\overline{BD} + \overline{CD} = 1$ yields $$\overline{CD} = \frac{1}{x+1}$$ and $$\overline{DB} = \frac{x}{x+1}.$$ MT on $\triangle ABC$ and line $FG$, with $\overline{AG} + \overline{CG} = 1$, gives you $$ \overline{AG} = \frac{x+1}{x+2}$$ and $$ \overline{CG} = \frac{1}{x+2}.$$ Finally, MT on $\triangle CGF$ with line $AP$ yields $$ \frac{\overline{GD}}{\overline{DF}} = \frac{1}{x(x+2)}.$$

Areas Computation

Now we only need to compute areas of triangles with fixed altitude and a given ratio between bases.

Firstly we have $$[ACF] = [ABC](1+x).$$

Then $$[AFE] = [ACF] \frac{\overline{AE}}{\overline{AC}},$$ that is $$[AFE] = [ABC]\frac{(x+1)^2}{x}.$$

We also have $$[GFE] = [AFE]\frac{\overline{GE}}{\overline{AE}},$$ yielding $$ [GFE] =2[ABC]\frac{(x+1)^2}{x(x+2)}.$$

Finally observe that $$\frac{[DFE]}{[GDE]} = \frac{\overline{GD}}{\overline{DF}} = \frac{1}{x(x+2)}.$$

Thus we have the system of equations $$ \begin{cases} \frac{[DFE]}{[GDE]} = \frac{1}{x(x+2)}\\ [DFE] + [GDE] =2[ABC]\frac{(x+1)^2}{x(x+2)}. \end{cases} $$

leading, once solved, to the desired result, i.e. $$\boxed{[DFE] = 2[ABC]} $$



enter image description here

Without loss of generality, let

\begin{align} O&=(0,0),\quad A=(1,0),\quad B=(-\tfrac12,\tfrac{\sqrt3}2),\quad C=(-\tfrac12,-\tfrac{\sqrt3}2),\quad P=(\cos\phi,\sin\phi) ,\\ &\text{and let }S\text{ be the area of }\triangle ABC . \end{align} Then \begin{align} |AB|&=|BC|=|CA|=\sqrt3,\quad \angle BAE=120^\circ,\quad \angle APB=120^\circ,\quad \angle EPA=60^\circ ,\\ \angle OAP&=\angle APO=\angle PCA=\tfrac\phi2,\quad \angle PAE=60^\circ+\tfrac\phi2,\quad \angle AEP=60^\circ-\tfrac\phi2 ,\\ S&=\tfrac{3\sqrt3}4 . \end{align}

\begin{align} D&=(-\tfrac12, \tfrac32\cot\tfrac\phi2) ,\\ |AD|^2&=|AM|^2+|DM|^2=\tfrac94\,(1+\cot^2\tfrac\phi2) =\frac9{4\sin^2\tfrac\phi2} ,\\ |AD|&=\frac3{2\sin\tfrac\phi2} ,\\ |AP|^2&=2\,(1-\cos\phi) ,\\ \triangle AEP:\quad |AE|^2&=\frac{|AP|^2\sin^2\angle EPA}{\sin^2\angle AEP} =\frac{3\,(1-\cos\phi)}{2\sin^2(60^\circ-\tfrac\phi2)} =\frac{3\,\sin^2\tfrac\phi2}{\sin^2(60^\circ-\tfrac\phi2)} ,\\ |AE|&= \frac{\sqrt3\,\sin\tfrac\phi2}{\sin(60^\circ-\tfrac\phi2)} ,\\ \triangle BCF,\triangle AFC:\quad |CF|&= \frac{2\,S}{\sqrt3\,(\sin\tfrac\phi2+\sin(60^\circ-\tfrac\phi2))} = \frac{3}{2\,(\sin\tfrac\phi2+\sin(60^\circ-\tfrac\phi2))} , \end{align}

which leads to \begin{align} E&=(1+|AE|\,\cos30^\circ,\,|AE|\,\sin30^\circ) = \left(1+ \frac{3\,\sin\tfrac\phi2}{2\sin(60^\circ-\tfrac\phi2)}, \, \frac{\sqrt3\,\sin\tfrac\phi2}{2\sin(60^\circ-\tfrac\phi2)} \right) ,\\ F&=\left( -\tfrac12+|CF|\cos(30^\circ+\tfrac\phi2) ,\, -\tfrac{\sqrt3}2+|CF|\sin(30^\circ+\tfrac\phi2) \right) \\ &= \left( -\tfrac12+ \frac{3\,\sin(60^\circ-\tfrac\phi2)}{2\,(\sin\tfrac\phi2+\sin(60^\circ-\tfrac\phi2))} ,\, -\tfrac{\sqrt3}2+ \frac{3\,\cos(60^\circ-\tfrac\phi2)}{2\,(\sin\tfrac\phi2+\sin(60^\circ-\tfrac\phi2))} \right) . \end{align}

The area of $\triangle DFE$ in terms of coordinates of $D,F,E$ is known to be expressed as

\begin{align} \tfrac12(D_xF_y+F_xE_y+E_xD_y-D_yF_x-F_yE_x-E_yD_x) . \end{align}

The following maxima code


gives this result:

\begin{align} \frac{3^{3/2}(\sin^2\tfrac\phi2-3\cos^2\tfrac\phi2)}{ 2\,(\sin\tfrac\phi2-\sqrt3\cos\tfrac\phi2) (\sin\tfrac\phi2+\sqrt3\cos\tfrac\phi2) }, \end{align}

trivially simplified further to get $\frac{3\sqrt3}2$, which is indeed, twice the area of $\triangle ABC$.

  • $\begingroup$ Ohh well! Thank you ! Now I get another solution. $\endgroup$ – November ft Blue Mar 22 '18 at 1:27

Proof for any point $P=(k,P(k)), \ \forall k \in[-\frac{\sqrt3}3s,\frac{\sqrt3}3s], \ s\in \mathbb{R}$

The area of triangles $\triangle DEF$ and $\triangle ABC$ can be written as: $\require{cancel}$ $$A_{\triangle ABC}=\pm\frac12 \det\begin{vmatrix} A_X&A_Y&1\\ B_X&B_Y&1\\ C_X&C_Y&1\\ \end{vmatrix}$$ $$ A_{\triangle DEF}=\pm\frac12 \det\begin{vmatrix} D_X&D_Y&1\\ E_X&E_Y&1\\ F_X&F_Y&1\\ \end{vmatrix}$$

For any circumcircle of an equilateral triangle, the radius is equal to $\frac{\sqrt3}3 s$, where $s$ is the side of the triangle, thus the circumcircle is defined by $$x^2+y^2=\frac13s^2$$ Which means, the $P$ is defined by $P(x)=\sqrt{\frac13s^2-x^2}$, or $\big(x,P(x)\big)$

For an equilateral triangle $\triangle ABC$ with side $s$, the vertices $A,B$ and $C$ are: $$\begin{align} A=&\Bigg(-\frac{\sqrt{-s^4+8 s^2-4}}{2 \sqrt{3} s},\frac{\sqrt{3} s^2-2 \sqrt{3}}{6 s}\Bigg)\\ B=&\Bigg(0,\frac{s}{\sqrt{3}}\Bigg)\\ C=&\Bigg(\frac{\sqrt{-s^4+8 s^2-4}}{2 \sqrt{3} s},\frac{\sqrt{3} s^2-2 \sqrt{3}}{6 s}\Bigg)\\ \end{align}$$

Then suppose we have a square matrix $M_1$, arranged as discussed above, that contains points $A,B$ and $C$, the area of the $\triangle ABC$, will be: $$A_{\triangle ABC}=\pm\frac12\det|M_1|=\frac{\left(s^2+2\right) \sqrt{-s^4+8 s^2-4}}{12 s^2}$$

However, we assume $s=1$, because it simplifies the process. Then we have: $$A_{\triangle ABC}=\frac{\sqrt{3}}{4}$$

Given $(k,P(k))$, $\forall k \in [-\frac{\sqrt3}3,\frac{\sqrt3}3]$, which defines point $P$, you get points $D,E$ and $F$ as: $$\begin{align} D=&\Bigg(\frac{-\sqrt{1-3 k^2}+3 k+1}{2 \sqrt{1-3 k^2}+6 k+4},\frac{5 \sqrt{1-3 k^2}-3 k+1}{2 \sqrt{3} \left(\sqrt{1-3 k^2}+3 k+2\right)}\Bigg)\\ E=&\Bigg(\frac{3 k}{2-2 \sqrt{1-3 k^2}},-\frac{1}{2 \sqrt{3}}\Bigg)\\ F=&\Bigg(\frac{\sqrt{1-3 k^2}+3 k-1}{2 \left(\sqrt{1-3 k^2}-3 k+2\right)},\frac{5 \sqrt{1-3 k^2}+3 k+1}{2 \sqrt{3} \left(\sqrt{1-3 k^2}-3 k+2\right)}\Bigg)\\ \end{align}$$

Suppose now we have a matrix $M_2$, arranged as discussed above, that contains the points $D,E$ and $F$, the area of $\triangle DEF$, for $s=1$, is: $$\begin{align} A_{\triangle DEF}=&\pm\frac12\det|M_2|\\ =&\frac{-2592 \sqrt{3} \sqrt{1-3 k^2} k^2+216 \sqrt{3} \sqrt{1-3 k^2}-216 \sqrt{3}}{216 \left(\sqrt{1-3 k^2}-1\right) \left(\sqrt{1-3 k^2}-3 k+2\right) \left(\sqrt{1-3 k^2}+3 k+2\right)} \end{align}$$ Simplifying that actually gives us: $$\begin{align} A_{\triangle DEF}=&\frac{-216 \sqrt{3} \left(12 \sqrt{1-3 k^2} k^2-\sqrt{1-3 k^2}+1\right)}{2\cdot-216\sqrt{3} \left(12 \sqrt{1-3 k^2} k^2-\sqrt{1-3 k^2}+1\right)}\\ =&\frac{\cancel{-216} \sqrt{3} \cancel{\left(12 \sqrt{1-3 k^2} k^2-\sqrt{1-3 k^2}+1\right)}}{2\cdot\cancel{-216} \cancel{(12 \sqrt{1-3 k^2} k^2-\sqrt{1-3 k^2}+1})}\\ =&\frac{\sqrt3}2 \end{align}$$

This goes to show that for any value of $k$, and thus, for any point $P$, the area will always be twice that of $\triangle ABC$.

Take note, however, the $k$ must be an element of $[-\frac{\sqrt3}3s,\frac{\sqrt3}3s]$

  • $\begingroup$ As you can see, there is no more need for a proof by induction $\endgroup$ – John Glenn Mar 21 '18 at 16:48
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    $\begingroup$ It must be a university solution which confuse me a lot because I am a high school student. They are all new for me. I don’t even understand how to get the coordinate of vertices. But thank you so much for your effort. I will ask my teacher to explain me about your solution. :) $\endgroup$ – November ft Blue Mar 21 '18 at 17:12
  • $\begingroup$ It's actually quite elementary, albeit computationally exhausting. It will help you a lot if you read up on linear algebra. As for getting coordinates of the vertices, you'll find that the distance formula, or simply, Pythagorean's theorem, to be very helpful. I know my answer is not at all intuitive, so I'm sure your professor has a clearer, more intuitive answer. $\endgroup$ – John Glenn Mar 21 '18 at 17:17
  • $\begingroup$ I will learn more about that:) thank you , I appreciate it. $\endgroup$ – November ft Blue Mar 21 '18 at 17:22
  • $\begingroup$ But one question , the coordinate of Vertices A,B,C could be change right ? it depends on how we set it? $\endgroup$ – November ft Blue Mar 21 '18 at 17:23

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