Let $ABC$ be of triangle with $\angle BAC = 60^\circ$ . Let $P$ be a point in its interior so that $PA=1, PB=2$ and $PC=3$. Find the maximum area of triangle $ABC$.

I took reflection of point $P$ about the three sides of triangle and joined them to vertices of triangle. Thus I got a hexagon having area double of triangle, having one angle $120$ and sides $1,1,2,2,3,3$. We have to maximize area of this hexagon. For that, I used some trigonometry but it went very complicated and I couldn't get the solution.


Let $\mathcal{A}$ be the area of $\triangle ABC$. Let $\theta$ and $\phi$ be the angles $\angle PAC$ and $\angle BAP$ respectively.
We have $\theta + \phi = \angle BAC = \frac{\pi}{3}$. As functions of $\theta$ and $\phi$, the side lengths $b$, $c$ and area $\mathcal{A}$ are:

$$ \begin{cases} c(\theta) &= \cos\theta + \sqrt{2^2-\sin^2\theta}\\ b(\phi) &= \cos\phi + \sqrt{3^2-\sin^2\phi} \end{cases} \quad\text{ and }\quad \mathcal{A}(\theta) = \frac{\sqrt{3}}{4} c(\theta)b\left(\frac{\pi}{3}-\theta\right) $$ In order for $\mathcal{A}(\theta)$ to achieve maximum has a particular $\theta$, we need

$$\frac{d\mathcal{A}}{d\theta} = 0 \iff \frac{1}{\mathcal{A}}\frac{d\mathcal{A}}{d\theta} = 0 \iff \frac{1}{c}\frac{dc}{d\theta} - \frac{1}{b}\frac{db}{d\phi} = 0 \iff \frac{\sin\theta}{\sqrt{2^2-\sin^2\theta}} - \frac{\sin\phi}{\sqrt{3^2-\sin^2\phi}} = 0$$ This implies $$\frac{\sin\theta}{2} = \frac{\sin\phi}{3} = \frac13 \sin\left(\frac{\pi}{3} - \theta\right) = \frac13 \left(\frac{\sqrt{3}}{2}\cos\theta - \frac12\sin\theta\right) \iff 4\sin\theta = \sqrt{3}\cos\theta$$ and hence $$\theta = \tan^{-1}\left(\frac{\sqrt{3}}{4}\right) \approx 0.4086378550975924 \;\;( \approx 23.41322444637054^\circ )$$

Furthermore, we have $\displaystyle\;\frac{\sin\theta}{2} = \frac{\sin\phi}{3} = \frac{\sqrt{3}}{2\sqrt{19}}\;$. Substitute this into the expression for side lengths and area, we get $$ \begin{cases} c &= \frac{4+\sqrt{73}}{\sqrt{19}}\\ b &= \frac{7+3\sqrt{73}}{2\sqrt{19}} \end{cases} \quad\implies\quad \mathcal{A} = \frac{\sqrt{3}}{8}(13+\sqrt{73}) \approx 4.664413635668018 $$ Please note that the condition $\displaystyle\;\frac{\sin\theta}{2} = \frac{\sin\phi}{3}$ is equivalent to $\angle ABP = \angle ACP$. If one can figure out why these two angles equal to each other when $\mathcal{A}$ is maximized, one should be able to derive all the result here w/o using any calculus.

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Let $\theta=\measuredangle PAB$ in the triangle you specify. Then $\measuredangle PAC=60°-\theta$.

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By the law of cosines,

$$2^2=1^2+c^2-2\cdot 1\cdot c\cdot \cos\theta$$ $$3^2=1^2+b^2-2\cdot 1\cdot b\cdot \cos(60°-\theta)$$

Solving those equations for $b$ and $c$,

$$c=\cos\theta+\sqrt{\cos^2\theta+3}$$ $$b=\cos(60°-\theta)+\sqrt{\cos^2(60°-\theta)+8}$$

Since the triangle's area is $\frac 12bc\sin A$ and $A=60°$, putting them all together we get

$$Area=\frac{\sqrt 3}{4}\left(\cos(60°-\theta)+\sqrt{\cos^2(60°-\theta)+8}\right)\left(\cos\theta+\sqrt{\cos^2\theta+3}\right)$$

where $0°<\theta<60°$. This looks like a bear to maximize analytically, so I did it numerically with a graph and got

$$Area_{max}\approx 4.66441363567 \quad\text{at}\quad \theta\approx 23.413°$$

I could get only five significant digits for $\theta$ but the maximum area should be accurate to all shown digits. I checked this with a construction in Geogebra and it checks. WolframAlpha timed out while trying to find an exact maximum value of the triangle's area from that my formula.

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  • $\begingroup$ This was a question asked in Olympiad so is there any way this could be solved in time without calculator or any simpler method to do this?? $\endgroup$ – Satvik Mashkaria Dec 27 '15 at 8:32
  • $\begingroup$ @SatvikMashkaria: I dont' see any better way than mine other than maximizing that area function by taking its derivative--which would not be easy. If I have time later today I'll look at the problem in another way and try the derivative. $\endgroup$ – Rory Daulton Dec 27 '15 at 12:16
  • $\begingroup$ Using Mathematica's FindRoot[...] in conjunction with the Inverse Symbolic Calculator, we find that $\theta=\arccos\dfrac4{\sqrt{19}}$ $\endgroup$ – Lucian Dec 27 '15 at 21:19
  • $\begingroup$ @Lucian: That agrees with achille lui's excellent answer of $\tan^{-1}\left(\frac{\sqrt 3}{4}\right)$. $\endgroup$ – Rory Daulton Dec 27 '15 at 21:22
  • $\begingroup$ @RoryDaulton: All roads lead to Rome. :-$)$ $\endgroup$ – Lucian Dec 27 '15 at 21:33

Let E be the point outside triangle such that AEPC is a parallelogram , let F be point outside the triangle such that CPBF is a parallelogram, and let D be a point outside ABC such that APBC is paralellogram. Now, you have hexagon AECFBD with sides EC=1,CF=2,FB=3,BD=1, AD=2, EA=3, and draw line EB. Then, ADBE and EBFC are the same(congruent) and now surface of a hexagon is double of a surface of ADEB. Now, in that, you have three sides which are determined, plus AD is parallel to EB. Now consider ADBE. There, ED=1,DA=2,AB=3. Let M be perpendicular to EB from D and let N be perpendicular to EB from A. Now let EM be x, and NB be y. Let DM be h, as well. We have to maximise the surface of ADEB. We can see that surface is sum of surfaces of NMDA and EMD and ANB. So, it is $2h+\frac{hx}{2}+\frac{hy}{2}$. Relations that we have imply from triangles EMD and ANB, Now, those are $1=h^2+x^2, 9 = h^2+y^2$ so $y^2-x^2=8$ ,and $h^2=1-x^2$.

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  • $\begingroup$ So you didn't make use of $\angle{BAC}=60$? $\endgroup$ – Satvik Mashkaria Dec 27 '15 at 7:00
  • $\begingroup$ You mean "APCD is a paralellogram". $\endgroup$ – user236182 Dec 27 '15 at 7:02
  • $\begingroup$ can you please prove that that ADBE and EBFC are the same(congruent) ? As much I know four sides are equal doesn't imply that the quadrilaterals are congruent. $\endgroup$ – Satvik Mashkaria Dec 27 '15 at 7:05

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