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A point $P$ is located inside an equilateral triangle and is at a distance of 5, 12, and 13 from its vertices. Compute the edge length of the triangle.

The answer is $\sqrt{169 + 60\sqrt(3)}$.

If $s$ is the edge length of the triangle, and if $x$ is the measure of the angle with vertex at $P$ and with sides of lengths 5 and 12, and if $y$ is the measure of the angle with vertex at $P$ and with sides of lengths 5 and 13, according to the Law of Cosines, \begin{equation*} s^{2} = 169 - 120\cos(x) \end{equation*} \begin{equation*} s^{2} = 194 - 130\cos(y) \end{equation*} and since $\cos(360 - (x + y)) = \cos(x + y)$, \begin{equation*} s^{2} = 313 - 312\cos(x+y) \end{equation*} So, \begin{align*} \cos{x} = - \frac{s^{2} - 169}{120} , \\ \cos{y} = - \frac{s^{2} - 194}{130} , \\ \cos(x+y) = - \frac{s^{2} - 313}{312} . \end{align*} For any real numbers $\theta$ and $\phi$, \begin{equation*} \bigl[\cos{\theta}\cos{\phi} - \cos(\theta + \phi)\bigr]^{2} = \sin^{2}\theta\sin^{2}\phi = \bigl(1 - \cos^{2}\theta\bigr)\bigl(1 - \cos^{2}\phi\bigr) . \end{equation*} So, \begin{align*} &\left[\frac{s^{2} - 169}{120} \cdot \frac{s^{2} - 194}{130} + \frac{s^{2} - 313}{312}\right]^{2} \\ &\qquad \qquad = \left(1 - \left(\frac{s^{2} - 169}{120}\right)^{2}\right)\left(1 - \left(\frac{s^{2} - 194}{130}\right)^{2}\right) \end{align*} or equivalently, by multiplying by $(120^{2})(130^{2})312$, \begin{align*} &\Bigl[312(s^{2} - 169)(s^{2} - 194) + (120^{2})(130^{2})(s^{2} - 313)\Bigr]^{2} \\ &\qquad \qquad = 312 \Bigl((120^{2})(130^{2}) - 130^{2}(s^{2} - 194)\Bigr) \Bigl((120^{2})(130^{2}) - 120^{2}(s^{2} - 194)\Bigr) . \end{align*}

How is the quartic equation in the variable $s$ to be solved?

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  • $\begingroup$ Have a look at page 19 of these notes. The trick is not to perform an algebraic bashing, it is to consider the symmetric of $P$ with respect to the triangle sides and to perform a slick decomposition of the resulting hexagon. $\endgroup$ – Jack D'Aurizio Apr 29 '18 at 17:55
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    $\begingroup$ Also see my answer to a similar question for distances $3,4,5$ instead. In general, if you have an equilateral triangle of side $\ell$ and the distances of a point inside to the three vertices are $a, b, c$, then $\ell^2 = \frac12(a^2+b^2+c^2) + 2\sqrt{3}\Delta$ where $\Delta$ is the area of a triangle with side $a,b,c$. $\endgroup$ – achille hui Apr 30 '18 at 11:12
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Let $ABC$ be an equilateral triangle. $P$ is a point inside $\triangle ABC$ such that $PA=5$, $PB=12$ and $PC=13$. Rotate $C$ and $P$ about $A$ through $60^\circ$ to $B$ and a point $X$. Rotate $A$ and $P$ about $B$ through $60^\circ$ to $C$ and a point $Y$. Rotate $B$ and $P$ through $60^\circ$ to $A$ and a point $Z$.

Since $\triangle ABX\cong\triangle ACP$, $\triangle BCY\cong\triangle BAP$ and $\triangle CAZ\cong\triangle CBP$,

$$[AXBYCZ]=2[\triangle ABC]$$

Note that $\triangle APX$, $\triangle BPY$ and $\triangle CPZ$ are equilateral triangles of sides $5$, $12$ and $13$ respectively. Also, $\triangle PBX$, $\triangle YPC$ and $\triangle AZP$ are right-angled triangles of sides $5$-$12$-$13$.

\begin{align*} [\triangle ABC]&=\frac{1}{2}\left[\frac{1}{2}(5)^2\sin60^\circ+\frac{1}{2}(12)^2\sin60^\circ+\frac{1}{2}(13)^2\sin60^\circ+3\times\frac{1}{2}(5)(12)\right]\\ &=\frac{169\sqrt{3}}{4}+45 \end{align*}

So, \begin{align*} \frac{1}{2}(AB)^2\sin60^\circ&=\frac{169\sqrt{3}}{4}+45\\ AB^2&=169+60\sqrt{3} \end{align*}

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  • $\begingroup$ My apologies. I did not see your reply. I will look at your solution soon - hopefully today. (I will delete this message.) $\endgroup$ – user74973 Jun 2 '18 at 20:34
  • $\begingroup$ I appreciate your time to offer this explanation. It is lucid and elegant! $\endgroup$ – Adelyn Oct 1 '18 at 2:49
  • $\begingroup$ What is the origin of it? $\endgroup$ – Adelyn Oct 1 '18 at 2:50
  • $\begingroup$ By the way, I posted the problem. $\endgroup$ – Adelyn Oct 1 '18 at 2:51
  • $\begingroup$ The Google account from which I posted it expired. $\endgroup$ – Adelyn Oct 1 '18 at 2:53
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Given $a,b,c$, the equilateral triangle with the internal point $P$, located at distances $a,b,c$ from its vertices, can be constructed as follows.

Start with $\triangle ABC$, $|BC|=a$, $|CA|=b$, $|AB|=c$, $\angle CAB=\alpha$, $\angle ABC=\beta$, $\angle BCA=\gamma$.

Construct a helper external equilateral triangle, based on any side of $\triangle ABC$, for example, $\triangle ADB$. Then $|CD|=u$ is the side of sought equilateral triangle, which can be constructed in-place as either $\triangle CFD$ with $P=A$ or $\triangle DEC$ with $P=B$.

Note that the three angles at $P$ are $\alpha+60^\circ$, $\beta+60^\circ$, $\gamma+60^\circ$, hence such equilateral triangle with internal point $P$ can be constructed only if angles of $\triangle ABC$ are less than $120^\circ$.

The length of the side, $u$, can be found by the cosine rule, for example: \begin{align} \triangle ADC:\quad u^2&= b^2+c^2-2bc\cos(\alpha+60^\circ) \\ &=b^2+c^2-bc(\cos\alpha-\sqrt3\sin\alpha) \\ &=b^2+c^2-bc\cos\alpha+bc\sqrt3\sin\alpha) \\ &=b^2+c^2-bc\cos\alpha+2\sqrt3 S_{\triangle ABC} ,\\ \triangle ABC:\quad \cos\alpha&=\frac1{2bc}\,(b^2+c^2-a^2) ,\\ u^2&= b^2+c^2-\tfrac12\,(b^2+c^2-a^2)+2\sqrt3 S_{\triangle ABC} \\ &=\tfrac12\,(a^2+b^2+c^2) +2\sqrt3 S_{\triangle ABC} , \end{align}
where $S_{\triangle ABC}$ is the area of $\triangle ABC$.

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  • $\begingroup$ Thanks for your reply. I will look at your solution soon - hopefully today. I appreciate the time that you took to write an explanation. (I will delete this message.) $\endgroup$ – user74973 Jun 2 '18 at 20:33
  • $\begingroup$ I have now carefully read your argument. I appreciate the construction! Do you know the origin of such a construction? $\endgroup$ – Adelyn Oct 1 '18 at 2:56
  • $\begingroup$ I posted the problem. The Google account from which I posted it expired. $\endgroup$ – Adelyn Oct 1 '18 at 2:56
  • $\begingroup$ @Adelyn: I'm not aware of the origin, it just came out naturally. $\endgroup$ – g.kov Oct 1 '18 at 7:00

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