The area of the smallest equilateral triangle with one vertex on each of the sides of the right triangle with side lengths $2\sqrt3$, $5$, and $\sqrt{37}$, as shown, is $\tfrac{m\sqrt{p}}{n}$, where $m$, $n$, and $p$ are positive integers, $m$ and $n$ are relatively prime, and $p$ is not divisible by the square of any prime. Find $m+n+p$.

I am trying to solve this problem just with complex numbers. I let the vertices of the large triangle to be $a,b,c$ for $b=0,a=2\sqrt{3}i,c=5.$ I let the vertices of the equilateral triangle to be $z_1,z_2,z_3$ for $z_1$ along $AB$, $z_2$ along $AC$, and $z_3$ along $BC$. In the first place, it is well-known that $z_1^2+z_2^2+z_3^2=z_1z_2+z_2z_3+z_3z_1.$

It is futile to consider the equations for $A,Z_1,B$ and $B,Z_3,C$ to be collinear because we set those as the imaginary and real axes. With the equation for $A,Z_2,C$ to be collinear, we have $$\frac{z_2-a}{z_2-c}=\frac{\overline{z_2}-\overline{a}}{\overline{z_2}-\overline{c}}=\frac{\overline{z_2}+a}{\overline{z_2}-c},$$

with that step from $\overline{a}=-a$ and $\overline{c}=c,$ as these are along the imaginary and real axes respectively. With some manipulation, we get to $$a(z_2+\overline{z_2})+c(z_2-\overline{z_2})-2ac=0$$ $$\Longleftrightarrow a\Re(z_2)+c\Im(z_2)i=10\sqrt{3}$$ $$\Longleftrightarrow 2\sqrt{3}x_2+5y_2=10\sqrt{3},$$ for $\Re(z)=x,\Im(z)=y.$

But I'm not quite sure what to do now, because I can't think of any synthetic simplifications.

Asymptote diagram of problem


1 Answer 1


Rewrite the equilateral condition $z_1^2+z_2^2+z_3^2=z_1z_2+z_2z_3+z_3z_1$ as $$z_3^2 -(z_1+z_2)z_3 + z_2^2+z_3^2-z_1z_2=0$$ which is quadratic in $z_3$, yielding $$z_3= e^{i\frac\pi3}z_1 + e^{-i\frac\pi3}z_2$$ Substitute $z_3$ into the line equation below for the hypotenuse

$$\frac{z+\bar z}{10}+ \frac{z-\bar z}{4\sqrt3 i}=1$$ obtained from the given side lengths, leading to $$\frac7{20}|z_1 |+ \frac{11\sqrt3}{60}|z_2 |=1 $$ where $z_1=\bar z_1$ and $z_2= -\bar z_2$ are used. Then, the area of the equilateral triangle is

$$A= \frac{\sqrt3}{4} |z_2-z_1|^2 = \frac{\sqrt3}{4}( |z_2|^2 +|z_1|^2 )\ge \frac{\sqrt3}{4} \frac{\left(\frac7{20}|z_1 |+ \frac{11\sqrt3}{60}|z_2 |\right)^2 }{\left(\frac7{20}\right)^2 + \left(\frac{11\sqrt3}{60}\right)^2 }= \frac{75\sqrt3}{67} $$ where the C-S inequality is applied.

  • 1
    $\begingroup$ Great answer, I realized that instead of using the criterion for an equilateral triangle, we can perform a spiral similarity and arrive at the same expression regarding the point on the hypotenuse: $z_3\mapsto e^{i\pi/3}(z_3-z_1)+z_1=z_2,$ where we take $z_2$ on the hypotenuse and $z_1$ on the imaginary axis. $\endgroup$
    – mpnm
    Dec 24, 2020 at 4:59

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