# Find the side of an equilateral triangle inscribed in a circle. Excuse the poor drawing.

$\triangle CDE$ is an equilateral triangle inscribed inside a circle, with side length $16$. Let $F$ be the midpoint of $DE$. Points $G$ and $H$ are on the circle so that $\triangle FGH$ is an equilateral. Find the side length of $\triangle FGH$ and express it as $a\sqrt{b}-c$ where $a, b, c$ are positive integers.

My attempt

$$\theta \text{ in the Large } \triangle = \theta \text{ in the Small } \triangle = 60^{\circ}$$

$$(\text{side of the Large } \triangle) \cdot \sin 60^{\circ} + (\text{side of the Small } \triangle )\cdot \sin 60^{\circ} = 2\cdot(\text{radius of circle})$$

Solving for the small side, I got $\dfrac{64}{3}-16$, which doesn't match the required form.

Thanks for any help.

• Hint: The problem with your method is that the diameter through C extends through the bottom of the small triangle, hence your "2r" claim is inaccurate. On the other hand, you can easily compute the angle $<AFH$. You know (or can compute) two of the sides of triangle AFH and can use the law of cosines. – lulu Jul 23 '15 at 20:39
• Have you tried applying the cosine rule in triangle AFG? – David Quinn Jul 23 '15 at 20:39
• I tried this way but I might have made an error I haven't done that law in a while and never was good at it. I got $4$$\sqrt{5}$$-$$1 The thing is the other stipulation I forgot to mention was b is not divisible by the square of a prime and I think 5 is. – HighSchool15 Jul 23 '15 at 21:06 • 5 is not divisible by the square of a prime: it is itself a prime. You did get right that b=5: see my answer for the details. Did the question really say that a and c are integers? – Rory Daulton Jul 23 '15 at 21:10 • Yes it said a, b, and c are positive integers, and b is not divisible by the square of a prime. – HighSchool15 Jul 23 '15 at 21:15 ## 3 Answers This answer may not be the simplest, but it is straightforward. I particularized the problem by making one side of the large equilateral triangle the segment between the points (-8,0) and (8,0). Then simple geometry tells us the third vertex is at (8\sqrt{3},0), the circumcenter of the triangle is at A(0,\dfrac{8\sqrt{3}}3), and the radius of the circumcircle is \dfrac{16\sqrt{3}}3. The equation of the circumcircle is then$$x^2+\left(y-\frac{8\sqrt{3}}3\right)^2=\left(\frac{16\sqrt{3}}3\right)^2x^2+\left(y-\frac{8\sqrt{3}}3\right)^2=\frac{256}3$$The one side of the small equilateral triangle, \overline{FH}, is on the line y=-\sqrt{3}x. That gives us two equations in two unknowns for the coordinates of point H which is on both the circle and the line. Substitute the expression for y in the linear equation into the quadratic equation and we get a quadratic equation for x:$$x^2+\left(-\sqrt 3x-\frac{8\sqrt 3}3\right)^2=\frac{256}34x^2+16x-64=0x^2+4x-16=0$$Solving this gives us this positive value for x:$$x=2\sqrt 5-2$$The side of the small equilateral triangle is twice the x-coordinate of point H, so the side of the triangle is$$4\sqrt 5-4$$The final answer to your problem, given the side is a\sqrt b-c, is$$a=4, \quad b=5, \quad c=4$$I checked this answer numerically with Geogebra, the source of my diagram above, and my answer checks. • If I am reading correctly, your triangle has side length 2 whereas the OP wanted 16. Of course everything scales without difficulty. – lulu Jul 23 '15 at 21:25 • So it would be the same as the one I got using your hint of 4$$\sqrt{5}$$-$$4$ but the $b$ wouldn't change and $5$ is divisible by the square of a prime. – HighSchool15 Jul 23 '15 at 21:45
• @lulu: Oops, how stupid of me. I'll eat dinner then correct that. Thanks for the correction! – Rory Daulton Jul 23 '15 at 21:49
• Yes this was right and a lot simpler to follow then using the law of cosines but both gave the same answer which is neat. Thanks for typing this up and helping out. – HighSchool15 Jul 23 '15 at 22:07

Let $R$ be the radius of circle. Draw a perpendicular to any of sides of large equilateral $\triangle CDE$ to obtain a right triangle by which we can calculate the side length $L$ of large equilateral triangle as follows $$L=2R\cos 30^\circ=16$$ $$\implies 2R\frac{\sqrt{3}}{2}=16 \implies \color{blue}{R=\frac{16}{\sqrt{3}}}$$ Let $l$ be side length of small equilateral $\triangle FGH$ & draw a perpendicular say $AM$, from center $A$ passing through $F$, to the opposite side $GH$ of small $\triangle FGH$ Then we have $$AM=AF+FM=R\sin 30^\circ+l\sin 60^\circ=\frac{16}{\sqrt{3}}\frac{1}{2}+l\frac{\sqrt{3}}{2}=\color{red}{\frac{16+3l}{2\sqrt{3}}}$$ in right $\triangle AMG$ applying pythagorean theorem as follows $$AG^2=AM^2+GM^2$$ $$\left(\frac{16}{\sqrt{3}}\right)^2=\left(\frac{16+3l}{2\sqrt{3}}\right)^2+\left(\frac{l}{2}\right)^2$$ $$12l^2+96l-768=0$$ Solving for $l$ as follows $$l=\frac{-96\pm\sqrt{(96)^2-4(12)(-768)}}{2(12)}$$ $$l=\frac{-96\pm96\sqrt{5}}{24}=-4\pm 4\sqrt{5}$$ but $l>0$ hence the side of the small equilateral $\triangle FGH$ $$l=\color{red}{4\sqrt{5}-4}$$ Now, comparing the side $l$ with $\color{red}{a\sqrt{b}-c}$ we get $$\color{blue}{a=4, b=5, c=4}$$

Extending $\overline{FG}$ crosses $\overline{CE}$ at it midpoint, $F^\prime$ and hits the circle again at $G^\prime$. Clearly, $\overline{F^\prime G^\prime}$ is the side of a triangle congruent to $\triangle FGH$. Define $p := |\overline{DF}| = |\overline{EF}| = |\overline{FF^\prime}|$ (where the last equality follows from the Triangle Midsegment Theorem) and $q := |\overline{FG}| = |\overline{F^\prime G^\prime}|$. By the Power of a Point Theorem applied to point $F$, we have

$$|\overline{DF}|\;|\overline{EF}| = |\overline{GF}|\;|\overline{G^\prime F}|$$ $$\to \quad p^2 = q ( p + q ) \quad \to \quad q^2 + p q - p^2 = 0 \quad \to \quad q = \frac{p}{2}\left( \sqrt{5} - 1 \right) \tag{\star}$$

(Note that we have taken the positive root of the quadratic equation.)

Then, since $p=8$,

$$q = 4(\sqrt{5}-1) = 4\sqrt{5}-4 \qquad\to\qquad (a,b,c) = (4,5,4)$$