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How can one go about solving the following problem?

Inscribe a circle in an arbitrary triangle. Call it's radius $r_1$. Inscribe three more circles so that each one is tangent to two sides of the triangle and the first circle (i.e., each at a different corner). Call the radii $r_2, r_3, r_4$. Find a relationship between $r_1, r_2, r_3$ and $r_4$.

The most promising method of attack for me was to consider the isosceles triangles at each corner: the base being the tangent line to the point of intersection of the angle bisector of the triangle and the first circle. But I'm stuck.

Any suggestions much appreciated.

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Have you considered the simpler case of inscribing a circle in an equilateral triangle of side-length $1$? In this case we will have $r_2 = r_3 = r_4$. I am not promising a general formula will pop out here, but when I am stuck on a problem, I always consider the simplest example first. –  JavaMan Dec 1 '11 at 18:08
    
ah, great idea. brb –  sasha Dec 1 '11 at 18:25
    
@David Er, it seems I misread the question. You're right, thanks for pointing it out. I removed the comment now. –  Srivatsan Dec 1 '11 at 19:25
    
This is just an (trivial) observation and may lead to nowhere: draw three tangents on the inscribed circle of radius $r_1$ that are perpendicular to the angle bisectors of the triangle (this is somewhat ill-stated, but I hope the meaning is clear). Then the other three circles are the incircles of the three triangles formed by those tangents and the original triangle. –  David Mitra Dec 1 '11 at 19:34
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This is an open problem at gogeometry.com (that contains a proposed answer) gogeometry.com/education/… –  David Mitra Dec 1 '11 at 20:13

1 Answer 1

up vote 4 down vote accepted

I'm using the same notation as in the link mentioned by David Mitra. By similarity we have $$ \frac{r}{OA}=\frac{r_a}{OA-r-r_a}=\sin\left(\frac{\alpha}{2}\right) $$ where $\alpha=\angle A$. Then $$ \frac{r_a}{r} = \frac{1-\sin(\alpha/2)}{1+\sin(\alpha/2)} = \frac{1-\cos((\beta+\gamma)/2)}{1+\cos((\beta+\gamma)/2)} = \tan\left(\frac{\beta+\gamma}{4}\right)^2 $$ where we used that $\alpha+\beta+\gamma=\pi$. Similarly $$ \frac{r_b}{r} = \tan\left(\frac{\alpha+\gamma}{4}\right)^2 \qquad\qquad \frac{r_c}{r} = \tan\left(\frac{\alpha+\beta}{4}\right)^2 $$ Therefore, $\sqrt{r_ar_b}+\sqrt{r_ar_c}+\sqrt{r_br_c}$ can be written as $$ \begin{split} \frac{\sqrt{r_ar_b}+\sqrt{r_ar_c}+\sqrt{r_br_c}}{r} = \frac{\sin(\rho)\sin(\sigma)\cos(\tau)+\sin(\rho)\cos(\sigma)\sin(\tau) +\cos(\rho)\sin(\sigma)\sin(\tau)}{\cos(\rho)\cos(\sigma)\cos(\tau)} \end{split} $$ where $\rho=(\beta+\gamma)/4$, $\sigma=(\alpha+\gamma)/4$, and $\tau=(\alpha+\beta)/4$. Note that $\rho+\sigma+\gamma=\pi/2$. Consider $x$, $y$, $z$ such that $x+y+z=\pi/2$. Then $$ \begin{split} \sin(x)\sin(y)\cos(z) &= \frac{1}{2}(\cos(x-y)-\cos(x+y))\cos(z) \\&= \frac{1}{4}(\cos(x-y+z)+\cos(x-y-z) -\cos(x+y+z)-\cos(x+y-z)) \\&= \frac{1}{4}( \cos\left(\frac{\pi}{2}-2y\right) + \cos\left(-\frac{\pi}{2}+2x\right) - \cos\left(\frac{\pi}{2}\right)-\cos\left(\frac{\pi}{2}-2z\right) \\&=\frac{1}{4}\left( \sin(2y)+\sin(2x)-\sin(2z)\right) \end{split} $$ And similarly $$ \begin{split} \cos(x)\cos(y)\cos(z) &= \frac{1}{2}(\cos(x-y)+\cos(x+y))\cos(z) \\&= \frac{1}{4}(\cos(x-y+z)+\cos(x-y-z) +\cos(x+y+z)+\cos(x+y-z)) \\&=\frac{1}{4}\left( \sin(2y)+\sin(2x)+\sin(2z)\right) \end{split} $$ This means that $(\sqrt{r_ar_b}+\sqrt{r_ar_c}+\sqrt{r_br_c})/r$ is equal to $$ \frac{\sqrt{r_ar_b}+\sqrt{r_ar_c}+\sqrt{r_br_c}}{r} = \frac{\sin(2\rho)+\sin(2\sigma)+\sin(2\tau)}{\sin(2\rho)+\sin(2\sigma)+\sin(2\tau)}=1 $$

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Wow, this is a great effort--thanks. I've been away from my computer for a few days so I couldn't respond earlier. I have something similar to what you have which I will be posting shortly. –  sasha Dec 5 '11 at 17:10

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