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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.2011-12-01
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    ah, great idea. brb2011-12-01
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    @David Er, it seems I misread the question. You're right, thanks for pointing it out. I removed the comment now.2011-12-01
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    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.2011-12-01
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    @DavidMitra, thanks. I saw that earlier and reported it in the OP. I'm still stuck.2011-12-01
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    @sasha Ah, silly me. I misread that...2011-12-01
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    @sasha Would you mind telling us where this problem comes from?2011-12-01
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    This is an open problem at gogeometry.com (that contains a proposed answer) http://gogeometry.com/education/p454_triangle_incircle_inradius_tangent_circle_radii.htm2011-12-01
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    @DavidMitra The formula provided there doesn't seem right. (Is that subtraction under each radical?) I ask because, after trying some integral values for each of the inradii, the $r$ always comes out smaller than them. That can't be.2011-12-01
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    @sasha It looks like multiplication to me. Beyond that, I can't say.2011-12-01
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    @DavidMitra It looks to be about 2 pixels wide (and distinct from the period next to it), so that's the reason I thought it to be subtraction at first. Must be multiplication though else the dimensions wouldn't come out right! BTW, what's the site's purpose? I can't seem to find an "About" or a FAQ.2011-12-01
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    @Phira It was part of another problem I was working on which I thought it might be easier to tackle. Apparently not! So, in a way, I made up the OP for my purposes.2011-12-01
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    @sasha It must be mult.; their subtraction signs are obvious looking at the other pages. I just stumbled on the site after googling "four circles triangle". Looks like a good site, despite the lack of detail... BTW, thanks for the question, I've found it very interesting :) I'll think about it some more...2011-12-01

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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.2011-12-05