Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

inradius

Inside triangle ABC there are three circles with radius $r_1$, $r_2$, and $r_3$ each of which is tangent to two sides of the triangle and to its incircle with radius r. All of $r$, $r_1$, $r_2$, and $r_3$ are distinct perfect square integers. Find the smallest value of inradius $r$.

share|improve this question
1  
Nice problem. Where is it from? –  Beni Bogosel Dec 27 '12 at 12:31
    
Saw it on a math group. :) –  John Chang Dec 27 '12 at 12:32
    
Is there a requirement that the circle with radius $r$ is tangent to all three sides of the triangle? –  Jay Dec 27 '12 at 12:33
    
It is the incircle so it must be tangent to all three sides. –  John Chang Dec 27 '12 at 12:38
    
A hint for a solution method (I could try it, but can't right now): Notice that each small circle is homothetic to the incircle so you should be able to find the ratios $r/r_i$ in terms of $a,b,c$. After that I don't know what you could do. I guess you may need some assumptions on $a,b,c$, also. –  Beni Bogosel Dec 27 '12 at 12:38

2 Answers 2

I managed to obtain some relations between $r,r_1,r_2,r_3$ and $a,b,c$. I used only standard geometry tools and I won't give the details for now...

The relations are: (I consider $r_1$ the circle which is not tangent to the side $a$, and the same for $r_2,b$ and $r_3,c$)

$$ p-a =2\sqrt{r_1r} \frac{r}{r-r_1}$$ (construct symilarly the other two)

From here you can construct a relatioin between $r,r_1,r_2,r_3$ in the following way:

$$ r=\frac{S}{p}=\sqrt{\frac{(p-a)(p-b)(p-c)}{p}} $$ and substituting we have $$ r= \sqrt{\displaystyle \frac{\frac{8\sqrt{r_1r_2r_3}r^4\sqrt{r}}{(r-r_1)(r-r_2)(r-r_3)}}{2\sqrt{r_1r}\frac{r}{r-r_1}+2\sqrt{r_2r}\frac{r}{r-r_2}+2\sqrt{r_3r}\frac{r}{r-r_3}}}. $$

Here you may be able to use your hypothesis on $r,r_1,r_2,r_3$ to get something out of this.

share|improve this answer
    
Is $p$ the semi-perimeter? –  John Chang Dec 28 '12 at 3:06
    
At this point, why not just plug in $r_1=1$, $r_2=4$, $r_3=9$ and solve the resulting quadratic? I get $r=27/2$ (the other root is $r=1$, but we need to toss that because it's impossible). –  Potato Dec 28 '12 at 5:31
    
Yes, $p$ is the semiperimeter. @Potato: I guess that's the way we should proceed. Plug in the smallest possible $r_1,r_2,r_3$ and find $r$. note that $r$ must also be a perfect square, so your result is not good. –  Beni Bogosel Dec 28 '12 at 19:05
    
@BeniBogosel Ah, I didn't see the requirement that $r$ be a perfect square. Anyway, if you square and multiply out, the equation simplifies considerably, and you get a quadratic in $r$... –  Potato Dec 28 '12 at 19:31

For typographic convenience (and reduced visual clutter) in the following, I write "$A_2$", "$B_2$", "$C_2$" for the half-angles "$A/2$", "$B/2$", "$C/2$".


Let $P$ be the incenter of the triangle with radius $r =: s^2$; and let $P_1$, $P_2$, $P_3$ be the centers of the circles with respective radii $r_1 =: s_1^2$, $r_2 =: s_2^2$, $r_3 =: s_3^2$.

If $Q$ is the point of tangency of the incircle with $AB$, then $\triangle APQ$ has a right angle at $Q$, and we have $|AP| = \frac{r}{\sin A_2}$. With $Q_1$ the corresponding point of tangency creating $\triangle AP_1Q_1$, we have $$\frac{r_1}{r} = \frac{|AP_1|}{|AP|} = \frac{|AP|-r-r_1}{|AP|} = \frac{r-(r+r_1)\sin A_2}{r}$$ so that $$\sin A_2 = \frac{r-r_1}{r+r_1} = \frac{s^2 - s_1^2}{s^2+s_1^2}$$ Likewise, $$\sin B_2 = \frac{s^2-s_2^2}{s+s_2^2} \qquad \sin C_2 = \frac{s^2-s_3^2}{s+s_3^2}$$ whence $$\cos B_2 = \frac{2 s s_2}{s^2+s_2^2} \qquad \cos C_2 = \frac{2 s s_3}{s^2+s_3^2}$$

(Note: We know the cosines of the half-angles must be non-negative.)

Now, $A_2 + B_2 + C_2 = \pi_2$, so that

$$\begin{align} \sin A_2 &= \cos(B_2+C_2) \\ \sin A_2 &= \cos B_2 \cos C_2 - \sin B_2 \sin C_2 \\ \frac{s^2-s_1^2}{s^2+s_1^2} &= \frac{4 s^2 s_2 s_3 - ( s^2 - s_2^2)( s^2 - s_3^2)}{(s^2+s_2^2)(s^2+s_3^2)} \end{align}$$

Thus,

$$2 s^2 \left( s^2 - s_1 s_2 - s_2 s_3 - s_3 s_1 \right) \left( s^2 + s_1 s_2 - s_2 s_3 + s_3 s_1 \right) = 0$$

and we have

$$s^2 = s_1 s_2 + s_2 s_3 + s_3 s_1 \qquad \text{or} \qquad - s_1 s_2 + s_2 s_3 - s_3 s_1$$

As the latter option is clearly less than $\max\{s_1^2, s_2^2, s_3^2\}$, whereas $s^2$ must exceed that value (the incircle is bigger than the other three), we are left to solve the Diophantine equation

$$s^2 = s_1 s_2 + s_2 s_3 + s_3 s_1$$

in distinct positive integers dominated by $s$. One solution (found via Mathematica's FindInstance function) is

$$(s,s_1,s_2,s_3) = (9, 7, 6, 3)$$

which (approximately) corresponds to an $106.26^\circ$-$45.24^\circ$-$28.5^\circ$ triangle. No guarantees that this minimizes $s$ (and thus $r$).

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.