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I got my exam on Thursday, and just got a few questions left. Anyway I would aprreciate help a lot! Can anyone please help me to solve this task? You can see the picture below. The need is to finde the size of the two radius. I thought about working with cords, like the cord AC is the same size like another one. Still couldn´t really find something usefull. enter image description here

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The are of $ABM_2C=\frac{1}{2}(12+r_2)(r_1+r_2)$ which is equal to area of $ABC+BCM_2$. – Babak S. Mar 25 '13 at 12:28
up vote 0 down vote accepted

The key insight is that $\angle ACB=90°$.

To show this, we draw a line passing through $C$ that is tangent to both the circles $k_1,k_2$ at $C$. This is possible since the two circles are tangent to each other. Let this tangent intersect the line $AB$ at $D$.

Now, we have $DA=DC$ since $DA,DC$ are lines tangent to circle $k_1$ at $A,C$ respectively. Similarly, $DB=DC$. Combining these two equations, we have $DA=DC=DB$; therefore, $D$ is the center of the circle passing through points $A,B,C$, that is, the circumcircle of triangle $ABC$. Furthermore, $AB$ must be the diameter of this circle, and thus $\angle ACB=90°$ (by Thales' Theorem).

Once we have shown that $\angle ACB=90°$, the rest of the problem can be solved by trigonometry. However, for a more elegant approach, you can consider the following.

Let the midpoint of $AC$ be $P_1$ and the midpoint of $CB$ be $P_2$. Then $P_1M_1C$ is similar to $CAB$, and $P_2M_2C$ is similar to $CBA$. (This can be shown by labeling one of the angles - say $\angle CAB=\alpha$ and then finding the size of the rest of the angles). Once you have these relations, you can compare the lengths via similar triangles to obtain the results

$$r_2=M_2C=P_2C\times \frac {BA}{CA}=\frac{1}{2}BC\times\frac {BA}{CA}=5.625$$

and similarly

$$r_1=M_1C=P_1C\times \frac {AB}{CB}=\frac{1}{2}AC\times\frac {AB}{CB}=10$$

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thank you vincent tjeng! – Sophia Mar 25 '13 at 13:37
@my pleasure. It was a nice problem, especially the fact that there was a right angle! – Vincent Tjeng Mar 25 '13 at 23:57

$AM_1M_2B$ is a right angle trapez from figure we can see that $$\frac{r_1+r_2}{2} AB=\frac {AB+r_1+r_2}{2}r_1=AM_1C+BM_2C+ABC $$ where $$AB=\sqrt{12^2+9^2}=15$$ $$AM_1C=6\sqrt{r_1^2-6^2}$$ $$BM_2C=9/2\sqrt{r_2^2-(9/2)^2}$$ $$ABC=54$$ so we get the system

$$\frac{r_1+r_2}{2} 15=\frac {15+r_1+r_2}{2}r_1$$ $$\frac {15+r_1+r_2}{2}r_1=6\sqrt{r_1^2-6^2}+9/2\sqrt{r_2^2-(9/2)^2}+54 $$

$$15(r_1+r_2)=(15+r_1+r_2)r_1$$ $$(15+r_1+r_2)r_1=12\sqrt{r_1^2-6^2}+9\sqrt{r_2^2-(9/2)^2}+108 $$

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$AC=12$, $CB=9$ and values of $AB$ can be $\{4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20\}$. How do you know its only $15$ unless you prove the angle is right angled.? – Inceptio Mar 25 '13 at 12:52
Since $ABC$ is right triangle only $AB=15$ is possible – Adi Dani Mar 25 '13 at 13:16
It isn't proved ! You need to prove it before you consider the length. – Inceptio Mar 25 '13 at 13:17
thank you for help! – Sophia Mar 25 '13 at 13:31
You are welcome! – Adi Dani Mar 25 '13 at 13:33

$\triangle MAC$ and $\triangle MBC$ are isosceles. enter image description here

Construct a tangent $CD$ which is common at $C$.

Now $\angle DCM= \angle M_2BD=90^0$, which means $BDCM_2$ is a cyclic Quadrilateral.

Similarly, prove $ADCM_1$ is a cyclic quadrilateral.

$$\angle BCM_2=\angle CBM_2=x$$(Isosceles)

$\angle BCM_2=BDM_2=x$(Why? Since it is cyclic, $M_2B$ projects equal angles)

$$\angle M_2DC= \angle M_2BC=x$$

$\angle CDB=2x$

In quadrilateral $ADCM_1$,

$\angle M_1AC=\angle M_1CA =y$(Isosceles)

$\angle M_1DA=M_1CA=y \implies \angle CDA=2y$

$2x+2y=90^0 \implies x+y=90^0$

In $\triangle ACB$,

$\angle CAB=90-y$ and $\angle CBA=90-x$

$\angle BCA= x+y$, But we have $x+y=90^0$. Therefore, $\triangle ACB$ is right angled.

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yes I saw that wioth isocleles. but even if it´s right angled, how can I get the radius? – Sophia Mar 25 '13 at 12:15
It is clear that $\sqrt{144+81}=15$ – Babak S. Mar 25 '13 at 12:18
@BabakS.: It doesn't work if we don't prove that. – Inceptio Mar 25 '13 at 12:19
okay i try to prove now.. – Sophia Mar 25 '13 at 12:25
i dont really know how to prove the right angle, because there is no angle you have any idea? – Sophia Mar 25 '13 at 12:28

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