1. ## Tangent circle questions

1.Two semicircles of radius 6 units are inscribed in a semicircle of radius 12 units. A circle is inscribed so that it is tangent to all three semicircles. Find the circumference of the circle

2. In triangle ABC, AB=260, AC=400, and BC=520. Point D is chosen on BC so that the circle inscribed in triangle ABD and ADC are tangent to AD at the same point. What is the length of BD?

2. Originally Posted by chengbin
1.Two semicircles of radius 6 units are inscribed in a semicircle of radius 12 units. A circle is inscribed so that it is tangent to all three semicircles. Find the circumference of the circle

...
1. Draw a rough sketch (see attachment)

2. Calculate the length of r using Pythagorean theorem:

$\displaystyle (12-r)^2+6^2=(6+r)^2~\implies~r=4$

3. Therefore the circumference is $\displaystyle c = 2\cdot \pi \cdot r = 8 \pi$

3. Originally Posted by chengbin
2. In triangle ABC, AB=260, AC=400, and BC=520. Point D is chosen on BC so that the circle inscribed in triangle ABD and ADC are tangent to AD at the same point. What is the length of BD?

We have $\displaystyle MD=ND=RD=x, \ BM=BQ=y, \ AQ=AR=AP=z, \ CP=CN=t$

Then $\displaystyle 2(x+y+z+t)=AB+AC+BC=1180\Rightarrow x+y+z+t=590$

But $\displaystyle x+y=BD, \ z+t=AC=400\Rightarrow BD+400=590\Rightarrow BD=190$

4. Originally Posted by earboth
1. Draw a rough sketch (see attachment)

2. Calculate the length of r using Pythagorean theorem:

$\displaystyle (12-r)^2+6^2=(6+r)^2~\implies~r=4$

3. Therefore the circumference is $\displaystyle c = 2\cdot \pi \cdot r = 8 \pi$
I don't get it. Do you mind showing me your logic? Thanks.

5. Originally Posted by chengbin
I don't get it. Do you mind showing me your logic? Thanks.
Not sure what exactly you don't get ....

I've marked the right triangle I use for my calculations (see attachment)

$\displaystyle (12-r)^2+6^2=(6+r)^2~\implies~ 144 - 24r +r^2 + 36 = 36+12r + r^2$

Collect like terms and separate the variable and the constants on different sides of the equation:

$\displaystyle r^2-r^2-24r-12r = -144 - 36 + 36~\implies -36r = -144~\implies~r=4$

The original question asked to calculate the circumference of the tangent circle. Since you know the radius you can use the formula:

$\displaystyle c = 2 \cdot \pi \cdot r$

Plug in the value of r and you'll get:

$\displaystyle c = 2 \cdot \pi \cdot 4 = 8 \pi$

6. I got it. You made a right triangle to find the length. I didn't see that before.

Thanks.