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The question says, "A circle is inscribed in a triangle whose sides are $40$ cm, $40$ cm and $48$ cm respectively. A smaller circle is touching two equal sides of the triangle and the first circle. Find the radius of smaller circle."

I can find the radius of the inscribed circle fairly easily by assuming the radius as $r$, and using the Heron's Formula: $$\frac{1}{2} * r * (40 + 40 + 48) = \sqrt{\left(\frac{40 + 40 + 48}{2}\right) \left(\frac{40 + 40 + 48}{2}-40\right)\left(\frac{40 + 40 + 48}{2}-40\right)\left(\frac{40 + 40 + 48}{2}-48\right)}$$

Which evaluates to give : $r = 12$, so The inscribed circle has a radius of $12$ cm.

But The smaller circle is only in touch with the other circle, and I can't get anything to work like constructions or etc. Trigonometry doesn't work too (maybe I'm doing it wrong, I'm a Grade 11 student anyway).

The most I can do is to find the area which is not occupied by the circle, but occupied by the triangle simply by subtracting the areas of both. [Which is $768 - \pi*(12)^2$ cm].

And this question was on a small scholarship paper I've attended, and it also had some more questions like it (I came to solve most of them).

rikusp2002
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    Drawing the common tangent of the two circles other than the two equal sides may be useful. – Yuta May 16 '19 at 03:55
  • @Yuta I've tried it, and it becomes parallel with the side which is opposite to the vertex the smaller circle is close to. This arises a situation of Thales' Theorem, but I can't determine the actual ratio of the segments, therefore, I can't determine the ratio of the two radii. Moreover, there is an extremely small segment which remains outside of the smaller circle, which gets included in the ratio we're trying to calculate. It's not accurate at all – rikusp2002 May 16 '19 at 04:05

3 Answers3

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See the figure below. One unit on the paper is six units in your problem. $AB=48,AC=40,BC=40$. Circle $D$ has radius $12$ as you say. $HI$ is tangent to both circles and parallel to $AB$, so $ABC$ is similar to the small triangle cut off by $HI$. $EC=32$ by Pythagoras, $EG=24$ from the circle, so $CG=8$ and the small triangle is $\frac 14$ the size of the large one. That says the radius of the small circle is $\frac 14 \cdot 12=3$

enter image description here

Ross Millikan
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  • How can we say that the small triangle is $\frac{1}{4}$th of the triangle? I mean I can clearly see the ratio of altitude of both the triangles is 1 : 4, but the base also shortens following the same ratio and that makes the triangle way shorter that what it would be if it was only one-fourth. – rikusp2002 May 16 '19 at 04:10
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    Having $HI$ parallel to $AB$ says all the angles are the same, which means the triangles are similar. Then any linear dimension is in proportion. The base of the small triangle is also $\frac 14$ of $AB$, so it is $12$. The figure shows it as $2$ units, which is correct. – Ross Millikan May 16 '19 at 04:13
  • Thanks a lot! I had the confusion of the ratio of the bases of the smaller triangle to the bigger triangle will be $1 : 5$ as I mistook $EG = 32$ cm. Now I'm good! – rikusp2002 May 16 '19 at 04:16
  • $EC=32$ is correct. The base of the large triangle is $48$, so the right triangle above the $x$ axis is $24-32-40$ – Ross Millikan May 16 '19 at 04:18
  • Yes, I got it! That was a typo on my end, I actually intended to type $EG$ haha – rikusp2002 May 16 '19 at 04:26
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enter image description here

$$\frac{R}{12}=\frac{2x}{48}=\frac{40-(24+x)}{40} \implies \frac{R}{12}=\frac{2x+2(16-x)}{48+2(40)}$$

CY Aries
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Let in $\Delta ABC$ we have $AB=AC=40$ and $BC=48.$

Also, let $(I,12)$ be a given circle, which touches to $AC$ and $BC$ in the point $E$ and $D$ respectively, $(O,x)$ be the little circle, which touched to $AC$ in the point $F$.

Thus, $$AE=\frac{40+40-48}{2}=16$$ and since $\Delta AIE\sim \Delta AOF,$ we obtain: $$\frac{AF}{AE}=\frac{OF}{IE}$$ or $$\frac{AF}{16}=\frac{x}{12},$$ which gives $$AF=\frac{4}{3}x,$$ $$FE=16-\frac{4}{3}x$$ and by the Pythagoras's theorem we obtain: $$FE=2\sqrt{IE\cdot OF}$$ or $$16-\frac{4}{3}x=2\sqrt{12x}.$$ Can you end it now?

Michael Rozenberg
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