Two equal circles inscribed in a triangle

Question

Two equal circles are inscribed in a triangle as shown, with AC = 20 cm, AB = 13 cm and BC = 21 cm. Find the radius of the circle. I have no clue how it can be done except the drawing made in the diagram as shown. [Only Pythagoras theorem and algebraic manipulations are allowed to be used]

Answer 1

With trig:

Hint: The cosine rule gives $$\cos A = \dfrac{b^2+c^2-a^2}{2bc} \ \text{ where }a=BC,b=CA,c=AB$$ so you know all trigonometric values like $f(z)$ or $f(z/2)$ where $z$ is the angle $A,B$ or $C$, $f$ is one of the six trigonometric functions.

Further hint: Note that $$BC = r\cot \frac{B}2 + 2r + r\cot \frac{C}2$$

Without trig:

Hint: Call the centers of the circles $B',C'$, with $B'$ closer to $B$.
$BB' \cap CC' = I =$ incenter of $\triangle ABC$
Lengths of $BI,CI$ are known from applications of internal angle bisector theorem. Use similarity, and you might even be able to use the fact that $ID$ (where $D$ is the midpoint of $BC$) passes through the mutual touchpoint of the two circles.

Answer 2

Observe that the side $BC$ is the combination of three segments. One of them is directly related to the radius $r$. The problem is how to get the other two segments through trigonometry.
Notice that this triangle ABC is a combination of two Pythagorean triangles. You can certainly get the values like $\sin$ and $\cos$ for these angles, then use the half-angle formula to express the segments in terms of $r$, then you are done, because you can express $BC$ as a multiple of the radius $r$.

Answer 3

Hint: Cut the circles vertically in half, then glue the figure back together as shown.

Now two circles merges into one, and the small triangle formed is similar to the original triangle, with a scale factor of $\frac {21-2r}{21}$ (why?)

The original triangle has area $126$ and perimeter $54$, hence its inradius is $\frac {14}3$.

Thus the new triangle has inradius

$$\frac {14}{3} \times \frac{21-2r}{21} = r$$

hence $r = \dfrac {42}{13}$.

Answer 4

The dimensions of the triangle were chosen to make this a little easier than it first appears. The triangle can be divided to two Pythagorean triples..

The slope of the line from the B to the center of the closer circle is the same as the slope of the line from B to the incenter of the 5,12,13 right triangle.

There are a few ways to find this, but I like this one.

$a+b = 12\\ a+c = 13\\ b+c = 5$

Solving this system: $a = 10, b = 2, c = 3$

The slope is $\frac 23$ Doing the same to the $12,16,20$ triangle gives a slope of $-\frac 13$

$\frac 32 r + 2r + 3r = 21\\ r = \frac {42}{13}$

Answer 5

Let $a,b, c$ be the sides and note that

\begin{align} a = 2r + r(\cot \frac{B}2 +\cot \frac{C}2)=2r + r\frac{\cos\frac A2}{\sin\frac B2 \sin\frac C2}\\ = 2r +r\frac{2R\sin A}{4R\sin\frac A2\sin\frac B2 \sin\frac C2} = 2r +r \frac{a}{r_i} \end{align} with $R$ and $r_i$ the circumradius and inradius, respectively. Then, $$ r = \frac{a}{2+\frac a{r_i}} = \frac{a}{2+\frac {a(a+b+c)}{2K}} $$ where $K$ is the area of the triangle. Apply the Heron’s area formula to get $$r= \frac a{ 2+\frac{a\sqrt{s}}{\sqrt{(s-a)(s-b)(s-c)}}},\>\>\>\>\> s=\frac{a+b+c}2$$ Substitute $a=21$, $b=20$ and $c=13$ to obtain $r=\frac{42}{13}$.

Answer 6

The area of the whole triangle can be found using Heron's formula: the semiperimeter $s$ is $\frac{1}{2}(13+20+21) = 27$, and $A = \sqrt{27(27-13)(27-20)(27-21)} = 126$. The altitude of the triangle from the $21$ cm base is thus $\frac{126}{1/2 \cdot 21} = 12$.

Now rearrange the area of the triangle to form three smaller triangles with bases $13, 20$ and $21-2r$ and heights of $r$, and a right-angled trapezoid in the centre. This has area $\frac{1}{2}r(13+20+21-2r) + \frac{1}{2}(2r)(12+r)$.

Equating the two forms of the area, we have $r(54 - 2r) + 2r(12+r) = 252$, or $-2r^2+54r+24r+2r^2=252$. Hence $r = \boxed{\frac{42}{13}}$.

Answer 7

Let the centres of the circles be $C_{1}$ and $C_{2}$. The segments $BC_{1}$ and $CC_{2}$,when extended, meet at $I$ , the incentre of $\triangle ABC$. Circle $C_{1}$ and $C_{2}$ touch the side $BC$ at points $E$ and $F$ respectively. The incircle of $\triangle ABC$ touches the side $BC$ at point $D$. Draw $C_{1}E$, $C_{2}F$, $C_{1}C_{2}$ and $ID$. Let $ID$ intersect $C_{1}C_{2}$ at point $L$. Let the radii of the similar circles and the inradius of $\triangle ABC$ be $r'$ and $r$ respectively.

$\triangle IC_{1}C_{2}\sim \triangle IBC$ because $C_{1}C_{2}\parallel BC$.

$\Rightarrow \frac{C_{1}C_{2}}{BC}=\frac{IL}{ID}$

$\Rightarrow \frac{2r'}{21}=\frac{r-r'}{r}$

$\Rightarrow r'=\frac{ar}{2r+a}=\frac{21r}{2r+21}$

Use $\Delta=rs$ to get $r$ and then plug it into the equation to get $r'$.