# Three circles in isosceles triangle

I need a little help with this geometry problem, that I need to solve synthetic (I know the analytic solution).

We write in an isosceles triangle those sides are 13 cm, 13 cm, 10 cm three circles. Every circle is tangent to two sides of the triangle and the other two circles. The radii of the circles that are tangent to the base are congruent. Find the radii of the three circles.

I also managed to prove synthetically that the two congruent radii are 2 cm.

• Hint: $(5,12,13)$ is a Pythagorean triple. If you depict the configuration on graph paper, you can get a good idea about the locations of the incenters involved. Mar 22 '18 at 19:49
• You may also consider that Steiner's construction of the Malfatti circles is pretty involved, but it greatly simplifies if the original triangle is an isosceles one. Mar 22 '18 at 20:00

This answer is going to be restyled according to the suggestions of Douglas Rogers.
The old version is preserved below; now there is a precious preliminary lemma.

Let us assume that two circles centered at $$A,B$$ are externally tangent at $$T$$, and $$\tau$$ is a common exterior tangent. Let $$A',B'$$ be the projections of $$A,B$$ on $$\tau$$ and let $$U$$ be the intersection of the tangent through $$T$$ with $$\tau$$. By powers of a point, $$UA'=UB'=UT$$, which leads to $$\widehat{AUB}=90^\circ$$. If $$M$$ is the midpoint of $$AB$$, $$U$$ is the projection of $$M$$ on $$\tau$$, hence the circle centered at $$M$$ through $$U$$ is tangent to $$\tau$$.

We are ready to solve the Malfatti problem in a isosceles triangle.

Let us assume that $$ABC$$ is isosceles on $$BC$$ and $$M$$ is the midpoint of $$BC$$. The Malfatti circles tangent to $$BC$$, centered at $$O_B,O_C$$, are simple to be located: their centers lie on the $$B,C$$ angle bisectors and on the lines from $$M$$ forming angles of $$45^\circ$$ with $$BC$$. Let $$T$$ be the tangency point of these circles. By the previous Lemma, a circle through $$T$$ and $$O_B$$ tangent to the $$AB$$-side meets the $$AM$$-line at the center $$O_A$$ of the last Malfatti circle. In order to construct this circle, we consider $$P=AB\cap O_B O_C$$ and draw a circle with diameter $$O_B T$$. The length of a tangent from $$P$$ gives the geometric mean of $$PO_B$$ and $$PT$$, thus the location of $$U$$ on $$AB$$. If we draw the perpendicular to $$UO_B$$ through $$U$$, its intersection with the $$AM$$-line is precisely $$O_A$$ and the problem is solved.

In the given configuration we have $$BM=MB=5$$, $$AM=12$$ and the radius of the Malfatti circles centered at $$O_B$$ and $$O_C$$ is $$2$$. In particular $$PT = \frac{AT}{AM} BM = \frac{25}{6},\quad PO_B=PT-2 = \frac{13}{6},\quad PU=\frac{5}{6}\sqrt{13}.$$ Let $$R$$ be the projection of $$O_B$$ on $$AB$$. Then $$BR=3$$, so $$RP=\frac{5}{6}$$ and $$RU=\frac{5}{6}(\sqrt{13}-1)$$. This is also the distance of $$U$$ from the $$O_A O_B$$ line, whose square equals the product of the radii of the circle centered at $$O_A$$ and the circle centered at $$O_B$$. It follows that the last unknown radius is $$\frac{1}{2}\left[\frac{5}{6}(\sqrt{13}-1)\right]^2 = \frac{25}{36}(7-\sqrt{13}) .$$

Old version: This is an implementation of Steiner's construction of the Malfatti circles for an isosceles triangle.

1. The red circles are symmetric and they are straighforward to locate;
2. The orange circles are constructed in such a way that they are tangent to the "oblique" sides and they go through a center of a red circle and the midpoint of the centers of the red circles;
3. The orange circles locate the center of the blue circle and the problem is solved.