# Inscribed kissing circles in an equilateral triangle

Triangle is equilateral (AB=BC=CA), I need to find AB and R. Any hints? I was trying to make another triangle by connecting centers of small circles but didn't found anything

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Draw tangent lines for the circles and make three smaller triangles. Then (prove for general case) that big radii is sum of smaller radii, i.e R= 4+4+4=12. So your side is $2\sqrt{3} R=24\sqrt{3}$ – Please Delete Account Mar 13 '11 at 18:38

Two hints:

• Draw the line segments from $A$ and $B$ through the center of the large circle and use the 30°-60°-90° triangles and similar triangles to get a relationship between $AB$ and $R$.
• Draw the line parallel to $\overline{BC}$ through the point of tangency of the large circle and the small circle closest to $A$ to get a small equilateral triangle, and use what you learned about the relationship between $AB$ and $R$ on the small triangle.

Above is the small triangle formed at the top of your diagram. $DF=DE=4$, since both are radii of the small circle. You can use the 30°-60°-90° triangles to find $AD$ and $AE$. In particular, what do you find is $\frac{AD}{AF}$?

edit 2: Since the homework problem is now done, here's how I would actually have done the problem myself, though it is not the solution I would expect from a geometry student: In an equilateral triangle, the median, altitude, angle bisector, perpendicular bisector, etc. are all the same segment. The point of concurrency of the angle bisectors is the center of the inscribed circle and the point of concurrency of the medians divides the medians in the ratio $2:1$, so the height of the smaller triangle shown above is $R=3\cdot 4=12$ and the height of the large triangle is $3R=36$. That height is the $\sqrt{3}$ side in a $1:\sqrt{3}:2$ right triangle (30°-60°-90°), where $AB$ is the $2$ side, so $AB=\frac{2}{\sqrt{3}}\cdot 36=\frac{72}{\sqrt{3}}=24\sqrt{3}$.

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@Isaac, I'm tempted to do this almost in reverse. Let $D$ be the center of the smallest circle at the top. Let $E$ be the point where the smallest circle is tangent to $\overline{AB}$. Then $ADE$ is a 30-60-90 triangle and so the length $AD$ is immediately calculated. Now, do as you suggest in your first hint and the answer drops out. – cardinal Mar 13 '11 at 18:20
@cardinal: That's almost certainly what I'd do if I were doing the problem myself, but I suspect that forming the smaller equilateral triangle is conceptually harder for most students who are just learning geometry and are not necessarily accustomed to adding lines/segments to diagrams beyond an occasional altitude. – Isaac Mar 13 '11 at 18:23
@Isaac @cardinal eh, I got $AB=2\sqrt3R$ I got $AD=4\sqrt3$ but can't get R... – Templar Mar 13 '11 at 19:00
@Templar: Can you find $AE$? There is a relationship between $AD$ and $AE$ that will also apply to the corresponding segments in the whole triangle. – Isaac Mar 13 '11 at 19:04
@Isaac you mean $\frac{AD}{AE}=\cos 30$ ? – Templar Mar 13 '11 at 19:09

Let $a$ be the side of the triangle. If $A$ denotes the area and $P$ denotes the perimeter, then the radius of the incircle is given by $R = \frac{2A}{P} = \frac{2\sqrt{3} a^2/4}{3a} = \frac{\sqrt{3} a}{6}$

Let $x$ be the distance of the center of the smaller circle to the nearest vertex.

The altitude is $x + 8 + 2R = \frac{\sqrt{3}a}{2}$.

From similar triangle, we also have $\frac{x+4}{4} = \frac{x+8+R}{R} \Rightarrow \frac{x}{4} = \frac{x+8}{R}$

You have three equations in $x$,$R$ and $a$ solve them to get your $R$ and $a$.

EDIT

$x+8 = \frac{\sqrt{3}a}{2} - 2R = \frac{\sqrt{3}a}{2} - \frac{\sqrt{3}a}{3} = \frac{\sqrt{3}a}{6} = R$. Hence, $\frac{x}{4} = 1 \Rightarrow x = 4$.

$R = x + 8 = 12 \Rightarrow a = 2 \sqrt{3} R = 24 \sqrt{3}$.

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you haven't defined $A$ or $P$ and so the meaning of each might not be clear to the OP (especially since another $A$ is already present in the diagram). This solution seems overly complicated to me, especially since all one needs is the basic fact about side-lengths of a 30-60-90 triangle and the ability to recognize symmetry. (Virtually) No algebra needed. – cardinal Mar 13 '11 at 18:45