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Polygon $ABC$ is an equilateral triangle. Three congruent semicircular arcs are drawn on the three sides and towards the interior of the triangle. The terminal points of the semicircular arcs are the three vertices of the triangle $ABC$. Semicircle $C_1$ of diameter $AB$ Semicircle $C_2$ of diameter $BC$ Semicircle $C_3$ of diameter $CA$

A small circle $F$ is drawn such that it is tangent internally to $C_1$ and $C_3$ and externally to $C_2$

What is the radius of circle $ F$?

Here is the picture

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enter image description here

Using coordinates . . .

For convenience of notation, let $h=\sqrt{3}$.

Let $B = (-1,0),\;C=(1,0),\;A=(0,h)$.

Let $P$ be the center of the required circle.

Then $P=(0,1+r)$, where $r$ is the unknown radius.

Let $c(P,r)$ denote the circle centered at $P$, with radius $r$.

Let $M$ be the midpoint of segment $CA$.

Then $M=\bigl({\large{\frac{1}{2}}},{\large{\frac{h}{2}}}\bigr)$.

Let $s(M,1)$ denote the semicircle centered at $M$, with radius $1$, as shown in the diagram.

Let $T$ be the point where $c(P,r)$ meets $s(M,1)$.

Since $c(P,r)$ and $s(M,1)$ are tangent to each other at $T$, it follows that the points $M,P,T$ are collinear.

enter image description here

Then since $MT=1$ and $PT=r$, we get $MP=1-r$, hence by the distance formula \begin{align*} &MP^2=(1-r)^2\\[4pt] \implies\;&\left({\small{\frac{1}{2}}}-0\right)^{\!2}+\left({\small{\frac{h}{2}}}-(1+r)\right)^{\!\!2}=(1-r)^2\\[4pt] \implies\;&r=\frac{4h-1-h^2}{4(4-h)}\\[4pt] &\phantom{r}=\frac{4\sqrt{3}-4}{4(4-\sqrt{3})}\\[4pt] &\phantom{r}=\frac{3\sqrt{3}-1}{13}\approx .3227809558\\[4pt] \end{align*}

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From the red triangle in the picture we infer $$(1-r)^2=\left({1\over2}\right)^2+\left(r+\left(1-{\sqrt{3}\over2}\right)\right)^2\ ,$$ and this leads to $$r={3\sqrt{3}-1\over13}\ .$$

enter image description here

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Although I think using coordinates is the best way to solve these types of problems, it's fun to find other solution.
The figure below is pretty apparent and gives us the construction for the circle. Let the side of the triangle be $2R$

Using the Law of Cosines we get $MH=R\sqrt{2-\sqrt{3}},\ MD=R\sqrt{5-2\sqrt{3}}\ $ (note that we are using $\cos 105^{\circ}=\dfrac{\sqrt{2}+\sqrt{6}}{4}$). We also have $DE\cdot MD=AD\cdot DH$, which gives us $$\frac{R^2(\sqrt{3}-1)}{MD^2}=\frac{DE}{MD}\Rightarrow \frac{\sqrt{3}-1}{5-2\sqrt{3}}=\frac{IE}{R-IE}$$ That last equation simplified to $$IE=\frac{\sqrt{3}-1}{4-\sqrt{3}}R=\frac{3\sqrt{3}-1}{13}R$$

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