We use without proof the fact that if a circle with centre $O$ and radius $r$ touches (that is, is tangent to) line segments $A B, B C$ and $C D$ at $X, Y$ and $Z$, respectively, then $\angle O B X=\angle O B Y=\frac{1}{2}(\angle A B C)$ and $\angle O C Y=\angle O C Z=\frac{1}{2}(\angle B C D)$.

Suppose that $\angle A B C=\theta$ and $\angle B C D=\alpha$.

Since $O Y$ is perpendicular to $B C$, then $\tan (\angle O B Y)=\frac{O Y}{B Y}$ and $\tan (\angle O C Y)=\frac{O Y}{Y C}$.

![](https://cdn.mathpix.com/cropped/2024_04_20_33a8a6559d675f59baa2g-142.jpg?height=412&width=460&top_left_y=195&top_left_x=1445)

Thus, $B Y=\frac{O Y}{\tan (\angle O B Y)}=\frac{r}{\tan (\theta / 2)}$ and $Y C=\frac{O Y}{\tan (\angle O C Y)}=\frac{r}{\tan (\alpha / 2)}$.

Since $B C=B Y+Y C$, then

$$
\begin{aligned}
B C & =\frac{r}{\tan (\theta / 2)}+\frac{r}{\tan (\alpha / 2)} \\
B C & =r\left(\frac{1}{\tan (\theta / 2)}+\frac{1}{\tan (\alpha / 2)}\right) \\
B C & =r\left(\frac{\tan (\theta / 2)+\tan (\alpha / 2)}{\tan (\theta / 2) \tan (\alpha / 2)}\right) \\
r & =\frac{B C \tan (\theta / 2) \tan (\alpha / 2)}{\tan (\theta / 2)+\tan (\alpha / 2)}
\end{aligned}
$$

Consider now the given quadrilateral $Q R S T$. We know that $T Q=3$. Since $\triangle P Q R$ is equilateral, then $Q R=P Q$ and so $Q R=P T+T Q=1+3=4$.

Since $P R=Q R=4$ and $S$ is the midpoint of $P R$, then $R S=2$. Since $\triangle P S T$ has $P T=1$ and $P S=\frac{1}{2} P R=2$ and $\angle S P T=60^{\circ}$, then it is a $30^{\circ}-60^{\circ}-90^{\circ}$ triangle, so we have $S T=\sqrt{3}$.

Also, we have $\angle P T S=90^{\circ}$ and $\angle P S T=30^{\circ}$, which give $\angle S T Q=90^{\circ}$ and $\angle R S T=150^{\circ}$.

To summarize, quadrilateral $Q R S T$ has $Q R=4, R S=2$, $S T=\sqrt{3}, T Q=3, \angle T Q R=60^{\circ}, \angle Q R S=60^{\circ}, \angle R S T=150^{\circ}$,

![](https://cdn.mathpix.com/cropped/2024_04_20_33a8a6559d675f59baa2g-142.jpg?height=463&width=510&top_left_y=1362&top_left_x=1450)
and $\angle S T Q=90^{\circ}$.

Next, we determine the radius of a circle tangent to each set of three consecutive sides, ignoring the fact that the fourth side might restrict the size of the circle. After determining these radii, we examine how the fourth side comes into play.

Suppose that we have a circle tangent to line segments $T Q, Q R$ and $R S$.

By the formula above, the radius of this circle would equal

$$
\frac{4 \tan \left(60^{\circ} / 2\right) \tan \left(60^{\circ} / 2\right)}{\tan \left(60^{\circ} / 2\right)+\tan \left(60^{\circ} / 2\right)}=\frac{4 \tan \left(30^{\circ}\right) \tan \left(30^{\circ}\right)}{\tan \left(30^{\circ}\right)+\tan \left(30^{\circ}\right)} \approx 1.1547
$$

Suppose that we have a circle tangent to line segments $Q R, R S$ and $S T$.

By the formula above, the radius of this circle would equal

$$
\frac{2 \tan \left(60^{\circ} / 2\right) \tan \left(150^{\circ} / 2\right)}{\tan \left(60^{\circ} / 2\right)+\tan \left(150^{\circ} / 2\right)}=\frac{2 \tan \left(30^{\circ}\right) \tan \left(75^{\circ}\right)}{\tan \left(30^{\circ}\right)+\tan \left(75^{\circ}\right)}=1
$$

Suppose that we have a circle tangent to line segments $R S, S T$ and $T Q$. By the formula above, the radius of this circle would equal

$$
\frac{\sqrt{3} \tan \left(150^{\circ} / 2\right) \tan \left(90^{\circ} / 2\right)}{\tan \left(150^{\circ} / 2\right)+\tan \left(90^{\circ} / 2\right)}=\frac{\sqrt{3} \tan \left(75^{\circ}\right) \tan \left(45^{\circ}\right)}{\tan \left(75^{\circ}\right)+\tan \left(45^{\circ}\right)} \approx 1.3660
$$

Suppose that we have a circle tangent to line segments $S T, T Q$ and $Q R$. By the formula above, the radius of this circle would equal

$$
\frac{3 \tan \left(90^{\circ} / 2\right) \tan \left(60^{\circ} / 2\right)}{\tan \left(90^{\circ} / 2\right)+\tan \left(60^{\circ} / 2\right)}=\frac{3 \tan \left(45^{\circ}\right) \tan \left(30^{\circ}\right)}{\tan \left(45^{\circ}\right)+\tan \left(30^{\circ}\right)} \approx 1.0981
$$

We need to determine the radius of the largest circle that can be drawn inside quadrilateral QRST.

The largest such circle will be touching at least two adjacent sides of $Q R S T$. Why is this? If a circle were touching zero sides or one side of $Q R S T$, it could be slid until it was touching two consecutive sides and then expanded a little bit and so cannot be the largest such circle. If it were touching two opposite sides of $Q R S T$, then it could be slid to touch a side adjacent to one of these two sides, perhaps losing contact with one of the two opposite sides.

Consider now a circle touching two adjacent sides of $Q R S T$. Such a circle can be expanded while maintaining contact to these two sides until it touches a third side. Once it touches the third side, it can't be expanded any further because its radius is fixed by the calculation that we did at the beginning of the solution.

Therefore, the largest circle will be touching three of the sides of QRST.

In order to complete the solution, we need to determine which of the circles that touch three sides actually lie completely inside $Q R S T$.

We do this by examining each of the four pairs of consecutive sides, and determining what the largest circle is that can be drawn touching these sides.

- We consider a circle tangent to $S T$ and $T Q$. We expand the circle, keeping it tangent to $S T$ and $T Q$, until it touches either $Q R$ or $R S$. From the previous calculations, the circle that also touches $Q R$ has radius about 1.0981, and the circle that also touches $R S$ has radius about 1.3660. Thus, the circle will first touch $Q R$. In this case, the largest circle that is completely inside the quadrilateral has radius 1.0981.
- We consider a circle tangent to $T Q$ and $Q R$. We expand the circle, keeping it tangent to $T Q$ and $Q R$, until it touches either $S T$ or $R S$. From the previous calculations, the circle that also touches $S T$ has radius about 1.0981, and the circle that also touches $R S$ has radius about 1.1547. Thus, the circle will first touch $S T$. In this case, the largest circle that is completely inside the quadrilateral has radius 1.0981.
- We consider a circle tangent to $Q R$ and $R S$. We expand the circle, keeping it tangent to $Q R$ and $R S$, until it touches either $S T$ or $T Q$. From the previous calculations, the circle that also touches $S T$ has radius 1, and the circle that also touches $T Q$ has radius about 1.1547. Thus, the circle will first touch $S T$. In this case, the largest circle that is completely inside the quadrilateral has radius 1 .
- We consider a circle tangent to $R S$ and $S T$. We expand the circle, keeping it tangent to $R S$ and $S T$, until it touches either $Q R$ or $T Q$. From the previous calculations, the circle that also touches $Q R$ has radius about 1, and the circle that also touches $T Q$ has radius about 1.3660. Thus, the circle will first touch $Q R$. In this case, the largest circle that is completely inside the quadrilateral has radius 1 .

Finally, comparing the four cases, we see that the largest circle that we can obtain has radius about 1.0981, which is closest to 1.10 .

ANsWER: (B)
In equilateral triangle $P Q R, S$ is the midpoint of $P R$, $T$ is on $P Q$ so that $P T=1$ and $T Q=3$. Many circles can be drawn inside quadrilateral $Q R S T$ so that no part extends outside of $Q R S T$. The radius of the largest such circle is closest to
(A) 1.00
(B) 1.10
(C) 1.15
(D) 1.05
(E) 1.37

![](https://cdn.mathpix.com/cropped/2024_04_20_ac36362783317e0251fdg-071.jpg?height=447&width=507&top_left_y=706&top_left_x=1226)