3. Alice draws a heart on her math notebook as follows: first, she draws two circles with radius $1 \, \text{cm}$ and centers $O_{1}, O_{2}$, which are externally tangent. Let $r$ be the common tangent to the two circles passing through the point of tangency. She then chooses a point $P$ on $r$ such that $\widehat{O_{1} P O_{2}}=60^{\circ}$ and draws the tangents to the two circles passing through $P$. What is the area of the heart obtained (i.e., the shaded area in the figure) in $\text{cm}^{2}$?

![](https://cdn.mathpix.com/cropped/2024_04_17_c5c03d45c82457ddb3cag-03.jpg?height=306&width=325&top_left_y=761&top_left_x=1525)
(A) $2 \sqrt{3}+\frac{4}{3} \pi$
(B) $\sqrt{3}+\pi$
(C) $\frac{7}{3} \pi$
(D) $\frac{3}{2}+2 \pi$
(E) None of the above
3. The answer is (A). Let $T_{1}, T_{2}, T_{3}$ be the points where the tangents from $P$ touch the two circles, as shown in the figure. We observe that, by known properties of tangents, the angles $\widehat{O_{1} T_{1} P}, \widehat{O_{1} T_{2} P}$, $\widehat{O_{2} T_{2} P}, \widehat{O_{2} T_{3} P}$ are right angles. We also note that the segments $P T_{1}, P T_{2}, P T_{3}$ all have the same length: $\overline{P T_{1}}=\overline{P T_{2}}$ because they are the two tangent segments from $P$ to the same circle, and $\overline{P T_{2}}=\overline{P T_{3}}$ for the same reason, applied to the other circle.

The triangles $P T_{1} O_{1}, P T_{2} O_{1}, P T_{2} O_{2}, P T_{3} O_{2}$ are then all congruent, as each of them has one side equal to $1 \mathrm{~cm}$ (the radii of the circles), one side equal to the length $\overline{P T_{1}}=\overline{P T_{2}}=\overline{P T_{3}}$, and the angle between them equal to $90^{\circ}$. Moreover, the angle $\widehat{T_{2} P O_{1}}$, given the symmetry of the figure, is half of $\widehat{O_{2} P O_{1}}=60^{\circ}$, so $\widehat{T_{2} P O_{1}}=\widehat{T_{1} P O_{1}}=30^{\circ}$. In the triangle $P T_{1} O_{1}$, we then have angles of $30^{\circ}, 60^{\circ}, 90^{\circ}$, which allows us to calculate $\overline{P O_{1}}=2 \overline{T_{1} O_{1}}=2 \mathrm{~cm}, \overline{P T_{1}}=\frac{\sqrt{3}}{2} \overline{P O_{1}}=\sqrt{3} \mathrm{~cm}$, and thus

![](https://cdn.mathpix.com/cropped/2024_04_17_c5c03d45c82457ddb3cag-12.jpg?height=294&width=354&top_left_y=1686&top_left_x=1505)
also the area of the triangle $P O_{1} T_{1}$, which is $\frac{1}{2} \overline{P T_{1}} \cdot \overline{T_{1} O_{1}}=\frac{1}{2} \cdot 1 \mathrm{~cm} \cdot \sqrt{3} \mathrm{~cm}=$ $\frac{\sqrt{3}}{2} \mathrm{~cm}^{2}$.

The portion of the circle contained in the union of the triangles $P T_{1} O_{1}$ and $P T_{2} O_{1}$ is a circular segment with central angle $\widehat{T_{1} O_{1} T_{2}}=180^{\circ}-\widehat{T_{2} P T_{1}}=120^{\circ}=\frac{1}{3} 360^{\circ}$, so this area is $\frac{1}{3} \pi(1 \mathrm{~cm})^{2}$. The complement of this circular segment within the circle is another circular segment, with a central angle of $240^{\circ}$ (and thus an area of $\frac{2}{3} \pi \mathrm{cm}^{2}$).

The total area of the heart is then given by the area of the circular segments with endpoints $T_{2} T_{1}$ and $T_{3} T_{2}$ and a central angle of $240^{\circ}$ (total area $2 \cdot \frac{2}{3} \pi \mathrm{cm}^{2}$), plus the area of four triangles congruent to $P T_{1} O_{1}$ (i.e., $P T_{1} O_{1}, P T_{2} O_{1}, P T_{2} O_{2}, P T_{3} O_{2}$), which, as already seen, each have an area of $\frac{\sqrt{3}}{2} \mathrm{~cm}^{2}$. The answer is therefore $\frac{4}{3} \pi+4 \frac{\sqrt{3}}{2}=\frac{4}{3} \pi+2 \sqrt{3}$.