3B. In an equilateral triangle $ABC$ with area 1, an equilateral triangle $MNP$ is inscribed such that $M$ lies on side $BC$, $N$ lies on side $AC$, $P$ lies on side $AB$, and $MP$ is perpendicular to $AB$. What is the area of triangle $MNP$?
Solution. The triangles $A P N, B M P$ and $C N M$ are right-angled with angles $30^{\circ}$ and $60^{\circ}$, and since they all have the side of triangle $M N P$ as a leg, they are congruent. If the side of triangle $A B C$ is $a$, and the side of $M N P$ is $x$, and if $\overline{M B}=y$, then $\overline{P B}=\frac{y}{2}$ and thus $\overline{C M}=\frac{y}{2}$. The following hold: $y+\frac{y}{2}=a, x=\frac{y}{2} \sqrt{3}$, and from this

![](https://cdn.mathpix.com/cropped/2024_06_05_e56627a2e774dea33268g-06.jpg?height=318&width=368&top_left_y=836&top_left_x=1114)

$$
x=\frac{\frac{2 a}{3}}{2} \sqrt{3}=\frac{a}{3} \sqrt{3}
$$

The area of triangle MNP is

$$
\frac{x^{2}}{4} \sqrt{3}=\frac{3 a^{2}}{4 \cdot 9} \sqrt{3}=\frac{1}{3} \frac{a^{2} \sqrt{3}}{4}=\frac{1}{3}
$$

## III year