## Task A-3.5.

In a scalene triangle $ABC$, the lengths of two altitudes are equal to the lengths of two medians. What is the ratio of the lengths of the remaining altitude and the remaining median?
## Solution.

Let $a, b$ and $c$ be the lengths of the sides $\overline{B C}, \overline{C A}$ and $\overline{A B}$, respectively. Denote by $v_{a}, v_{b}$ and $v_{c}$ the lengths of the altitudes from vertices $A, B$ and $C$, and by $t_{a}$, $t_{b}$ and $t_{c}$ the lengths of the medians from vertices $A, B$ and $C$, respectively. Furthermore, let $P_{A}, P_{B}$ and $P_{C}$ be the midpoints of the segments $\overline{B C}, \overline{C A}$ and $\overline{A B}$, and let $N_{A}, N_{B}$ and $N_{C}$ be the feet of the altitudes from vertices $A, B$ and $C$, respectively.

Since the altitude is the shortest segment connecting a vertex of a triangle to the points on the opposite side, the length of the altitude is less than or equal to the length of the median from the same vertex, with equality holding only if the triangle is isosceles. Therefore, the length of the altitude and the median from the same vertex cannot be equal due to the problem's assumption. Let us assume without loss of generality that $t_{a}=v_{b}$. Then it cannot be true that $t_{b}=v_{a}$, otherwise we would have

$$
t_{a}+t_{b}=v_{b}+v_{a}<t_{b}+t_{a}
$$

Let us therefore assume without loss of generality that $t_{b}=v_{c}$.

![](https://cdn.mathpix.com/cropped/2024_05_30_27ac938d21197d2216d9g-15.jpg?height=708&width=786&top_left_y=277&top_left_x=635)

Let $Q$ be the midpoint of the segment $\overline{N_{B} C}$. Then $\overline{Q P_{A}}$ is the median of the right triangle $B N_{B} C$ and is perpendicular to the segment $\overline{A C}$. We have

$$
\left|P_{A} Q\right|=\frac{1}{2}\left|B N_{B}\right|=\frac{1}{2} v_{b}=\frac{1}{2} t_{a}=\frac{1}{2}\left|P_{A} A\right|
$$

The triangle $P_{A} Q A$ is a right triangle in which one leg is half the length of the hypotenuse, so $\varangle C A P_{A}=30^{\circ}$. Similarly, it can be shown that $\varangle P_{B} B A=30^{\circ}$.

Let $T$ be the centroid of triangle $A B C$. In triangles $B P_{B} A$ and $A P_{B} T$, we have $\varangle P_{B} B A=\varangle P_{B} A T$, and they share the angle at vertex $P_{B}$, so they are similar by the $\mathrm{K}-\mathrm{K}$ similarity theorem. Therefore,

$$
\frac{\left|B P_{B}\right|}{\left|A P_{B}\right|}=\frac{\left|P_{B} A\right|}{\left|P_{B} T\right|} \Longrightarrow \frac{t_{b}}{b / 2}=\frac{b / 2}{t_{b} / 3} \Longrightarrow t_{b}=\frac{\sqrt{3}}{2} b
$$

Since $t_{b}=v_{c}$, from the right triangle $C N_{C} A$ we conclude that $\varangle C A N_{C}=60^{\circ}$. Depending on whether $N_{C}$ lies on the side $\overline{A B}$ or not, we have $\varangle C A B=60^{\circ}$ or $\varangle C A B=120^{\circ}$. Since the first case implies that the median from vertex $A$ is also the angle bisector of that angle (which contradicts the assumption that the triangle is scalene), we conclude that $\varangle C A B=120^{\circ}$.

![](https://cdn.mathpix.com/cropped/2024_05_30_27ac938d21197d2216d9g-15.jpg?height=708&width=811&top_left_y=1876&top_left_x=628)

In triangle $B A P_{B}$, we have

$$
\varangle B P_{B} A=180^{\circ}-\varangle B A P_{B}-\varangle A B P_{B}=180^{\circ}-120^{\circ}-30^{\circ}=30^{\circ}=\varangle A B P_{B}
$$

so this triangle is isosceles and we have $c=|A B|=\left|A P_{B}\right|=\frac{b}{2}$, i.e., $b=2 c$. From the Law of Cosines applied to $A B C$, we get

$$
a=\sqrt{4 c^{2}+c^{2}-4 c^{2} \cos 120^{\circ}}=c \sqrt{7}
$$

and from the Law of Cosines applied to $A P_{C} C$, we get

$$
t_{c}=\sqrt{\frac{1}{4} c^{2}+4 c^{2}-2 c^{2} \cos 120^{\circ}}=\frac{c \sqrt{21}}{2}
$$

The area of triangle $A B C$ is

$$
P=\frac{1}{2} c \cdot v_{c}=\frac{1}{2} c \cdot t_{b}=\frac{1}{2} c \cdot \frac{\sqrt{3}}{2} b=\frac{c^{2} \sqrt{3}}{2}
$$

so the length of the altitude from vertex $A$ is $v_{a}=\frac{2 P}{a}=c \sqrt{\frac{3}{7}}$. Finally,

$$
\frac{v_{a}}{t_{c}}=\frac{\sqrt{\frac{3}{7}}}{\frac{\sqrt{21}}{2}}=\frac{2}{7}
$$

## NATIONAL MATHEMATICS COMPETITION

## 4th grade - high school - A variant

## May 11, 2021.