(a) Right-angled $\triangle P Q R$ has $\angle P Q R=90^{\circ}, P Q=6$ and $Q R=8$. If $M$ is the midpoint of $Q R$ and $N$ is the midpoint of $P Q$, determine the lengths of the medians $P M$ and $R N$.

(b) $\triangle D E F$ has two medians of equal length. Prove that $\triangle D E F$ is isosceles.

(c) $\triangle A B C$ has its vertices on a circle of radius $r$. If the lengths of two of the medians of $\triangle A B C$ are equal to $r$, determine the side lengths of $\triangle A B C$.
(a) Since $P Q=6$ and $N$ is the midpoint of $P Q$, then $P N=N Q=3$.

Since $Q R=8$ and $M$ is the midpoint of $Q R$, then $Q M=M R=4$.

![](https://cdn.mathpix.com/cropped/2024_04_17_6308d40ff1469d7b994ag-08.jpg?height=304&width=417&top_left_y=775&top_left_x=954)

Since $\triangle P Q M$ is right-angled at $Q$ and $P M>0$, we can apply the Pythagorean Theorem to obtain

$$
P M=\sqrt{P Q^{2}+Q M^{2}}=\sqrt{6^{2}+4^{2}}=\sqrt{52}=2 \sqrt{13}
$$

Since $\triangle N Q R$ is right-angled at $Q$ and $R N>0$, we can apply the Pythagorean Theorem to obtain

$$
R N=\sqrt{N Q^{2}+Q R^{2}}=\sqrt{3^{2}+8^{2}}=\sqrt{73}
$$

Therefore, the two medians have lengths $P M=2 \sqrt{13}$ and $R N=\sqrt{73}$.

(b) Solution 1

Suppose that $D M$ and $E N$ are the two medians of $\triangle D E F$ that are equal in length.

Place $\triangle D E F$ in the coordinate plane so that $D E$ lies along the $x$-axis and so that $F$ is on the positive $y$-axis. (We can do this by rotating and translating the triangle as necessary.) We may assume that the rightmost of $D$ and $E$ is on the positive $x$-axis (since we can reflect the triangle in the $y$-axis if necessary), and that it is vertex $E$ that is the rightmost of $D$ and $E$ (since we can switch the labels if necessary).

Give $F$ coordinates $(0,2 c), E$ coordinates $(2 b, 0)$, and $D$ coordinates $(2 a, 0)$ for some real numbers $a, b, c$ with $b>a$ and $b>0$ and $c>0$. The number $a$ might be positive, negative or zero. (The cases $a0$ are illustrated below.)
![](https://cdn.mathpix.com/cropped/2024_04_17_6308d40ff1469d7b994ag-08.jpg?height=372&width=1428&top_left_y=2101&top_left_x=454)

Since $D M$ and $E N$ are medians, then $M$ is the midpoint of $E F$ and $N$ is the midpoint of $D F$.

Since $E$ has coordinates $(2 b, 0)$ and $F$ has coordinates $(0,2 c)$, then $M$ has coordinates $\left(\frac{1}{2}(2 b+0), \frac{1}{2}(0+2 c)\right)=(b, c)$.

Since $D$ has coordinates $(2 a, 0)$ and $F$ has coordinates $(0,2 c)$, then $N$ has coordinates $\left(\frac{1}{2}(2 a+0), \frac{1}{2}(0+2 c)\right)=(a, c)$.

Since $D$ has coordinates $(2 a, 0)$ and $M$ has coordinates $(b, c)$, then

$$
D M=\sqrt{(2 a-b)^{2}+(0-c)^{2}}=\sqrt{(2 a-b)^{2}+c^{2}}
$$

Since $E$ has coordinates $(2 b, 0)$ and $N$ has coordinates $(a, c)$, then

$$
E N=\sqrt{(2 b-a)^{2}+(0-c)^{2}}=\sqrt{(2 b-a)^{2}+c^{2}}
$$

Since $D M$ and $E N$ are equal in length, then

$$
\begin{aligned}
D M^{2} & =E N^{2} \\
(2 a-b)^{2}+c^{2} & =(2 b-a)^{2}+c^{2} \\
(2 a-b)^{2} & =(2 b-a)^{2} \\
4 a^{2}-4 a b+b^{2} & =4 b^{2}-4 a b+a^{2} \\
3 a^{2} & =3 b^{2} \\
a^{2} & =b^{2}
\end{aligned}
$$

Therefore, $a= \pm b$.

Since $D$ and $E$ are distinct points, then $a \neq b$.

Thus, $a=-b$.

Therefore, the vertices of $\triangle D E F$ are $D(-2 b, 0), E(2 b, 0)$, and $F(0,2 c)$.

But then sides $D F$ and $E F$ are equal in length (each has length $\sqrt{4 b^{2}+4 c^{2}}$ ), which means that $\triangle D E F$ is isosceles, as required.

Solution 2

Suppose that $D M$ and $E N$ are the two medians of $\triangle D E F$ that are equal in length, and that they cross at $G$.

Join $M$ to $N$.

![](https://cdn.mathpix.com/cropped/2024_04_17_6308d40ff1469d7b994ag-09.jpg?height=268&width=464&top_left_y=1899&top_left_x=931)

Since $D M$ and $E N$ are medians, then $M$ is the midpoint of $F E$ (which gives $F M=M E$ ) and $N$ is the midpoint of $F D$ (which gives $F N=N D$ ).

Since $\frac{F D}{F N}=\frac{F E}{F M}=2$ and the triangles share a common angle at $F$, then $\triangle F D E$ and $\triangle F N M$ are similar, with similarity ratio $2: 1$. This tells us that $\angle F D E=\angle F N M$. Thus, $N M$ is parallel to $D E$.

Also, $D E=2 N M$.

Consider next $\triangle G N M$ and $\triangle G E D$.

Since $N M$ is parallel to $D E$, then $\angle G N M=\angle G E D$ and $\angle G M N=\angle G D E$.

This means that $\triangle G N M$ is similar to $\triangle G E D$. Since $N M=\frac{1}{2} D E$, then the similarity ratio is $1: 2$.

Therefore, $G E=2 G N$ and $G D=2 G M$.

Since $D M=E N$, then $G D+G M=G E+G N$ or $2 G M+G M=2 G N+G N$. This gives $3 G M=3 G N$ or $G M=G N$.

Therefore, $\triangle G N M$ is isosceles with $\angle G N M=\angle G M N$.

Finally, consider $\triangle M N E$ and $\triangle N M D$.

Since $\angle M N E=\angle N M D, N M$ is a common side, and $E N=D M$, then the triangles are congruent (SAS).

Thus, $M E=N D$.

Since $M E=\frac{1}{2} F E$ and $N D=\frac{1}{2} F D$, then $F E=F D$, and so $\triangle D E F$ is isosceles, as required.

(c) Suppose that $\triangle A B C$ has equal medians $A M=C N$, with $M$ the midpoint of $B C$ and $N$ the midpoint of $A B$. Suppose also that $O$ is the centre of the circle.

We are given that $A M=C N=r$.

From (b), $\triangle A B C$ must be isosceles with $A B=B C$.

We consider three cases: $\angle A B C=90^{\circ}, \angle A B C90^{\circ}$.

Case 1: $\angle A B C=90^{\circ}$

Since $\angle A B C=90^{\circ}$, then $A C$ is a diameter of the circle with centre $O$ and radius $r$. Thus, $A C=2 r$.

Further, since $\triangle A B C$ is right-angled and isosceles, then $A B=B C=\frac{1}{\sqrt{2}} A C=\sqrt{2} r$. We draw median $A M$, which has length $r$.

![](https://cdn.mathpix.com/cropped/2024_04_17_6308d40ff1469d7b994ag-10.jpg?height=417&width=456&top_left_y=1407&top_left_x=932)

Now $\triangle M B A$ is right-angled at $B$, which means that $A M$ is its longest side.

Therefore, $A M>A B=\sqrt{2} r$.

But $A M=r$, which is a contradiction.

Therefore, $\angle A B C \neq 90^{\circ}$.

Case 2: $\angle A B C90^{\circ}$

Here, $\angle A B C$ is obtuse and $A B=B C$.

Since $\angle A B C$ is obtuse, then $B$ is on the same side of $O$ as $A C$.

Let $P$ be the midpoint of $A C$.

Draw $B P$ and $O P$.

Since $\triangle A B C$ is isosceles, then $B P$ is perpendicular to $A C$.

Since $P$ is the midpoint of $A C$, then $O P$ is perpendicular to $A C$ as well.

Therefore, $B P O$ is a straight line segment.

Draw a perpendicular from $M$ to $Q$ on $A C$.

![](https://cdn.mathpix.com/cropped/2024_04_17_6308d40ff1469d7b994ag-12.jpg?height=415&width=439&top_left_y=156&top_left_x=949)

Suppose that $O P=x$.

Since $O A=r$ (it is a radius) and $O P=x$, then $A P=\sqrt{O A^{2}-O P^{2}}=\sqrt{r^{2}-x^{2}}$.

Now $\triangle M Q C$ is similar to $\triangle B P C$, since each is right-angled and they share a common angle at $C$.

Since $B C=2 M C$, then $P C=2 Q C$ and $B P=2 M Q$.

Since $Q C=\frac{1}{2} P C$, then $P Q=P C-Q C=\frac{1}{2} P C=\frac{1}{2} A P=\frac{1}{2} \sqrt{r^{2}-x^{2}}$.

Since $B O=r$ and $O P=x$, then $M Q=\frac{1}{2} B P=\frac{1}{2}(r-x)$.

Finally, $\triangle A M Q$ is right-angled at $Q$, so by the Pythagorean Theorem, we have

$$
\begin{aligned}
A Q^{2}+M Q^{2} & =A M^{2} \\
(A P+P Q)^{2}+M Q^{2} & =A M^{2} \\
\left(\sqrt{r^{2}-x^{2}}+\frac{1}{2} \sqrt{r^{2}-x^{2}}\right)^{2}+\left(\frac{1}{2}(r-x)\right)^{2} & =r^{2} \\
\left(\frac{3}{2} \sqrt{r^{2}-x^{2}}\right)^{2}+\left(\frac{1}{2}(r-x)\right)^{2} & =r^{2} \\
\frac{9}{4}\left(r^{2}-x^{2}\right)+\frac{1}{4}\left(r^{2}-2 r x+x^{2}\right) & =r^{2} \\
9\left(r^{2}-x^{2}\right)+\left(r^{2}-2 r x+x^{2}\right) & =4 r^{2} \\
6 r^{2}-2 r x-8 x^{2} & =0 \\
3 r^{2}-r x-4 x^{2} & =0 \\
(3 r-4 x)(r+x) & =0
\end{aligned}
$$

Therefore, $r=-x$ (which is impossible since $r$ and $x$ are both positive) or $3 r=4 x$.

In this last case, $x=\frac{3}{4} r$.

Then $A C=2 \sqrt{r^{2}-x^{2}}=2 \sqrt{r^{2}-\frac{9}{16} r^{2}}=2 \sqrt{\frac{7}{16} r^{2}}=\frac{2 \sqrt{7}}{4} r=\frac{\sqrt{7}}{2} r$.

Also, $A B=B C=\sqrt{A P^{2}+B P^{2}}=\sqrt{\left(\sqrt{\frac{7}{16} r^{2}}\right)^{2}+\left(r-\frac{3}{4} r\right)^{2}}=\sqrt{\frac{7}{16} r^{2}+\frac{1}{16} r^{2}}=\frac{\sqrt{2}}{2} r$.

Therefore, having examined all of the possible cases, the side lengths of $\triangle A B C$ must be $A C=\frac{\sqrt{7}}{2} r$ and $A B=B C=\frac{\sqrt{2}}{2} r$.