Subfield: Geometry


Question: Triangle $A B C$ is right-angled at $B$ and has side lengths which are integers. A second triangle, $P Q R$, is located inside $\triangle A B C$ as shown, such that its sides are parallel to the sides of $\triangle A B C$ and the distance between parallel lines is 2 . Determine the side lengths of all possible triangles $A B C$, such that the area of $\triangle A B C$ is 9 times that of $\triangle P Q R$.

![](https://cdn.mathpix.com/cropped/2023_12_21_eacf83d29c256cdd2ca3g-1.jpg?height=412&width=480&top_left_y=604&top_left_x=1362)

Solution:
Let the sides of $\triangle A B C$ be $A B=c, B C=a, A C=b, a, b, c$ are all integers.

Since the sides of $\triangle P Q R$ are all parallel to the sides of $\triangle A B C$, then $\triangle A B C$ is similar to $\triangle P Q R$.

Now the ratio of areas of $\triangle A B C$ to $\triangle P Q R$ is $9=3^{2}$ to 1 , so the ratio of side lengths will be 3 to 1 .

So the sides of $\triangle P Q R$ are $P Q=\frac{c}{3}, Q R=\frac{a}{3}, P R=\frac{b}{3}$.



So we can label the diagram as indicated.

We join the corresponding vertices of the two triangles as

Area of trapezoid $B Q R C$

Area of trapezoid CRPA

Area of trapezoid $A P Q B$

$+\quad$ Area of $\triangle P Q R$

Area of $\triangle A B C$.

![](https://cdn.mathpix.com/cropped/2023_12_21_0e6430f266c4e3edca52g-1.jpg?height=450&width=534&top_left_y=282&top_left_x=1381)

Doing so gives,

$2\left(\frac{2}{3} a\right)+2\left(\frac{2}{3} b\right)+2\left(\frac{2}{3} c\right)+\frac{a c}{18}=\frac{a c}{72}$

Or upon simplifying $a c=3 a+3 b+3 c$ (Note that this relationship can be derived in a variety of ways.)

$$
\begin{array}{rlrl}
a c & =3 c+3 b+3 a & & \\
a c-3 c-3 a & =3 b & \\
a c-3 c-3 a & =3 \sqrt{a^{2}+c^{2}} & & \left(\text { since } b=\sqrt{a^{2}+c^{2}}\right) \\
a^{2} c^{2}+9 c^{2}+9 a^{2}-6 a c^{2}-6 a^{2} c+18 a c & =9\left(a^{2}+c^{2}\right) & & \text { (squaring both sides) } \\
a c(a c-6 c-6 a+18) & =0 & \\
a c-6 c-6 a+18 & =0 & \\
c(a-6) & =6 a-18 \\
c & =\frac{6 a-18}{a-6} \\
c & =6+\frac{18}{a-6} .
\end{array}
$$

Since $a$ is a side of a triangle, $a>0$. We are now looking for positive integer values such that $\frac{18}{a-6}$ is also an integer.

The only possible values for $a$ are 3, 7, 8, 9, 12, 15 and 24 .

Tabulating the possibilities and calculating values for $b$ and $c$ gives,

| $a$ | 3 | 7 | 8 | 9 | 12 | 15 | 24 |
| :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: |
| $c$ | 0 | 24 | 15 | 12 | 9 | 8 | 7 |
| $b$ | - | 25 | 17 | 15 | 15 | 17 | 25 |

Thus the only possibilities for the triangle are $(7,24,25),(8,15,7)$ and $(9,12,15)$.
The two triangles are similar with areas in the ratio 1:9.

Therefore the sides are in the ratio 1:3.

Let $a=B C, b=C A, c=B A$.



Then $\frac{a}{3}=P Q, \frac{b}{3}=Q R, \frac{c}{3}=P R$.

Locate points $K, L$ on $B C ; M, N$ on $C A$; and $T, S$ on $A B$ as shown.

$$
\begin{gathered}
B C=B K+K L+L C \\
a=B K+\frac{a}{3}+2
\end{gathered}
$$

![](https://cdn.mathpix.com/cropped/2023_12_21_e2a5fb5a1f25f531825cg-1.jpg?height=504&width=594&top_left_y=274&top_left_x=1362)

Therefore $B K=\frac{2}{3} a-2$.

In a similar way, $A N=\frac{2}{3} b-2$.

Now $\triangle B K P \cong \triangle B T P$ and $\triangle A N R \cong \triangle A S R$, both by $H L$.

Therefore $B T=B K=\frac{2}{3} a-2$ and $A S=A N=\frac{2}{3} b-2$.

Now, $A B=A S+S T+B T$

$$
\begin{aligned}
c & =\frac{2}{3} b-2+\frac{c}{3}+\frac{2}{3} a-2 \\
\frac{2}{3} c & =\frac{2}{3} b+\frac{2}{3} a-4 \\
c & =b+a-6 \\
b & =c+(6-a) .
\end{aligned}
$$

By Pythagoras, $a^{2}+b^{2}=c^{2}$

$$
\begin{aligned}
& a^{2}+[c+(6-a)] 2=c^{2} \\
& a^{2}+\not \not^{2}+2 c(6-a)+(6-a) 2=\not \not^{2} \\
& a^{2}+\quad(6-a)^{2}=-2 c(6-a) \\
& 2 a^{2}-\quad 12 a+36=2 c(a-6) \\
& a^{2}-\quad 6 a+18=c(a-6) \\
& c=\frac{a^{2}-6 a+18}{a-6} \\
& c=\frac{a(a-6)+18}{a-6} \\
& c=a+\frac{18}{a-6} \text {. }
\end{aligned}
$$

Since $a$ and $c$ are integers, $a-6$ is a divisor of 18 .

Also since $b<c$ and $b=c+(6-a)$, we conclude that $6-a<0$ so $a-6>0$.

Thus $a-6$ can be $1,2,3,6,9,18$.

The values of $a$ are: 7, 8, 9, 12, 15, 24 .

Matching values for $c: 25,17,15,15,17,25$

Matching values for $b: 24,15,12,9,8,7$



The distinct triangles are $(7,24,25),(8,15,17)$ and $(9,12,15)$.

Final answer: $(7,24,25),(8,15,7),(9,12,15)$