Let $t$ be TNYWR.

Point $O$ is at the origin and points $P(a, b)$ and $Q(c, d)$ are in the first quadrant, as shown. The slope of $O P$ is $\frac{12}{5}$ and the length of $O P$ is $13 t$. The slope of $O Q$ is $\frac{3}{4}$ and the length of $O Q$ is $10 t$. What is $a+c$ ?

![](https://cdn.mathpix.com/cropped/2024_04_17_ea957df7444903777e65g-2.jpg?height=444&width=444&top_left_y=2247&top_left_x=835)

## The CENTRE for EDUCATION in MATHEMATICS and COMPUTING cemc.uwaterloo.ca 2023 Canadian Team Mathematics Contest 
The line segment $O P$ passes through the origin and $P$, so its slope is equal to $\frac{b}{a}$.

Since the slope is $\frac{12}{5}$ and the line passes through the origin, there is some $k$ for which $b=12 k$ and $a=5 k$.

By the Pythagorean theorem, $a^{2}+b^{2}=O P^{2}$. Substituting $b=12 k, a=5 k$, and $O P=13 t$, we get the following equivalent equations.

$$
\begin{aligned}
a^{2}+b^{2} & =O P^{2} \\
(5 k)^{2}+(12 k)^{2} & =(13 t)^{2} \\
25 k^{2}+144 k^{2} & =169 t^{2} \\
169 k^{2} & =169 t^{2} \\
k^{2} & =t^{2}
\end{aligned}
$$

Since $P(a, b)$ is in the first quadrant, $a=5 k$ is positive, so $k$ is positive. As well, since $13 t$ is given to be a length, we can assume $t$ is positive, which means $k=t$, so $a=5 t$.

By reasoning similar to earlier, there is some $m$ so that $c=4 m$ and $d=3 m$.

By the Pythagorean theorem, $c^{2}+d^{2}=O Q^{2}$, so we get the following equivalent equations.

$$
\begin{aligned}
c^{2}+d^{2} & =O Q^{2} \\
(4 m)^{2}+(3 m)^{2} & =(10 t)^{2} \\
16 m^{2}+9 m^{2} & =100 t^{2} \\
25 m^{2} & =100 t^{2} \\
m^{2} & =4 t^{2}
\end{aligned}
$$

Since $c$ and $t$ are both positive, $m=\sqrt{4 t^{2}}=2 t$, so $c=8 t$.

Therefore, $a+c=5 t+8 t=13 t$.

Substituting $t=3$, we get $a+c=13(3)=39$.

Answer: $(6,3,39)$