CEMC Cayley, 2025

Solution 1:

Plot the points $A$, $B$, and $C$ as well as $D(0,5)$ and $E(r,7)$ so that $\triangle ABD$ and $\triangle BCE$ have right angles at $D$ and $E$, respectively.

These two right-angled triangles each have a horizontal side and a
vertical side, as shown.

Since $A$, $B$, and $C$ are all on the same line, $AD$ is parallel to $BE$, and $BD$ is parallel to $CE$, we must have that $\angle DAB=\angle EBC$ and $\angle DBA=\angle ECB$.

Hence, $\triangle ABD$ is similar to
$\triangle BCE$.

Using common ratios, we get $\dfrac{BC}{AB}=\dfrac{EC}{DB}=\dfrac{BE}{AD}$.

Each of $EC$ and $DB$ is vertical, so their lengths are the
difference between the $y$-coordinates of the two points.

Thus, $EC=t-7$ and $DB=7-5=2$.

Similarly, the lengths of $BE$ and
$AD$ are each the different between
the $x$-coordinates of the two
points, giving $BE=r$ and $AD=3$.

It is given that $BC=4AB$, so $\dfrac{BC}{AB}=4$. Therefore, $4=\dfrac{EC}{DB}=\dfrac{t-7}{2}$ and $4=\dfrac{BE}{AD}=\dfrac{r}{3}$.

Rearranging these two equations gives $8=t-7$ or $t=15$ and $12=r$, so $r+t=27$.

Solution 2:

Using the distance formula, $$AB = \sqrt{(7-5)^2+(0-(-3))^2} = \sqrt{4+9}=\sqrt{13}$$ and $$BC = \sqrt{(t-7)^2+(r-0)^2}=\sqrt{(t-7)^2+r^2}$$ It is given that
$BC=4AB$, so $4\sqrt{13}=\sqrt{(t-7)^2+r^2}$. Squaring
both sides gives $16\times 13 = (t-7)^2+r^2$.

It is also given that $A$, $B$, and $C$ are on a common line. This implies that
the slope of the segment $AB$ is the
same as the slope of the segment $BC$.

These slopes are $\dfrac{7-5}{0-(-3)}=\dfrac{2}{3}$ and
$\dfrac{t-7}{r-0}=\dfrac{t-7}{r}$,
respectively.

Setting the computed slopes equal, we have $\dfrac{t-7}{r}=\dfrac{2}{3}$, or $t-7=\dfrac{2}{3}r$.

Substituting into $16\times13 = (t-7)^2+r^2$, we get the following equivalent equations. $$\begin{align*} 16\times 13 &= \left(\dfrac{2}{3}r\right)^2 + r^2 \\ 16\times 13 &= \dfrac{4}{9}r^2+r^2 \\ 16\times 13 &= \dfrac{13}{9}r^2 \\ 16\times 9 &= r^2 \\ \sqrt{16}\sqrt{9} &= \sqrt{r^2} \\ 12 &= r\end{align*}$$ where the final equality is because
$r$ is assumed to be positive.

Therefore, we have $r=12$, from which
we get $t-7=\dfrac{2}{3}\times12=8$,
or $t=15$.

The answer to the question is $r+t=12+15=27$.