CEMC Euclid, 2025

Solution 1:

Since $p + q = 5$ and $q + r = 9$, then $p + q + q + r = 5 + 9$ or $p + 2q + r = 14$.

Since $p + 2q + r = 14$ and $p + q + r = 18$, then subtracting the two
equations gives $$(p+2q+r) - (p+q+r) = 14 - 18$$ and so $q = -4$.

Solution 2:

Since $p + q + r = 18$ and $p + q = 5$, then $5 + r = 18$ and so $r = 13$.  Since $q + r = 9$ and $r = 13$, then $q + 13 = 9$ and so $q = -4$.
To find the $x$-intercept of
the line with equation $6x + y = 24$,
we set $y = 0$ to obtain $6x = 24$ which gives $x = 4$.

To find the $y$-intercept of the line
with equation $6x + y = 24$, we set
$x = 0$ to obtain $y = 24$.

Since the parabola whose equation we want to determine has only one
$x$-intercept (namely $x = 4$), we can write its equation as
$y = a(x-4)^2$ for some real number
$a$.

Additionally, we know that the $y$-intercept of the parabola is $y = 24$, so it passes through the point
$(0, 24)$.

Substituting $(x,y) = (0, 24)$ into
$y = a(x-4)^2$, we obtain $24 = a(0-4)^2$ which gives $24 = 16a$ and so $a = \frac{3}{2}$.

Therefore, an equation of the parabola is $y = \frac{3}{2}(x-4)^2$.
Since $\dfrac{1}{2w} = \dfrac{1}{3y} = \dfrac{1}{4z}$ and $\dfrac{1}{2w} + \dfrac{1}{3y} + \dfrac{1}{4z} = \dfrac{1}{24}$, then each of
$\dfrac{1}{2w}$ and $\dfrac{1}{3y}$ and $\dfrac{1}{4z}$ is equal to one-third of
the total, which gives $\dfrac{1}{2w} = \dfrac{1}{3y} = \dfrac{1}{4z} = \dfrac{1}{3} \cdot \dfrac{1}{24} = \dfrac{1}{72}$.

Therefore, $2w = 3y = 4z = 72$ and so
$w = 36$ and $y = 24$ and $z = 18$.

Thus, $w + y + z = 36 + 24 + 18 = 78$.