2025 AMC 10A Problem 20

Place the centre of the silo at the origin, with east the positive $x$-direction and north the positive $y$-direction. The diameter is $20$, so the silo is the circle $x^2 + y^2 = 100$. The two farmers are at
$$M = (-20, -15), \qquad G = (20, -g).$$

The line $MG$ has slope $\dfrac{-g - (-15)}{20 - (-20)} = \dfrac{15-g}{40}$, so its equation is $40(y + 15) = (15-g)(x+20)$, that is
$$(15-g)\,x - 40y - (300 + 20g) = 0.$$

Tangency to the silo means the distance from the origin to this line equals the radius $10$:
$$\frac{300 + 20g}{\sqrt{(15-g)^2 + 1600}} = 10$$
(the numerator is positive because $g > 0$). Dividing by $10$ and squaring,
$$(30 + 2g)^2 = (15-g)^2 + 1600,$$
$$900 + 120g + 4g^2 = g^2 - 30g + 1825,$$
$$3g^2 + 150g - 925 = 0.$$

By the quadratic formula,
$$g = \frac{-150 \pm \sqrt{150^2 + 4 \cdot 3 \cdot 925}}{6} = \frac{-150 \pm \sqrt{33600}}{6} = \frac{-150 \pm 40\sqrt{21}}{6}.$$
Since $\sqrt{33600} = \sqrt{1600 \cdot 21} = 40\sqrt{21}$ and only the $+$ sign gives $g > 0$,
$$g = \frac{40\sqrt{21} - 150}{6} = \frac{20\sqrt{21} - 75}{3} \approx 5.55.$$

The normalisation in the problem picks out this last form: with $a = 20$, $b = 21$, $c = 75$, $d = 3$ we have $\gcd(a,c) = 5$ and $\gcd(d, 5) = 1$, as required, while the unreduced $\frac{40\sqrt{21}-150}{6}$ fails because $\gcd(6, \gcd(40,150)) = 2$. Hence
$$a + b + c + d = 20 + 21 + 75 + 3 = 119.$$

The answer is $\boxed{\textbf{(A)}\ 119}$.

"Tangent to a circle" is one of the few geometric conditions that translates into a single scalar equation, namely distance from centre to line equals radius. Once the centre is at the origin the distance formula is as short as it can be, and the only unknown in it is the coordinate being asked for. This beats chasing the tangent point, which would introduce a second unknown and require an extra perpendicularity condition. The awkward-looking coprimality clause exists only to make the intended form unique, so it is worth reading it as an instruction to reduce the fraction as far as possible before adding.

Use tangent lengths. The tangent point $T$ lies between $M$ and $G$ on the line of sight, so $MG = MT + TG$. From $MO^2 = 20^2 + 15^2 = 625$ the tangent length from MacDonald is $MT = \sqrt{625 - 100} = 5\sqrt{21}$, and from $GO^2 = 400 + g^2$ the tangent length from McGregor is $TG = \sqrt{300 + g^2}$. With $MG = \sqrt{40^2 + (15-g)^2}$ this gives
$$\sqrt{1600 + (15-g)^2} = 5\sqrt{21} + \sqrt{300 + g^2},$$
which after squaring twice reduces to the same quadratic $3g^2 + 150g - 925 = 0$. It is more work, but it is the route to take if the distance-to-a-line formula is not to hand.

• Using 20 as the radius because the problem says diameter 20, which replaces 10 by 20 in the distance equation and produces g = 80/3, a value that cannot be written in the required form.
• Writing $g = \frac{40\sqrt{21}-150}{6}$ and summing $40 + 21 + 150 + 6 = 217$; the condition on $d$ is exactly what rules this form out.
• Leaving $\sqrt{33600}$ as $20\sqrt{84}$, which violates the requirement that $b$ be squarefree.