2025 AMC 10A Problem 20

The only fact given about the line of sight is that it is tangent to the silo, which means it just touches the circle. For a circle, "a line just touches it" means the line's distance from the center equals the radius. So put the center at the origin and use that.

1. Set up coordinates. Put the center of the silo at $(0, 0)$, with east as the positive $x$-direction and north as the positive $y$-direction. The diameter is $20$, so the radius is $10$. The two farmers are at
$$M = (-20, -15) \quad \text{(MacDonald)}, \qquad G = (20, -g) \quad \text{(McGregor)}.$$

2. Find the line through $M$ and $G$. Its slope is
$$\frac{-g - (-15)}{20 - (-20)} = \frac{15 - g}{40}.$$
Through $M$, the line is $y + 15 = \frac{15-g}{40}(x + 20)$. Multiply both sides by $40$:
$$40y + 600 = (15 - g)x + 300 - 20g.$$
Move everything to one side:
$$(15 - g)x - 40y - (300 + 20g) = 0.$$
(Check with $M$: $(15-g)(-20) + 600 - 300 - 20g = -300 + 20g + 300 - 20g = 0$. Good.)

3. Distance from the center to the line. For a line $Ax + By + C = 0$, the distance from $(0, 0)$ is $\dfrac{|C|}{\sqrt{A^2 + B^2}}$. Here that is
$$\frac{300 + 20g}{\sqrt{(15 - g)^2 + 40^2}}.$$
(The top is positive because $g > 0$.) Set it equal to the radius:
$$\frac{300 + 20g}{\sqrt{(15-g)^2 + 1600}} = 10.$$

4. Solve for $g$. Divide the top by $10$ and multiply across:
$$30 + 2g = \sqrt{(15-g)^2 + 1600}.$$
Square both sides:
$$900 + 120g + 4g^2 = 225 - 30g + g^2 + 1600.$$
Collect everything on the left:
$$3g^2 + 150g - 925 = 0.$$

5. Quadratic formula.
$$g = \frac{-150 \pm \sqrt{150^2 + 4 \cdot 3 \cdot 925}}{2 \cdot 3} = \frac{-150 \pm \sqrt{22500 + 11100}}{6} = \frac{-150 \pm \sqrt{33600}}{6}.$$
Since $33600 = 1600 \times 21$, we get $\sqrt{33600} = 40\sqrt{21}$. We need $g > 0$, so take the plus sign:
$$g = \frac{40\sqrt{21} - 150}{6}.$$

6. Put it in the required form. The problem says $d$ must share no factor with the greatest common divisor of $a$ and $c$. Here $a = 40$, $c = 150$ have greatest common divisor $10$, and $d = 6$ shares the factor $2$ with it, so this form is not allowed. Divide top and bottom by $2$:
$$g = \frac{20\sqrt{21} - 75}{3} \approx \frac{91.65 - 75}{3} \approx 5.55.$$
Now $a = 20$, $c = 75$ have greatest common divisor $5$, which shares nothing with $3$. Also $21 = 3 \times 7$ has no square factor. So $a = 20$, $b = 21$, $c = 75$, $d = 3$.

1. Add: $20 + 21 + 75 + 3 = 119$.

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

"Tangent to a circle" turns into one equation: the distance from the center to the line equals the radius. Putting the center at the origin makes that distance formula as short as possible, and the only unknown left in it is $g$. Next time a line touches a circle, reach for this before trying to find the touching point, which would add a second unknown.

If you don't know the distance-to-a-line formula, use tangent lengths and Pythagoras. Let $T$ be the point where the line of sight touches the silo, and $O$ the center. The radius $OT$ meets the tangent line at a right angle.

1. From MacDonald: $MO^2 = 20^2 + 15^2 = 625$, so $MT = \sqrt{625 - 10^2} = \sqrt{525} = 5\sqrt{21}$.
2. From McGregor: $GO^2 = 20^2 + g^2$, so $GT = \sqrt{400 + g^2 - 100} = \sqrt{300 + g^2}$.
3. $T$ lies between the two farmers, so $MG = MT + GT$. Also $MG^2 = 40^2 + (15 - g)^2$. So
$$\sqrt{1600 + (15 - g)^2} = 5\sqrt{21} + \sqrt{300 + g^2}.$$
4. Square both sides: $1825 - 30g + g^2 = 825 + g^2 + 10\sqrt{21}\sqrt{300 + g^2}$, which simplifies to $100 - 3g = \sqrt{21}\sqrt{300 + g^2}$.
5. Square again: $10000 - 600g + 9g^2 = 6300 + 21g^2$, that is $12g^2 + 600g - 3700 = 0$. Dividing by $4$ gives the same equation $3g^2 + 150g - 925 = 0$ as before.

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 $\boxed{\textbf{(A)}\ 119}$.