2025 AMC 10A Problems/Problem 20

A silo (right circular cylinder) with diameter 20 meters stands in a field. MacDonald is located 20 meters west and 15 meters south of the center of the silo. McGregor is located 20 meters east and $g > 0$ meters south of the center of the silo. The line of sight between MacDonald and McGregor is tangent to the silo. The value of g can be written as $\frac{a\sqrt{b}-c}{d}$ , where $a,b,c,$ and $d$ are positive integers, $b$ is not divisible by the square of any prime, and $d$ is relatively prime to the greatest common divisor of $a$ and $c$ . What is $a+b+c+d$ ?

$\textbf{(A) } 118 \qquad\textbf{(B) } 119 \qquad\textbf{(C) } 120 \qquad\textbf{(D) } 121 \qquad\textbf{(E) } 122$

Solution 1

import olympiad;
size(350);

// scale: 1 unit = 1 meter
pair O = (0,0);
real R = 10;

// Points per problem statement
pair B = (-20,0);
pair A = (-20,-15);
pair C = (20,0);

// exact g and point D
real g = (20*sqrt(21)-75)/3;      // exact expression used for positioning
pair D = (20, -g);

// circle (silo)
draw(circle(O,R), heavyblue);

// ground line (BC, extended)
draw((-60,0)--(60,0), gray+0.6);

// tangent line AD and its extension
path lineAD = A--D;
draw(lineAD, black);

// tangent point T = intersection of line AD with circle
pair[] inters = intersectionpoints(lineAD, circle(O,R));
pair T;
if (length(inters) > 0) {
  // choose the intersection closer to A (tangent gives one point but numerical returns one)
  T = inters[0];
} else {
  T = projection(O, A--D);
}

// intersection with ground line (y=0)
pair F = intersectionpoint(lineAD, (-100,0)--(100,0));

// draw segments and helper lines
draw(A--B, dashed+gray);   // vertical from A to B
draw(B--O, dashed+gray);   // B to center O

// mark points
dot(O); label("$O$", O, NE);
dot(A); label("$A$", A, SW);
dot(B); label("$B$", B, N);
dot(C); label("$C$", C, N);
dot(D); label("$D$", D, SE);
dot(T); label("$T$", T, NE);
dot(F); label("$F$", F, S);

// right-angle marker at B for triangle A-B-O
draw(rightanglemark(A,B,O,8));

// annotate radius and small helper
draw(O--T, medium+darkblue); label("$10$", (O+T)/2 + (0.7,0.2));

// annotate x = CF
real xdist = F.x - C.x;
pair xm = (C + F)/2;
draw(C--F, Arrow(size=5));
label("$x =", xm + (0,-0.5));
label(string(xdist), xm + (0,-1.0)); // numeric for quick reference

// show algebraic g near D (use LaTeX in label)
label("$g=<br />dfrac{20<br />sqrt{21}-75}{3}$", D + (6,0));

// small view adjustments
xlimits(-40, 60);
ylimits(-20, 6);
shipout(bbox(5mm));
 (Error making remote request. Unknown error_msg)

Let the silo center be $O$ , let the point MacDonald is situated at be $A$ , and let the point $20$ meters west of the silo center be $B$ . $ABO$ is then a right triangle with side lengths $15, 20,$ and $25$ .

Let the point $20$ meters east of the silo center be $C$ , and let the point McGregor is at be $D$ with $CD=g>0$ . Also let $AD$ be tangent to circle $O$ at $T$ .

Extend $BC$ and $AD$ to meet at point $F$ . This creates $3$ similar triangles, $ABF\sim DCF \sim OEF$ . Let the distance between point $C$ and $F$ be $x$ . The similarity ratio between triangles $ABF$ and $DCF$ is then $\frac{longer\;leg}{shorter\;leg} = \frac{40+x}{15} = \frac{x}{g}$

This is currently unsolvable so we bring in triangle $OEF$ . The hypotenuse of triangle $OEF$ is $OF=20+x$ and its shorter leg is the radius of the silo $=10$ . We can then establish a second similarity relationship between triangles $OEF$ and $ABF$ with $\frac{shorter\; leg}{hypotenuse}=\frac{10}{20+x}=\frac{15}{AF}$

Now we find the hypotenuse of $ABF$ in terms of x using the Pythagorean theorem. $AF^2=15^2+(40+x)^2$ . Which simplifies to $AF^2=225+1600+80x+x^2=1825+80x+x^2$ So $AF=\sqrt{x^2+80x+1825}$

Plugging back in we get $\frac{10}{20+x}=\frac{15}{\sqrt{x^2+80x+1825}}$ . Now we can begin to break this down by multiplying both sides by both denominators. $10(\sqrt{x^2+80x+1825})=15(20+x)$ Dividing both sides by $5$ then squaring yields, $4x^2+320x+7300=9x^2+360x+3600$ This furthermore simplifies to $5x^2+40x-3700=0$ At which point we can divide off a $5$ and then apply the quadratic formula on $x^2+8x-740=0$ which we take the positive root of.

$\frac{-8+\sqrt{64+2960=3024=144*21}}{2}$

Simplifying yields that $x=6\sqrt{21}-4$

Then to solve for $g$ we simply plug $6\sqrt{21}-4$ back into the first similarity ratio to get $\frac{36+6\sqrt{21}}{15}=\frac{6\sqrt{21}-4}{g}$

Multiply both sides by $15g$ and dividing by $36+6\sqrt{21}$ will let us solve for $g=\frac{15(6\sqrt{21}-4)}{36+6\sqrt{21}}$ and after rationalizing the denominator we get $\frac{20\sqrt{21}-75}{3}$ . $20+21+75+3=\boxed{\textbf{(A)}~119}$

$\sim$ Neoronean

Video Solution (In 5 Mins)

https://youtu.be/1Xc1blWnWSo?si=pBd85mIZwYjbOEdG ~ Pi Academy

Video Solution

https://youtu.be/gWSZeCKrOfU

~MK

2025 AMC 10A (Problems • Answer Key • Resources)
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2025 AMC 10A Problems/Problem 20

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Solution 1

Video Solution (In 5 Mins)

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