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2025 AMC 10A Problems/Problem 20: Difference between revisions

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Now we find the hypotenuse of <imath>ABF</imath> in terms of x using the Pythagorean theorem. <imath>AF^2=15^2+(40+x)^2</imath>. Which simplifies to <imath>AF^2=225+1600+80x+x^2=1825+80x+x^2</imath>  So <imath>AF=\sqrt{x^2+80x+1825}</imath>
Now we find the hypotenuse of <imath>ABF</imath> in terms of x using the Pythagorean theorem. <imath>AF^2=15^2+(40+x)^2</imath>. Which simplifies to <imath>AF^2=225+1600+80x+x^2=1825+80x+x^2</imath>  So <imath>AF=\sqrt{x^2+80x+1825}</imath>
Plugging back in we get <imath>\frac{10}{20+x}=\frac{15}{\sqrt{x^2+80x+1825}}</imath>. Now we can begin to break this down by multiplying both sidse by both denominators. <imath>10(\sqrt{x^2+80x+1825})=15(20+x)</imath> Dividing both sides by <imath>5</imath> then squaring yields, <imath>4x^2+320x+7300=9x^2+360x+3600</imath>

Revision as of 15:29, 6 November 2025

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 light 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

Let the silo center be $O$, let the point MacDnoald 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 it 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 sidse by both denominators. $10(\sqrt{x^2+80x+1825})=15(20+x)$ Dividing both sides by $5$ then squaring yields, $4x^2+320x+7300=9x^2+360x+3600$

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