AMC10 2025 A

AMC10 2025 A · Q20

AMC10 2025 A · Q20. It mainly tests Circle theorems, Coordinate geometry.

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 $a+b+c+d$?

一个直径 $20$ 米的筒仓（右圆柱体）矗立在田野中。MacDonald 位于筒仓中心西 $20$ 米、南 $15$ 米处。McGregor 位于筒仓中心东 $20$ 米、南 $g > 0$ 米处。MacDonald 和 McGregor 之间的视线与筒仓相切。$g$ 的值为 $\frac{a\sqrt{b}-c}{d}$，其中 $a,b,c,d$ 为正整数，$b$ 无任何质数的平方因子，$d$ 与 $a$ 和 $c$ 的最大公因数互质。求 $a+b+c+d$？

(A) 119 119

(B) 120 120

(D) 122 122

(E) 123 123

Answer

Correct choice: (A)

正确答案：（A）

Solution

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 $E$. Extend $BC$ and $AD$ to meet at point $F$. This creates $3$ similar triangles, $\triangle ABF\sim \triangle DCF \sim \triangle OEF$. 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. \[x = \frac{-8+\sqrt{64+2960}}{2} = \frac{-8+\sqrt{3024}}{2} =\frac{-8+\sqrt{144 \cdot 21}}{2}.\] 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 $\frac{20\sqrt{21}-75}{3}$. $20+21+75+3=\boxed{\textbf{(A)}~119}$

设筒仓中心为 $O$，MacDonald 位置为 $A$，筒仓中心西 $20$ 米点为 $B$。则 $ABO$ 为直角三角形，边长 $15,20,25$。设筒仓中心东 $20$ 米点为 $C$，McGregor 位置为 $D$，$CD=g>0$。设 $AD$ 在 $E$ 点与圆 $O$ 相切。延长 $BC$ 和 $AD$ 相交于点 $F$。这产生 $3$ 个相似三角形，$\triangle ABF\sim \triangle DCF \sim \triangle OEF$。设点 $C$ 到 $F$ 距离为 $x$。三角形 $ABF$ 和 $DCF$ 的相似比为 $\frac{40+x}{15} = \frac{x}{g}$。目前无法求解，因此引入三角形 $OEF$。$OEF$ 的斜边为 $OF=20+x$，较短腿为筒仓半径 $=10$。于是建立三角形 $OEF$ 和 $ABF$ 的第二相似关系：$\frac{10}{20+x}=\frac{15}{AF}$。现在用勾股定理求 $ABF$ 的斜边关于 $x$：$AF^2=15^2+(40+x)^2=225+1600+80x+x^2=1825+80x+x^2$，所以 $AF=\sqrt{x^2+80x+1825}$。代入得 $\frac{10}{20+x}=\frac{15}{\sqrt{x^2+80x+1825}}$。两边乘以分母：$10(\sqrt{x^2+80x+1825})=15(20+x)$。两边除以 $5$ 后平方：$4x^2+320x+7300=9x^2+360x+3600$。进一步化简为 $5x^2+40x-3700=0$。除以 $5$ 得 $x^2+8x-740=0$，取正根。 \[x = \frac{-8+\sqrt{64+2960}}{2} = \frac{-8+\sqrt{3024}}{2} =\frac{-8+\sqrt{144 \cdot 21}}{2}.\] 化简得 $x=6\sqrt{21}-4$。然后将 $6\sqrt{21}-4$ 代入第一相似比：$\frac{36+6\sqrt{21}}{15}=\frac{6\sqrt{21}-4}{g}$。两边乘以 $15g$ 并除以 $36+6\sqrt{21}$ 得 $g=\frac{15(6\sqrt{21}-4)}{36+6\sqrt{21}}$，分母有理化后得 $\frac{20\sqrt{21}-75}{3}$。$20+21+75+3=\boxed{\textbf{(A)}~119}$。

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