Revision as of 14:54, 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

Since all the sides are multiples of $5$ , dialating the triangle by a factor of $\frac {1}{5}$ makes calculations simpler. Apply Heron's formula to solve for the length of the altitude from $C$ to $AB$ . $\sqrt(20(4)(5)(11))$ = $20\sqrt11$ which is the area, The altitude is simply 2x(Area/base)= $5\sqrt11/2$

Now with 2 sides of a right triangle, we can solve for the third side BD using the Pythagorean theorem. BD^2=BC^2-CD^2= 9^2-(5sqrt11/2)^2 = 81-275/4=)324-275+/4 = 49/4. The side BD is 7/2.

The Angle Bisector Theorem on this triangle specifies that the ratio between DP and PC is (7/2)/9=7/18. Therefore DP=7/25(5sqrt11/2)=7sqrt11/10.

Applying the Pythagorean Theorem once again to find BP we get (7/2)^2+(7sqrt11/10)^2=441/25. Therefore BP=21/5. Multiplying by the original factor of 5, BP=21.

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

Revision as of 14:54, 6 November 2025

Solution 1

@@ Line 4: / Line 4: @@
 =Solution 1=
+Since all the sides are multiples of <imath>5</imath>, dialating the triangle by a factor of <imath>\frac {1}{5}</imath> makes calculations simpler. Apply Heron's formula to solve for the length of the altitude from <imath>C</imath> to <imath>AB</imath>. <imath>\sqrt(20(4)(5)(11))</imath>= <imath>20\sqrt11</imath> which is the area, The altitude is simply 2x(Area/base)=<imath>5\sqrt11/2</imath>
+Now with 2 sides of a right triangle, we can solve for the third side BD using the Pythagorean theorem. BD^2=BC^2-CD^2= 9^2-(5sqrt11/2)^2 = 81-275/4=)324-275+/4  = 49/4. The side BD is 7/2.
+The Angle Bisector Theorem on this triangle specifies that the ratio between DP and PC is (7/2)/9=7/18. Therefore DP=7/25(5sqrt11/2)=7sqrt11/10.
+Applying the Pythagorean Theorem once again to find BP we get (7/2)^2+(7sqrt11/10)^2=441/25. Therefore BP=21/5. Multiplying by the original factor of 5, BP=21.