2025 AMC 10A Problems/Problem 20: Difference between revisions
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=Solution 1= | =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\ | 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\sqrt{11}</imath> which is the area, The altitude is simply 2x(Area/base)=<imath>\frac {5\sqrt{11}}{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. | 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. | ||
Revision as of 14:56, 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 
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 
, where 
and 
are positive integers, 
is not divisible by the square of any prime, and is relatively prime to the greatest common divisor of 
and 
. What is 
?
Solution 1
Since all the sides are multiples of 
, dialating the triangle by a factor of 
makes calculations simpler. Apply Heron's formula to solve for the length of the altitude from 
to 
. 
= 
which is the area, The altitude is simply 2x(Area/base)=
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.
