2025 AMC 12A Problems/Problem 20: Difference between revisions

Latest revision as of 23:03, 11 November 2025

Problem

The base of the pentahedron shown below is a $13 \times 8$ rectangle, and its lateral faces are two isosceles triangles with base of length $8$ and congruent sides of length $13$ , and two isosceles trapezoids with bases of length $7$ and $13$ and nonparallel sides of length $13$ .

$[asy] import graph3; size(200); real l = 13; real w = 8; real offset = (l - 7)/2; // 3 real midy = w/2; // 4 real h = 12; triple O1 = (0,0,0); triple O2 = (l,0,0); triple O3 = (l,w,0); triple O4 = (0,w,0); triple T1 = (offset, midy, h); triple T2 = (l - offset, midy, h); currentprojection=orthographic((-4,-6,3)); draw(O4--O1--O2, linewidth(1)); draw(O2--O3--O4, dashed + linewidth(1)); draw(O3--T2, dashed + linewidth(1)); draw(O1--T1, linewidth(1)); draw(O4--T1, linewidth(1)); draw(O2--T2, linewidth(1)); draw(T1--T2, linewidth(1)); label("13", (O1+O2)/2, 3*-Y); // Bottom length label("13", (O2+T2)/2, 1.5*X); label("13", (O4+T1)/2, 2*-X); label("8", (O1+O4)/2, 2*-X); // Width label("7", (T1+T2)/2, 1.5*Z); // Top length [/asy]$

What is the volume of the pentahedron?

$\textbf{(A) } 416 \qquad \textbf{(B) } 520 \qquad \textbf{(C) } 528 \qquad \textbf{(D) } 676 \qquad \textbf{(E) } 832$

Solution 1 (Split Into Three Parts)

Notice that the triangular faces have a slant height of $\sqrt{13^2-4^2}=\sqrt{153}$ and that the height is therefore $\sqrt{153-(\frac{13-7}{2})^2} = 12$ . Then we can split the pentahedron into a triangular prism and two pyramids. The pyramids each have a volume of $\frac{1}{3}(3)(8)(12) = 96$ and the prism has a volume of $\frac{1}{2}(8)(12)(7) = 336$ . Thus the answer is $336+96 \cdot 2 = \boxed{\text{(C) } 528}$

~ Shadowleafy

Solution 2

Note that the height is $12$ from the previous method \n Note that as you go up, the length and width both decrease linearly and reach $0$ at the end

So the answer is $\int_0^{12} (8 - \tfrac{2}{3}x)(13 - \tfrac{1}{2}x) dx = \boxed{528}$

Solution (3D Pythagoras)

Let $H$ be the height of the solid. By 3D Pythagoras (recursion of 2D Pythagoras), $\left( \frac{8}{2} \right)^2+\left( \frac{13-7}{2} \right)^2+H^2=13^2=4^2+3^2+H^2=5^2+H^2$ This gives the $5-12-13$ triple, so $H=12$ . Continue as in other solutions.

~imosilver

Note

It just reminds about one of old aime problem {https://artofproblemsolving.com/wiki/index.php?title=1983_AIME_Problems/Problem_11 1983 AIME Problem 11}. ~DRA777

Video Solution 1 by OmegaLearn

https://youtu.be/WwMDpnuZrJ4

Video Solution by SpreadTheLove

https://www.youtube.com/watch?v=BadjXLma84M

@@ Line 32: / Line 32: @@
 What is the volume of the pentahedron?
+<imath>\textbf{(A) } 416 \qquad \textbf{(B) } 520 \qquad \textbf{(C) }  528  \qquad  \textbf{(D) } 676 \qquad \textbf{(E) } 832</imath>
 == Solution 1 (Split Into Three Parts) ==
@@ Line 45: / Line 47: @@
 So the answer is <imath>\int_0^{12} (8 - \tfrac{2}{3}x)(13 - \tfrac{1}{2}x) dx = \boxed{528}</imath>
-== Solution 3 ==
+== Solution (3D Pythagoras) ==
+Let <imath>H</imath> be the height of the solid. By 3D Pythagoras (recursion of 2D Pythagoras),
+<cmath>\left( \frac{8}{2} \right)^2+\left( \frac{13-7}{2} \right)^2+H^2=13^2=4^2+3^2+H^2=5^2+H^2</cmath>
+This gives the <imath>5-12-13</imath> triple, so <imath>H=12</imath>. Continue as in other solutions.
+~imosilver
+== Note ==
 It just reminds about one of old aime problem {https://artofproblemsolving.com/wiki/index.php?title=1983_AIME_Problems/Problem_11 1983 AIME Problem 11}.
 ~DRA777

2025 AMC 12A (Problems • Answer Key • Resources)
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