Difference between revisions of "2024 AIME I Problems/Problem 14"

Revision as of 21:05, 2 February 2024

Solution 1

Notice that $41=4^2+5^2$ , $89=5^2+8^2$ , and $80=8^2+4^2$ , let $A~(0,0,0)$ , $B~(4,5,0)$ , $C~(0,5,8)$ , and $D~(4,0,8)$ . Then the plane $BCD$ has a normal $\mathbf n:=\frac14\overrightarrow{BC}\times\overrightarrow{CD}=\frac14\begin{pmatrix}-4\\0\\8\end{pmatrix}\times\begin{pmatrix}4\\-5\\0\end{pmatrix}=\begin{pmatrix}10\\8\\5\end{pmatrix}.$ Hence, the distance from $A$ to plane $BCD$ , or the height of the tetrahedron, is $h:=\frac{\mathbf n\cdot\overrightarrow{AB}}{|\mathbf n|}=\frac{10\times4+8\times5+5\times0}{\sqrt{10^2+8^2+5^2}}=\frac{80\sqrt{21}}{63}.$ Each side of the tetrahedron has the same area due to congruency by "S-S-S", and we call it $S$ . Then by the volume formula for cones, \begin{align*} \frac13Sh&=V_{D\text-ABC}=V_{I\text-ABC}+V_{I\text-BCD}+V_{I\text-CDA}+V_{I\text-DAB}\\ &=\frac13Sr\cdot4. \end{align*} Hence, $r=\tfrac h4=\tfrac{20\sqrt{21}}{63}$ , and so the answer is $20+21+63=\boxed{104}$ .

Solution by Quantum-Phantom

@@ Line 1: / Line 1: @@
-==Solution 1==
-<asy>
-import three;
-currentprojection = orthographic(1,1,1);
-triple O = (0,0,0);
-triple A = (0,2,0);
-triple B = (0,0,1);
-triple C = (3,0,0);
-triple D = (3,2,1);
-triple E = (3,2,0);
-triple F = (0,2,1);
-triple G = (3,0,1);
-draw(A--B--C--cycle, red);
-draw(A--B--D--cycle, red);
-draw(A--C--D--cycle, red);
-draw(B--C--D--cycle, red);
-draw(E--A--O--C--cycle);
-draw(D--F--B--G--cycle);
-draw(O--B);
-draw(A--F);
-draw(E--D);
-draw(C--G);
-label("$O$", O, SW);
-label("$A$", A, NW);
-label("$B$", B, W);
-label("$C$", C, S);
-label("$D$", D, NE);
-label("$E$", E, SE);
-label("$F$", F, NW);
-label("$G$", G, NE);
-</asy>
-Inscribe tetrahedron <math>ABCD</math> in an rectangular prism as shown above.
-By the Pythagorean theorem, we note
-<cmath>OA^2 + OB^2 = AB^2 = 41,</cmath>
-<cmath>OA^2 + OC^2 = AC^2 = 80, \text{and}</cmath>
-<cmath>OB^2 + OC^2 = BC^2 = 89.</cmath>
-Solving yields <math>OA = 4, OB = 5,</math> and <math>OC = 8.</math>
-Since each face of the tetrahedron is congruent, we know the point we seek is the center of the circumsphere of <math>ABCD.</math> We know all rectangular prisms can be inscribed in a circumsphere, therefore the circumsphere of the rectangular prism is also the circumsphere of <math>ABCD.</math>
-We know that the distance from all <math>4</math> faces must be the same, so we only need to find the distance from the center to plane <math>ABC</math>.
-Let <math>O = (0,0,0), A = (4,0,0), B = (0,5,0),</math> and <math>C = (0,0,8).</math> We obtain that the plane of <math>ABC</math> can be marked as <math>\frac{x}{4} + \frac{y}{5} + \frac{z}{8} = 1,</math> or <math>10x + 8y + 5z - 40 = 0,</math> and the center of the prism is <math>(2,\frac{5}{2},4).</math>
-Using the Point-to-Plane distance formula, our distance is
-<cmath>d = \frac{|10\cdot 2 + 8\cdot \frac{5}{2} + 5\cdot 4 - 40|}{\sqrt{10^2 + 8^2 + 5^2}} = \frac{20}{\sqrt{189}} = \frac{20\sqrt{21}}{63}.</cmath>
-Our answer is <math>20 + 21 + 63 = \boxed{104}.</math>
-- spectraldragon8
 ==Solution 1==
 Notice that <math>41=4^2+5^2</math>, <math>89=5^2+8^2</math>, and <math>80=8^2+4^2</math>, let <math>A~(0,0,0)</math>, <math>B~(4,5,0)</math>, <math>C~(0,5,8)</math>, and <math>D~(4,0,8)</math>. Then the plane <math>BCD</math> has a normal

2024 AIME I (Problems • Answer Key • Resources)
Preceded by Problem 13	Followed by Problem 15
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15
All AIME Problems and Solutions

Page

Toolbox

Search

Difference between revisions of "2024 AIME I Problems/Problem 14"

Revision as of 21:05, 2 February 2024

Solution 1

See also