Difference between revisions of "2024 AIME I Problems/Problem 14"
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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}. | 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}. | ||
\end{equation*} | \end{equation*} | ||
− | Each side of the tetrahedron has the same area due to congruency by "S-S-S", and we call it | + | Each side of the tetrahedron has the same area due to congruency by "S-S-S", and we call it |
\begin{align*} | \begin{align*} | ||
\frac13Sh&=V_{D\text-ABC}=V_{I\text-ABC}+V_{I\text-BCD}+V_{I\text-CDA}+V_{I\text-DAB}\ | \frac13Sh&=V_{D\text-ABC}=V_{I\text-ABC}+V_{I\text-BCD}+V_{I\text-CDA}+V_{I\text-DAB}\ | ||
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Our answer is <math>20 + 21 + 63 = \boxed{104}.</math> | Our answer is <math>20 + 21 + 63 = \boxed{104}.</math> | ||
− | - spectraldragon8 | + | - [https://artofproblemsolving.com/wiki/index.php/User:Spectraldragon8 spectraldragon8] |
==Solution 3(Formula Abuse)== | ==Solution 3(Formula Abuse)== | ||
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~AtharvNaphade | ~AtharvNaphade | ||
− | ==Solution | + | ==Solution 4== |
Let <math>AH</math> be perpendicular to <math>BCD</math> that meets this plane at point <math>H</math>. | Let <math>AH</math> be perpendicular to <math>BCD</math> that meets this plane at point <math>H</math>. | ||
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~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com) | ~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com) | ||
− | ==Solution | + | ==Solution 5(A quicker method to compute the height from <math>A</math> to plane <math>BCD</math>)== |
We put the solid to a 3-d coordinate system. Let <math>B = \left( 0, 0, 0 \right)</math>, <math>D = \left( \sqrt{80}, 0, 0 \right)</math>. | We put the solid to a 3-d coordinate system. Let <math>B = \left( 0, 0, 0 \right)</math>, <math>D = \left( \sqrt{80}, 0, 0 \right)</math>. | ||
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Because <math>AB = \sqrt{89}</math>, we have | Because <math>AB = \sqrt{89}</math>, we have | ||
− | + | <math>x^2 + y^2 + z^2 = 89 </math> | |
− | x^2 + y^2 + z^2 = 89 | ||
− | |||
Because <math>AD = \sqrt{41}</math>, we have | Because <math>AD = \sqrt{41}</math>, we have | ||
− | + | <cmath> | |
\left( x - \sqrt{80} \right)^2 + y^2 + z^2 = 41 \hspace{1cm} (2) | \left( x - \sqrt{80} \right)^2 + y^2 + z^2 = 41 \hspace{1cm} (2) | ||
− | + | </cmath> | |
Because <math>AC = \sqrt{80}</math>, we have | Because <math>AC = \sqrt{80}</math>, we have | ||
− | + | <cmath> | |
\left( x - \frac{4}{\sqrt{5}} \right)^2 + \left( y - \frac{3 \sqrt{21}}{\sqrt{5}} \right)^2 | \left( x - \frac{4}{\sqrt{5}} \right)^2 + \left( y - \frac{3 \sqrt{21}}{\sqrt{5}} \right)^2 | ||
+ z^2 = 80 \hspace{1cm} (3) | + z^2 = 80 \hspace{1cm} (3) | ||
− | + | </cmath> | |
Now, we compute <math>x</math>, <math>y</math> and <math>z</math>. | Now, we compute <math>x</math>, <math>y</math> and <math>z</math>. | ||
Taking <math>(1)-(2)</math>, we get | Taking <math>(1)-(2)</math>, we get | ||
− | + | <cmath> | |
2 \sqrt{80} x = 128 . | 2 \sqrt{80} x = 128 . | ||
− | + | </cmath> | |
Thus, <math>x = \frac{16}{\sqrt{5}}</math>. | Thus, <math>x = \frac{16}{\sqrt{5}}</math>. | ||
Taking <math>(1) - (3)</math>, we get | Taking <math>(1) - (3)</math>, we get | ||
− | + | <cmath> | |
2 \cdot \frac{4}{\sqrt{5}} x + 2 \cdot \frac{3 \sqrt{21}}{\sqrt{5}} y | 2 \cdot \frac{4}{\sqrt{5}} x + 2 \cdot \frac{3 \sqrt{21}}{\sqrt{5}} y | ||
= 50 . | = 50 . | ||
− | + | </cmath> | |
Thus, <math>y = \frac{61}{3 \sqrt{5 \cdot 21}}</math>. | Thus, <math>y = \frac{61}{3 \sqrt{5 \cdot 21}}</math>. | ||
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~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com) | ~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com) | ||
+ | ~numerophile (formatting edits) | ||
+ | |||
+ | ==Solution 6 (Different Perspective)== | ||
+ | Consider the following construction of the tetrahedron. Place <math>AB</math> on the floor. Construct an isosceles vertical triangle with <math>AB</math> as its base and <math>M</math> as the top vertex. Place <math>CD</math> on the top vertex parallel to the ground with midpoint <math>M.</math> Observe that <math>CD</math> can rotate about its midpoint. At a certain angle, we observe that the lengths satisfy those given in the problem. If we project <math>AB</math> onto the plane of <math>CD</math>, let the minor angle <math>\theta</math> be this discrepancy. | ||
+ | |||
+ | By Median formula or Stewart's theorem, <math>AM = \frac{1}{2}\sqrt{2AC^2 + 2AD^2 - CD^2} = \frac{3\sqrt{33}}{2}.</math> Consequently the area of <math>\triangle AMB</math> is <math>\frac{\sqrt{41}}{2} \left (\sqrt{(\frac{3\sqrt{33}}{2})^2 - (\frac{\sqrt{41}}{2})^2} \right ) = 4\sqrt{41}.</math> Note the altitude <math>8</math> is also the distance between the parallel planes containing <math>AB</math> and <math>CD.</math> | ||
+ | |||
+ | By Distance Formula, | ||
+ | <cmath>\begin{align*} | ||
+ | (\frac{\sqrt{41}}{2} - \frac{1}{2}CD \cos{\theta})^2 + (\frac{1}{2}CD \sin{\theta}^2) + (8)^2 &= AC^2 = 80 \ | ||
+ | (\frac{\sqrt{41}}{2} + \frac{1}{2}CD \cos{\theta})^2 + (\frac{1}{2}CD \sin{\theta}^2) + (8)^2 &= AD^2 = 89 \ | ||
+ | \implies CD \cos{\theta} \sqrt{41} &= 9 \ | ||
+ | \sin{\theta} &= \sqrt{1 - (\frac{9}{41})^2} = \frac{40}{41}. | ||
+ | \end{align*}</cmath> | ||
+ | Then the volume of the tetrahedron is given by <math>\frac{1}{3} [AMB] \cdot CD \sin{\theta} = \frac{160}{3}.</math> | ||
+ | |||
+ | The volume of the tetrahedron can also be segmented into four smaller tetrahedrons from <math>I</math> w.r.t each of the faces. If <math>r</math> is the inradius, i.e the distance to the faces, then <math>\frac{1}{3} r([ABC] + [ABD] + [ACD] + [BCD])</math> must the volume. Each face has the same area by SSS congruence, and by Heron's it is <math>\frac{1}{4}\sqrt{(a + b + c)(a + b - c)(c + (a-b))(c -(a - b))} = 6\sqrt{21}.</math> | ||
+ | |||
+ | Therefore the answer is, <math>\dfrac{3 \frac{160}{3}}{24 \sqrt{21}} = \frac{20\sqrt{21}}{63} \implies \boxed{104}.</math> | ||
+ | |||
+ | ~Aaryabhatta1 | ||
+ | |||
+ | ==Solution 7 (simplest way to calculate [ABCD])== | ||
+ | After finding that <math>ABCD</math> can be inscribed in a <math>4\times5\times8</math> box, note that <math>[ABCD]=\mathbf{box}-\frac{4}{3}(4)(5)(8)=\frac{160}{3}</math>, where the third term describes the area of 4 congruent right tetrahedrons whose right angle is at the apex and whose apex lengths are 4, 5, and 8. Then proceed as in Solution 3. | ||
+ | |||
+ | ~clarkculus | ||
+ | |||
+ | ==Solution 8 (fast)== | ||
+ | Observe that <math>ABCD</math> can be inscribed in a <math>4\times5\times8</math> box, which gives the coordinates of the points: we have <math>A=(8,0,0)</math>, <math>B=(8,5,4)</math>, <math>C=(0,0,4)</math> and <math>D=(0,5,0)</math>. Recall the formula <math>V=\frac{1}{3}RS</math>, where <math>V</math> is the volume of the tetrahedron, <math>S</math> is the surface area, and <math>R</math> is the insphere radius. Using the fact that the faces of <math>ABCD</math> have equal area and that <math>V=\frac{1}{3}[ACD] \times h</math>, where <math>h</math> is the distance from plane <math>ACD</math> to <math>B</math>, we can rearrange to get <math>R=\frac{h}{4}</math>. We compute that the the equation of plane <math>ACD</math> is <math>5x+8y+10z-40=0</math>, and applying point-to-plane gives <math>h=\frac{\vert 5(8)+8(5)+10(4)-40 \vert}{\sqrt{5^2+8^2+10^2}}=\frac{80}{\sqrt{189}} \implies R=\frac{20\sqrt{21}}{63} \implies \boxed{104}</math>. | ||
+ | |||
+ | ~bomberdoodles | ||
+ | |||
+ | ==Video Solution== | ||
+ | https://youtu.be/tq6lraC5prQ?si=5q1NX80POeR949qs | ||
+ | |||
+ | ~MathProblemSolvingSkills.com | ||
+ | |||
+ | |||
==Video Solution 1 by OmegaLearn.org== | ==Video Solution 1 by OmegaLearn.org== | ||
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==See also== | ==See also== | ||
{{AIME box|year=2024|n=I|num-b=13|num-a=15}} | {{AIME box|year=2024|n=I|num-b=13|num-a=15}} | ||
+ | |||
+ | [[Category:Intermediate Geometry Problems]] | ||
+ | [[Category:3D Geometry Problems]] | ||
{{MAA Notice}} | {{MAA Notice}} |
Revision as of 19:25, 17 November 2024
Contents
[hide]- 1 Problem
- 2 Solution 1
- 3 Solution 2
- 4 Solution 3(Formula Abuse)
- 5 Solution 4
- 6 Solution 5(A quicker method to compute the height from to plane )
- 7 Solution 6 (Different Perspective)
- 8 Solution 7 (simplest way to calculate [ABCD])
- 9 Solution 8 (fast)
- 10 Video Solution
- 11 Video Solution 1 by OmegaLearn.org
- 12 Video Solution 2
- 13 See also
Problem
Let be a tetrahedron such that , , and . There exists a point inside the tetrahedron such that the distances from to each of the faces of the tetrahedron are all equal. This distance can be written in the form , where , , and are positive integers, and are relatively prime, and is not divisible by the square of any prime. Find .
Solution 1
Notice that
Solution by Quantum-Phantom
Solution 2
Inscribe tetrahedron in an rectangular prism as shown above.
By the Pythagorean theorem, we note
Solving yields and
Since each face of the tetrahedron is congruent, we know the point we seek is the center of the circumsphere of We know all rectangular prisms can be inscribed in a circumsphere, therefore the circumsphere of the rectangular prism is also the circumsphere of
We know that the distance from all faces must be the same, so we only need to find the distance from the center to plane .
Let and We obtain that the plane of can be marked as or and the center of the prism is
Using the Point-to-Plane distance formula, our distance is
Our answer is
Solution 3(Formula Abuse)
We use the formula for the volume of iscoceles tetrahedron.
Note that all faces have equal area due to equal side lengths. By Law of Cosines, we find .
From this, we find and can find the area of as
Let be the distance we want to find. By taking the sum of (equal) volumes We have Plugging in and simplifying, we get for an answer of
~AtharvNaphade
Solution 4
Let be perpendicular to that meets this plane at point . Let , , and be heights to lines , , and with feet , , and , respectively.
We notice that all faces are congruent. Following from Heron's formula, the area of each face, denoted as , is .
Hence, by using this area, we can compute , and . We have , , and .
Because , we have . Recall that . Hence, . Hence, .
Analogously, and .
We introduce a function for that is equal to 1 (resp. -1) if point and the opposite vertex of side are on the same side (resp. opposite sides) of side .
The area of is
Denote .
The above equation can be organized as
This can be further reorganized as
Taking squares on both sides and reorganizing terms, we get
Taking squares on both sides and reorganizing terms, we get
Taking squares on both sides, we finally get
Now, we plug this solution to Equation (1). We can see that , . This indicates that is out of . To be specific, and are on opposite sides of , and are on the same side of , and and are on the same side of .
Now, we compute the volume of the tetrahedron , denoted as . We have .
Denote by the inradius of the inscribed sphere in .
Denote by the incenter.
Thus, the volume of can be alternatively calculated as
From our two methods to compute the volume of and equating them, we get
Therefore, the answer is .
~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com)
Solution 5(A quicker method to compute the height from to plane )
We put the solid to a 3-d coordinate system. Let , . We put on the plane. Now ,we compute the coordinates of .
Applying the law of cosines on , we get . Thus, . Thus, .
Denote with .
Because , we have
Because , we have
Because , we have
Now, we compute , and .
Taking , we get
Thus, .
Taking , we get
Thus, .
Plugging and into Equation (1), we get .
~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com) ~numerophile (formatting edits)
Solution 6 (Different Perspective)
Consider the following construction of the tetrahedron. Place on the floor. Construct an isosceles vertical triangle with as its base and as the top vertex. Place on the top vertex parallel to the ground with midpoint Observe that can rotate about its midpoint. At a certain angle, we observe that the lengths satisfy those given in the problem. If we project onto the plane of , let the minor angle be this discrepancy.
By Median formula or Stewart's theorem, Consequently the area of is Note the altitude is also the distance between the parallel planes containing and
By Distance Formula, Then the volume of the tetrahedron is given by
The volume of the tetrahedron can also be segmented into four smaller tetrahedrons from w.r.t each of the faces. If is the inradius, i.e the distance to the faces, then must the volume. Each face has the same area by SSS congruence, and by Heron's it is
Therefore the answer is,
~Aaryabhatta1
Solution 7 (simplest way to calculate [ABCD])
After finding that can be inscribed in a box, note that , where the third term describes the area of 4 congruent right tetrahedrons whose right angle is at the apex and whose apex lengths are 4, 5, and 8. Then proceed as in Solution 3.
~clarkculus
Solution 8 (fast)
Observe that can be inscribed in a box, which gives the coordinates of the points: we have , , and . Recall the formula , where is the volume of the tetrahedron, is the surface area, and is the insphere radius. Using the fact that the faces of have equal area and that , where is the distance from plane to , we can rearrange to get . We compute that the the equation of plane is , and applying point-to-plane gives .
~bomberdoodles
Video Solution
https://youtu.be/tq6lraC5prQ?si=5q1NX80POeR949qs
~MathProblemSolvingSkills.com
Video Solution 1 by OmegaLearn.org
Video Solution 2
~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com)
See also
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.