Difference between revisions of "DVI exam"
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<i><b>Solution</b></i> | <i><b>Solution</b></i> | ||
| − | Let us consider the uniform triangular prism <math>ABCA'B'C'.</math> Let <math>M</math> be the midpoint of <math>AB, M'</math> be the midpoint of <math>A'B', K</math> be the midpoint of <math>CM, L</math> be the midpoint of <math>C'M'.</math> | + | Let us consider the uniform triangular prism <math>ABCA'B'C'.</math> Let <math>M</math> be the midpoint of <math>AB, M'</math> be the midpoint of <math>A'B', K</math> be the midpoint of <math>CM, L</math> be the midpoint of <math>C'M', 2 FG = AB.</math> |
The area <math>[KED]</math> of <math>\triangle KED</math> in the sum with the areas of triangles <math>[KEL], [EDM'], [KDM]</math> is half the area of rectangle <math>CC'M'M,</math> so | The area <math>[KED]</math> of <math>\triangle KED</math> in the sum with the areas of triangles <math>[KEL], [EDM'], [KDM]</math> is half the area of rectangle <math>CC'M'M,</math> so | ||
| − | <cmath> \frac {[KED]}{[CC'M'M]} = \frac {1}{2} - \frac {1}{12}-\frac {1}{8} = \frac {5}{24}.</cmath> | + | <cmath> \frac {[KED]}{[CC'M'M]} = \frac {1}{2} - \frac {1}{12}- \frac {1}{12}-\frac {1}{8} = \frac {5}{24}.</cmath> |
| − | <cmath>FG \perp ED | + | <cmath>FG \perp ED.</cmath> Denote the distance between these lines <math>h.</math> The volume of the tetrahedron is <math>U = \frac {ED \cdot h \cdot FG}{6}.</math> |
| − | <cmath>\frac {ED \cdot | + | <cmath>\frac {ED \cdot h}{2} = \frac {5}{24} \cdot CC' \cdot CM \implies U =\frac{5}{24 \cdot 3} \cdot CC' \cdot CM \cdot FG.</cmath> |
| − | The volume of the prism is <math>V = \frac{CM \cdot AB}{2} \cdot | + | The volume of the prism is <math>V = \frac{CM \cdot AB}{2} \cdot CC' =CC' \cdot CM \cdot FG = 72.</math> |
| − | <cmath>\frac {U}{V} = \frac {\frac {5}{72} CC' \cdot CM \cdot FG}{\ | + | <cmath>\frac {U}{V} = \frac {\frac {5}{72} CC' \cdot CM \cdot FG}{CC' \cdot CM \cdot FG} = \frac {5}{24} \implies U = 5.</cmath> |
An arbitrary prism is obtained from a regular one as a result of an affine transformation. | An arbitrary prism is obtained from a regular one as a result of an affine transformation. | ||
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All points on the tetrahedron are defined affinely, which means that the volume ratio will be preserved. | All points on the tetrahedron are defined affinely, which means that the volume ratio will be preserved. | ||
| − | <i><b>Answer: | + | <i><b>Answer: 5.</b></i> |
==2022 222 problem 7== | ==2022 222 problem 7== | ||
Revision as of 05:56, 28 January 2024
2022 221 problem 7
The volume of a triangular prism 
with base 
and side edges 
is equal to 
Find the volume of the tetrahedron 
where 
is the centroid of the face 
is the point of intersection of the medians of 
is the midpoint of the edge 
and 
is the midpoint of the edge 
Solution
Let us consider the uniform triangular prism 
Let 
be the midpoint of 
be the midpoint of 
be the midpoint of 
be the midpoint of 
The area ![$[KED]$](http://latex.artofproblemsolving.com/f/b/a/fba57621447876aa5fd27d1e0dcafd32131a1bbd.png)
of 
in the sum with the areas of triangles ![$[KEL], [EDM'], [KDM]$](http://latex.artofproblemsolving.com/f/1/9/f19629b2cd5a7d03bc75888eb448b289d2031c3b.png)
is half the area of rectangle 
so
![\[\frac {[KED]}{[CC'M'M]} = \frac {1}{2} - \frac {1}{12}- \frac {1}{12}-\frac {1}{8} = \frac {5}{24}.\]](http://latex.artofproblemsolving.com/8/2/2/82288c2b393f8c31ad458f17922c8d8b09613d18.png)
![\[FG \perp ED.\]](http://latex.artofproblemsolving.com/1/d/0/1d04737c3f8ff9ca13b3b77a0d1953b3fe707c35.png)
Denote the distance between these lines 
The volume of the tetrahedron is 
![\[\frac {ED \cdot h}{2} = \frac {5}{24} \cdot CC' \cdot CM \implies U =\frac{5}{24 \cdot 3} \cdot CC' \cdot CM \cdot FG.\]](http://latex.artofproblemsolving.com/4/7/8/4788957d32d443cee46dc23c549a2567a6f5bd88.png)
The volume of the prism is 
![\[\frac {U}{V} = \frac {\frac {5}{72} CC' \cdot CM \cdot FG}{CC' \cdot CM \cdot FG} = \frac {5}{24} \implies U = 5.\]](http://latex.artofproblemsolving.com/b/f/a/bfa9050f204fa05927897118b7f5fe8a09a2cb15.png)
An arbitrary prism is obtained from a regular one as a result of an affine transformation.
All points on the tetrahedron are defined affinely, which means that the volume ratio will be preserved.
Answer: 5.
2022 222 problem 7
![\[r = 0.5, h = 3/2, KM = \frac {\sqrt{3}}{2},\]](http://latex.artofproblemsolving.com/1/d/c/1dca112ae88f73b3097142899ae597d9184f3a30.png)
![\[AM = KM \cdot \tan \alpha, BM = \frac {KM}{\tan \alpha},\]](http://latex.artofproblemsolving.com/2/b/c/2bc88738c2efa898feb0c22b93dc8754aa27f8d8.png)
![\[AM + BM = AB \implies\]](http://latex.artofproblemsolving.com/6/6/b/66bd3dd3f4588d2579c1569d2c36491417e20a8c.png)
![\[\tan \alpha + \frac {1}{\tan \alpha} = 2 \sqrt {2} \implies \tan \alpha = \sqrt {2} - 1 \implies \tan 2 \alpha = 1 \implies 2 \alpha = \frac {\pi}{4}.\]](http://latex.artofproblemsolving.com/a/d/9/ad94bbf8346d00722305ab7c3dca276520748cb6.png)
