Difference between revisions of "DVI exam"
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Answer: <math>\frac {1}{\sqrt{3}}</math> | Answer: <math>\frac {1}{\sqrt{3}}</math> | ||
| + | |||
| + | ==2020 215 problem 7== | ||
| + | The sphere touches all edges of the tetrahedron <math>ABCD.</math> It is known that the products of the lengths of crossing edges are equal. It is also known that <math>AB = 3, BC = 1.</math> Find <math>AC.</math> | ||
| + | |||
| + | <i><b>Solution</b></i> | ||
| + | |||
| + | The tangent segments from the common point to the sphere are equal. | ||
| + | |||
| + | Let us denote the segments from the vertex <math>A</math> to the sphere by <math>a.</math> | ||
| + | |||
| + | Similarly, we define <math>b, c, d.</math> | ||
| + | <cmath>AB = a + b = 3, BC = b + c = 1, a - c = (a+b) - (b+c) = 3 - 1 = 2.</cmath> | ||
| + | <cmath>AB \cdot CD = AD \cdot BC \implies 3(c+d) = 1(a + d) \implies a = 3c + 2d</cmath> | ||
| + | <cmath>a = c + 2 \implies c + d = 1 \implies b = d.</cmath> | ||
| + | <cmath>AD = AB = 3, AD \cdot BC = 3 \cdot 1 = 3 = (a+c)(b+ d) = (3 - b + 1 - b) \cdot 2b.</cmath> | ||
| + | If <math>b = \frac {3}{2}</math> then <math>c < 0.</math> | ||
| + | |||
| + | If <math>b = \frac {1}{2} = d = c, a = \frac {5}{2}, AC = 3.</math> | ||
| + | |||
| + | The tetrahedron <math>ABCD</math> is a regular pyramid with a regular triangle with side <math>\frac {1}{2}</math> at the base and side edges equal to <math>3.</math> | ||
| + | |||
| + | Answer: 3. | ||
==2022 221 problem 7== | ==2022 221 problem 7== | ||
Revision as of 06:39, 1 February 2024
Contents
2020 201 problem 6
Let a triangular prism 
with a base 
be given, 
Find the ratio in which the plane 
divides the segment 
if 
![\[BE : EC' = 1 : 2, CF : FA' = 1 : 3.\]](http://latex.artofproblemsolving.com/f/4/3/f43cb9bbd1a096e931b12e9632120ee427300518.png)
Solution
Let 
be the parallel projections of 
on the plane 
We use and get
![\[\frac {F'H'}{H'D'} = \frac {2}{k(m+1)} = 3 = \frac {FH}{HD}.\]](http://latex.artofproblemsolving.com/a/7/3/a7339a65a369eb4ea857429bfa777efe54924c51.png)
![\[\frac {E'H'}{AH'} = \frac {mk + 1}{k+1} = \frac {7}{9} = \frac {EH}{GH}.\]](http://latex.artofproblemsolving.com/b/5/3/b53ff0a64a3f63889ebd2818ce63cab0675b377b.png)
Let 
Similarly 
Answer: 
2020 202 problem 6
Let a tetrahedron 
be given, 
Find the cosine of the angle 
between the edges 
and 
Solution
Let us describe a parallelepiped 
around a given tetrahedron 
and 
are equal rectangles.
and 
are equal rectangles.
Denote 
![\[a^2 + b^2 = 5^2 = 25, a^2 + c^2 = 6^2 = 36.\]](http://latex.artofproblemsolving.com/6/8/e/68e00f3dd3d7b9960f5a1117ad518577f6566147.png)
![\[4AC'^2 = 4 a^2 = 5^2 + 6^2 - 2 \cdot 5 \cdot 6 \cos \varphi,\]](http://latex.artofproblemsolving.com/3/b/f/3bf851368d562ff90604bcdc9936f164bd45c32a.png)
![\[4AB'^2 = 4 c^2 = 5^2 + 6^2 + 2 \cdot 5 \cdot 6 \cos \varphi,\]](http://latex.artofproblemsolving.com/7/8/2/782e7d911530bc1fb1dec37d7ace8a0ec981194a.png)
![\[4(c^2 - a^2) = 4(6^2 - 5^2) = 4 \cdot 5 \cdot 6 \cos \varphi \implies \cos \varphi = \frac {6^2 - 5^2}{5 \cdot 6} = \frac {11}{30}.\]](http://latex.artofproblemsolving.com/d/b/5/db5a58aa1718780fd619d548fadeae49a807e6f3.png)
Answer: 
2020 203 problem 6
Let a cube 
with the base and side edges 
be given. Find the volume of a polyhedron whose vertices are the midpoints of the edges 
Solution
Denote the vertices of polyhedron 
Triangles 
and 
are equilateral triangles with sides 
and areas ![$[EFG] = \frac {\sqrt{3}}{8}.$](http://latex.artofproblemsolving.com/7/d/4/7d4ecf0a789889230c0130d1cb0ba081a2186a62.png)
This triangles lies in parallel planes, which are normal to cube diagonal 
The distance 
between this planes is ![\[\sqrt{3} - 2 \cdot \frac{\sqrt{3}}{6} = \frac {2}{\sqrt{3}}.\]](http://latex.artofproblemsolving.com/3/d/5/3d5120122dd5dc8bd7361d9c38eb574bffc6ae25.png)
So the volume of the regular prism with base and height is
![\[V_0 = \frac {\sqrt{3}}{8} \cdot \frac {2}{\sqrt{3}} = \frac {1}{4}.\]](http://latex.artofproblemsolving.com/f/8/1/f81d0faa250dc44fc6e7123ec718a5dfa7fac7e5.png)
Let the area ![$[A(x)]$](http://latex.artofproblemsolving.com/6/3/2/632d250f69754d75be9cb7db132aab278ecec113.png)
be the quadratic function of 
Let
![\[A_1 = A[x_1], A_2 = A[x_2], d = x_2 - x_1,\]](http://latex.artofproblemsolving.com/2/f/5/2f54b7b969882d991543f67e6b2457af7b9b122b.png)
![\[x_0 = \frac{x_1 + x_2}{2}, A_0 = A[x_0] \implies\]](http://latex.artofproblemsolving.com/d/8/d/d8d7b39efca4e4b11a2278d766d5aaf93cb76427.png)
![\[V = \frac{d}{6} \cdot \left(A_1 + A_2 + 4 A_0 \right).\]](http://latex.artofproblemsolving.com/b/5/6/b56228431f252f26490bffff50a15bc4c5d1c3a0.png)
Suppose, we move point 
along axis 
and cross the solid by plane contains and normal to axis. Distance from to each crosspoint this plane with the edge change proportionally position along axes, so the area is quadratic function from position.
![\[\frac {OE''}{ME} = \frac {\sqrt{3}}{2} \implies \frac {[E''F''G'']}{[EFG]} = 2 \left (\frac {OE''}{ME} \right)^2 = \frac {3}{2}.\]](http://latex.artofproblemsolving.com/0/5/5/055ea1c40b675f8e3d097088e0ec6f4bf9e661bd.png)
![\[V = \frac{d}{6} \cdot ([EFG] + {[E'F'G']} + 4 [E''F''G'']) = d \cdot [EFG] \cdot \frac {4}{3} = \frac {1}{4} \cdot \frac {4}{3} = \frac {1}{3}.\]](http://latex.artofproblemsolving.com/4/d/e/4dea25d061864455f1677c2995553e30a7a069a4.png)
Answer: 
2020 204 problem 6
Let a regular triangular pyramid be given. The circumcenter of the sphere 
is equidistant from the edge and from the plane of the base of the pyramid. Find the radius of the sphere inscribed in this pyramid if the length of the edge of its base is 
Solution
![\[OP = ON \implies BP = BN, BS = 2 BP = 2 BN,\]](http://latex.artofproblemsolving.com/7/3/7/7378ff38a089c8b8339dd754cb2de0737133f96b.png)
![\[AB = \sqrt{3}BN, \angle BSN = 30^\circ \implies\]](http://latex.artofproblemsolving.com/3/b/b/3bb9b1bc0fd3f60ed1cc2c11d816302a9fbb6cb2.png)
![\[SN = \frac {3}{2} SO = AB.\]](http://latex.artofproblemsolving.com/a/c/6/ac63a2af02c75fd4e61f265ab3c1339c10050191.png)
![\[NM = \frac {BN}{2} = \frac {AB}{2 \sqrt{3}} \implies\]](http://latex.artofproblemsolving.com/4/4/1/441ad0c3a3d25a5ab22298768f70f8a2f71f865d.png)
![\[\tan \angle MSN = \frac {1}{2\sqrt{3}} \implies\]](http://latex.artofproblemsolving.com/d/6/f/d6f3a390144733f5bd669f6a60939b4b8dabce84.png)
![\[\sin \angle MSN = \frac {1}{\sqrt{13}} = \frac {ID}{SN - IN}, IN = ID = \frac {AB}{1 + \sqrt{13}}.\]](http://latex.artofproblemsolving.com/b/c/9/bc9db5821870cf3d76c1aee0fce5156faa030642.png)
Answer: 
2020 205 problem 6
Let the quadrangular pyramid 
with the base parallelogram be given.
Point 
Point 
Find the ratio in which the plane 
divides the volume of the pyramid.
Solution
Let plane cross edge 
at point 
We make the central projection from point 
The images of points 
are 
respectively.
The image of is the crosspoint of 
and 
So lines 
and 
are crossed at point 
![\[\frac {2 OH}{SH} = \frac {BE}{SE} + \frac {DF}{SF} = \frac {AA}{SA}+ \frac {CG}{SG}.\]](http://latex.artofproblemsolving.com/4/6/e/46e205c8a523c15afcd5d4e57ecaa07ce365fd16.png)
![\[2 + \frac {1}{2} = 0 + \frac {CG}{SG} \implies \frac {CG}{SG} = \frac {5}{2}.\]](http://latex.artofproblemsolving.com/f/0/0/f00f280c4bde57c50b6e9b6fbb4d77678a2841a7.png)
Let’s compare volumes of some tetrachedrons, denote the volume of 
as ![$[X].$](http://latex.artofproblemsolving.com/a/9/5/a9518ed103cb6898857664595f950f87beba0328.png)
![\[\frac {[ABDS]}{[CBDS]} = \frac {[ABD]}{[CBD]} =1.\]](http://latex.artofproblemsolving.com/b/1/3/b13f5d2ff80c406713850c00666a47b6c8d2fcfd.png)
![\[\frac {[AEFS]}{[ABDS]} = \frac {[EFS]}{[BDS]} = \frac {SE \cdot SF}{SB \cdot SD} = \frac{2}{9}.\]](http://latex.artofproblemsolving.com/9/d/3/9d3458e407fba75e9c3664f909c5a87d8959e7fc.png)
![\[\frac {[GEFS]}{[CBDS]} = \frac {[EFS]}{[BDS]} \cdot \frac {GX}{CO} = \frac{2}{9} \cdot \frac {SG}{SC} = \frac{2}{9} \cdot \frac {2}{7} = \frac{4}{63}.\]](http://latex.artofproblemsolving.com/e/1/3/e13c2c3ca0963b2bbde92f31a9283f0315753b4f.png)
![\[\frac {[AEGFS]}{[ABDS]} = \frac {[AEFS]+[GEFS]}{[ABDS]} = \frac{2}{9} + \frac{4}{63} = \frac{2}{7} \implies \frac {[AEGFS]}{[ABCDS]} = \frac{1}{7}.\]](http://latex.artofproblemsolving.com/b/6/0/b6074f9baba7a26e618327a4a8a120ccf5d32c1e.png)
Answer: 1 : 6.
2020 206 problem 6
Given a cube with the base and side edges 
Find the distance between the line passing through the midpoints of the edges 
and 
and the line passing through the midpoints of the edges 
and 
Solution
Let points 
be the midpoints of 
respectively. We need to prove that planes 
and 
are parallel, perpendicular to 
Therefore, 
Point is the midpoint 
![\[B'O = \frac {\sqrt{3}}{2}, B'H = \frac {1}{2}, GH = \frac {\sqrt{2}}{2},\]](http://latex.artofproblemsolving.com/7/9/d/79d371f4ecadfb3d21c2df4c863601c68444713d.png)
![\[HQ = \frac{GH}{\sqrt{3}}, B'Q = \frac{\sqrt{3}}{6}, OQ = \frac {1}{\sqrt{3}} = IJ.\]](http://latex.artofproblemsolving.com/2/7/9/279e1d5a5cae56bf7e5e1ac505e21522accaf681.png)
For proof we can use one of the following methods:
1. Vectors: 
![\[\vec {B'G} = \vec e_y, \vec {B'H} = \vec e_z, \vec {HG} = \vec e_y - \vec e_z, \vec {B'D} = \vec e_x + \vec e_y + \vec e_z.\]](http://latex.artofproblemsolving.com/4/8/0/480b185749ed15d021d43f16976dcbfd9234f136.png)
Scalar product 
Similarly, ![\[\vec {B'E} = 2\vec e_y + \vec e_x, \vec {B'F} = 2\vec e_x+ \vec e_y, \vec {FE} = \vec e_y - \vec e_x.\]](http://latex.artofproblemsolving.com/3/b/7/3b77a07bcd4dea5e8ebc0e6f4ed50259538a68e0.png)
2. 
3. Rotating the cube around its axis 
we find that the point 
move to 
, then to 
then to
Answer: 
2020 215 problem 7
The sphere touches all edges of the tetrahedron It is known that the products of the lengths of crossing edges are equal. It is also known that 
Find 
Solution
The tangent segments from the common point to the sphere are equal.
Let us denote the segments from the vertex 
to the sphere by 
Similarly, we define 
![\[AB = a + b = 3, BC = b + c = 1, a - c = (a+b) - (b+c) = 3 - 1 = 2.\]](http://latex.artofproblemsolving.com/0/f/f/0ff73c50cb8f280ec69eaad380f6059223671a65.png)
![\[AB \cdot CD = AD \cdot BC \implies 3(c+d) = 1(a + d) \implies a = 3c + 2d\]](http://latex.artofproblemsolving.com/0/7/7/077efa9152659fbd266784dbd3bbec558493a371.png)
![\[a = c + 2 \implies c + d = 1 \implies b = d.\]](http://latex.artofproblemsolving.com/2/c/9/2c96c9d019eb6f72bf371f3829ef1e5796552e16.png)
![\[AD = AB = 3, AD \cdot BC = 3 \cdot 1 = 3 = (a+c)(b+ d) = (3 - b + 1 - b) \cdot 2b.\]](http://latex.artofproblemsolving.com/6/d/f/6dfcb7c9308b83e0a716329fa6fd71e6b877d6f4.png)
If 
then 
If 
The tetrahedron is a regular pyramid with a regular triangle with side 
at the base and side edges equal to 
Answer: 3.
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 
![\[\frac {U}{V} = \frac {\frac {5}{72} CC' \cdot CM \cdot FG}{CC' \cdot CM \cdot FG} = \frac {5}{72} \implies U = 5.\]](http://latex.artofproblemsolving.com/b/c/1/bc112e33f6cf11d8285c38145ecd3a0f9b2b4faa.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
A sphere of diameter 
is inscribed in a pyramid at the base of which lies a rhombus with an acute angle 
and side 
Find the angle if it is known that all lateral faces of the pyramid are inclined to plane of its base at an angle of 
Solution 1
Denote rhombus 
is the vertex of a pyramid 
is the center of the sphere, 
is the tangent point of 
and sphere, 
![\[IE = r, SK = SI + IK = \frac {3}{2}, KM = \frac {\sqrt{3}}{2}, SM = \sqrt{3}.\]](http://latex.artofproblemsolving.com/1/e/b/1ebd793d2268ad58c514225131e833858d042f23.png)
![\[AM = KM \cdot \tan \alpha, BM = \frac {KM}{\tan \alpha},\]](http://latex.artofproblemsolving.com/2/b/c/2bc88738c2efa898feb0c22b93dc8754aa27f8d8.png)
![\[AM + BM = AB = \sqrt{6}\implies\]](http://latex.artofproblemsolving.com/3/b/7/3b79cd17e4b558e64a8b1f514007ebe95da24557.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)
Solution 2
The area of the rhombus ![$[ABCD]= AB^2 \cdot \sin 2\alpha.$](http://latex.artofproblemsolving.com/7/1/9/71943de9a1d67756d77ec273ff85ee2cd1044898.png)
The area of the lateral surface is ![$[l]= 4 [SAB] = 2 \cdot AB \cdot SM.$](http://latex.artofproblemsolving.com/b/5/7/b57349997a507aab95d3f3c03357ea2065754ecb.png)
![\[[ABCD] = [l] \cdot \cos 60 ^\circ =[l] \cdot \frac{1}{2} \implies AB \cdot SM = AB^2 \cdot \sin 2\alpha \implies\]](http://latex.artofproblemsolving.com/5/6/9/5690be5735e2b7558af617c1fadf380ba8a1084e.png)
![\[\sin {2 \alpha} = \frac {SM}{AB} = \frac {\sqrt{3}}{\sqrt{6}} = \frac {1}{\sqrt{2}}.\]](http://latex.artofproblemsolving.com/5/7/9/57999904bddd7720f46679becac301321c7f3911.png)
Answer:
