Difference between revisions of "DVI exam"

Revision as of 03:45, 1 February 2024

2020 201 problem 6

Let a triangular prism $ABCA'B'C'$ with a base $ABC$ be given, $D \in AB', E \in BC', F \in CA'.$ Find the ratio in which the plane $DEF$ divides the segment $AA',$ if $AD : DB' = 1 : 1,$ $BE : EC' = 1 : 2, CF : FA' = 1 : 3.$

Solution

Let $E',D',F'$ be the parallel projections of $D,E,F (DD' || AA' || EE' || FF')$ on the plane $ABC, H' = AE' \cap D'F', HH' || AA'.$ $\frac {BD'}{AD'} = \frac {BD}{AD} = 1, \frac {BE'}{CE'} = \frac {BE}{C'E} = \frac {1}{2} = k, \frac {CF'}{AF'} = \frac {CF}{A'F} = \frac {1}{3} = m.$

We use and get $\frac {F'H'}{H'D'} = \frac {2}{k(m+1)} = 3 = \frac {FH}{HD}.$ $\frac {E'H'}{AH'} = \frac {mk + 1}{k+1} = \frac {7}{9} = \frac {EH}{GH}.$ Let $DD' = x, FF'= y, DH = u, FH = v \implies HH' = \frac{yu + vx}{u+v }= \frac {7}{16}.$

Similarly $HH' = \frac{AG \cdot EH + EE' \cdot HG}{EH+HG} \implies AG = \frac {4}{7} \implies \frac {AG}{GA'} = \frac {4}{3}.$

Answer: $AG : GA' = 4 : 3.$

2020 202 problem 6

Let a tetrahedron $ABCD$ be given, $AB = BC = CD = 5, CA = AD = DB = 6.$ Find the cosine of the angle $\varphi$ between the edges $BC$ and $AD.$

Solution

Let us describe a parallelepiped $AC'BD'B'DA'C$ around a given tetrahedron $ABCD.$

$AB = CD \implies AC'BD'$ and $B'DA'C$ are equal rectangles.

$AC = BD \implies AB'CD'$ and $C'DA'B$ are equal rectangles.

Denote $AC' = a, AD' = b, AB' = c \implies$ $a^2 + b^2 = 5^2 = 25, a^2 + c^2 = 6^2 = 36.$ $4AC'^2 = 4 a^2 = 5^2 + 6^2 - 2 \cdot 5 \cdot 6 \cos \varphi,$ $4AB'^2 = 4 c^2 = 5^2 + 6^2 + 2 \cdot 5 \cdot 6 \cos \varphi,$ $4(c^2 - a^2) = 4(6^2 - 5^2) = 4 \cdot 5 \cdot 6 \cos \varphi \implies \cos \varphi = \frac {6^2 - 5^2}{5 \cdot 6} = \frac {11}{30}.$ Answer: $\frac {11}{30}.$

2022 221 problem 7

The volume of a triangular prism $ABCA'B'C'$ with base $ABC$ and side edges $AA', BB', CC'$ is equal to $72.$ Find the volume of the tetrahedron $DEFG,$ where $D$ is the centroid of the face $ABC'A', E$ is the point of intersection of the medians of $\triangle A'B'C', F$ is the midpoint of the edge $AC$ and $G$ is the midpoint of the edge $BC.$

Solution

Let us consider the uniform triangular prism $ABCA'B'C'.$ Let $M$ be the midpoint of $AB, M'$ be the midpoint of $A'B', K$ be the midpoint of $CM, L$ be the midpoint of $C'M', 2 FG = AB.$

The area $[KED]$ of $\triangle KED$ in the sum with the areas of triangles $[KEL], [EDM'], [KDM]$ is half the area of rectangle $CC'M'M,$ so $\frac {[KED]}{[CC'M'M]} = \frac {1}{2} - \frac {1}{12}- \frac {1}{12}-\frac {1}{8} = \frac {5}{24}.$ $FG \perp ED.$ Denote the distance between these lines $h.$ The volume of the tetrahedron is $U = \frac {ED \cdot h \cdot FG}{6}.$ $\frac {ED \cdot h}{2} = \frac {5}{24} \cdot CC' \cdot CM \implies U =\frac{5}{24 \cdot 3} \cdot CC' \cdot CM \cdot FG.$ The volume of the prism is $V = \frac{CM \cdot AB}{2} \cdot CC' =CC' \cdot CM \cdot FG = 72.$ $\frac {U}{V} = \frac {\frac {5}{72} CC' \cdot CM \cdot FG}{CC' \cdot CM \cdot FG} = \frac {5}{72} \implies U = 5.$

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 $1$ is inscribed in a pyramid at the base of which lies a rhombus with an acute angle $2\alpha$ and side $\sqrt{6}.$ Find the angle $2\alpha$ if it is known that all lateral faces of the pyramid are inclined to plane of its base at an angle of $60^\circ.$

Solution 1

Denote rhombus $ABCD, K = AC \cap BD, S$ is the vertex of a pyramid $SK \perp ABC, I$ is the center of the sphere, $IK = r = \frac {1}{2}, M \in AB, SM \perp AB, E$ is the tangent point of $SM$ and sphere, $\angle SMK = 60 ^\circ.$ $IE = r, SK = SI + IK = \frac {3}{2}, KM = \frac {\sqrt{3}}{2}, SM = \sqrt{3}.$ $AM = KM \cdot \tan \alpha, BM = \frac {KM}{\tan \alpha},$ $AM + BM = AB = \sqrt{6}\implies$ $\tan \alpha + \frac {1}{\tan \alpha} = 2 \sqrt {2} \implies \tan \alpha = \sqrt {2} - 1 \implies \tan 2 \alpha = 1 \implies 2 \alpha = \frac {\pi}{4}.$ Solution 2

The area of the rhombus $[ABCD]= AB^2 \cdot \sin 2\alpha.$

The area of the lateral surface is $[l]= 4 [SAB] = 2 \cdot AB \cdot SM.$ $[ABCD] = [l] \cdot \cos 60 ^\circ =[l] \cdot \frac{1}{2} \implies AB \cdot SM = AB^2 \cdot \sin 2\alpha \implies$ $\sin {2 \alpha} = \frac {SM}{AB} = \frac {\sqrt{3}}{\sqrt{6}} = \frac {1}{\sqrt{2}}.$ Answer: $\frac {\pi}{4}.$

@@ Line 16: / Line 16: @@
 Answer: <math>AG : GA' = 4 : 3.</math>
+==2020 202 problem 6==
+[[File:2020 202 6.png|330px|right]]
+Let a tetrahedron <math>ABCD</math> be given, <math>AB = BC = CD = 5, CA = AD = DB = 6.</math> Find the cosine of the angle <math>\varphi</math> between the edges <math>BC</math> and <math>AD.</math>
+<i><b>Solution</b></i>
+Let us describe a parallelepiped <math>AC'BD'B'DA'C</math> around a given tetrahedron <math>ABCD.</math>
+<math>AB = CD \implies AC'BD'</math> and <math>B'DA'C</math> are equal rectangles.
+<math>AC = BD \implies AB'CD'</math> and <math>C'DA'B</math> are equal rectangles.
+Denote <math>AC' = a, AD' = b, AB' = c \implies</math>
+<cmath>a^2 + b^2 = 5^2 = 25,  a^2 + c^2 = 6^2 = 36.</cmath>
+<cmath>4AC'^2 = 4 a^2 = 5^2 + 6^2 - 2 \cdot 5 \cdot 6 \cos \varphi,</cmath>
+<cmath>4AB'^2 = 4 c^2 = 5^2 + 6^2 + 2 \cdot 5 \cdot 6 \cos \varphi,</cmath>
+<cmath>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}.</cmath>
+Answer: <math>\frac {11}{30}. </math>
 ==2022 221 problem 7==

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Difference between revisions of "DVI exam"

Revision as of 03:45, 1 February 2024

Contents

2020 201 problem 6

2020 202 problem 6

2022 221 problem 7

2022 222 problem 7