DVI exam

DVI is an exam in mathematics at the Moscow State University named after M.V. Lomonosov. The first four problems have a standard level. Problem 5 is advanced level of geometry. Problem 6 is an advanced level equation or inequality. Problem 7 is advanced level of stereometry.

Below are the most difficult problems of this exam in recent years. The headings indicate the year when the problem was used, the variant option of the exam, and the number of the problem.

2011 Problem 8

Solve the system of equations $\left\{\begin{array}{l} 2 x^2 + 4xy + 11 y^2 \le 1,\\4x + 7y \ge 3.\end{array}\right.$ Standard Solution $\left\{\begin{array}{l} 2 (x + y)^2 + 9 y^2 \le 1,\\4(x + y) +3y \ge 3.\end{array}\right.$ Denote $\left\{\begin{array}{l} u = \sqrt {2} (x + y),\\v = 3y.\end{array}\right.$ We get $\left\{\begin{array}{l} u^2 + v^2 \le 1,\\2 \sqrt{2} u +v \ge 3.\end{array}\right.$ First equation define inner points of the circle with radius $1$ and the circle. The distance from the straight line to the origin of the coordinate system $d$ is $\frac {1}{d^2} = \frac {1}{3^2} + \frac {(2 \sqrt{2})^2}{3^2} = 1 \implies d = 1,$ so the system of the equations define the only tangent point of the circle and the line. $u = \frac {2 \sqrt{2}}{3}, v = \frac {1}{3} \implies x = \frac {5}{9}, y = \frac {1}{9}.$ Short Solution $9(2 x^2 + 4xy + 11 y^2) \le 9 \le (4x + 7y)^2 \implies 2(x - 5y)^2 \le 0 \implies x = 5y \implies$ $81 y^2 \le 1, 27 y \ge 3 \implies y = \frac {1}{9}, x = \frac {5}{9}.$

2012 Problem 8

Let the tetrahedron $SABC, AC = BC = 5, AB = 6, AS = BS = 7, CS = 4$ be given.

A right circular cylinder is located so that the circle of its upper base touches each of the faces which contains vertex $S.$

The circle of the lower base lies in the $ABC$ plane and touches straight lines $AC$ and $BC.$

Find the height $h$ of the cylinder.

Solution

Denote $M$ the midpoint $AB.$ Plane $SCM$ is the bisector plane of segment $AB, SCM \perp ABC.$ $CM = \sqrt{AC^2 - \frac {AB^2}{4}}= 4 = SC.$

The inradius of $\triangle ABC$ equal to $\frac {3}{2},$ distance from incenter $I$ to vertex $C$ is $IC = 4 - \frac {3}{2} = \frac {5}{2}.$

Denote $D$ the foot from $S$ to $\overline{MC} \implies SD = \sqrt{15}, CD = 1.$

Denote $KK'L$ the crosssection of $SABC$ by plane of the upper base of cylinder, $O'$ is the incenter $\triangle KK'L, F'$ is the point of tangency incircle of $\triangle KK'L$ and $KK'.$

Denote $F$ and $O$ the foots from $F'$ and $O'$ to $\overline{MC}.$ Denote the radius $OF = r = O'F'.$

The circle of the lower base inscribed in angle equal to $\angle ACB,$ so $\frac {CO}{FO} = \frac{5}{3} \implies CO = \frac{5r}{3}, CF = \frac{2r}{3}.$ $\triangle MF'F \sim \triangle MSD \implies \frac {MF}{MD} = \frac {F'F}{SD}.$ Projection from the point $S$ maps $Q'$ onto $I \implies$ $\triangle IO'O \sim \triangle ISD \implies \frac {IO}{ID} = \frac {O'O}{SD}.$ $h = O'O = F'F \implies \frac {O'O}{SD} = \frac {F'F}{SD} =\frac {MF}{MD} = \frac {IO}{ID},$ $\frac {4 + \frac {2r}{3} } {4+1} = \frac {\frac{5}{2} + \frac {5r}{3} } { \frac {5}{2}+ 1} \implies r = \frac {1}{4} \implies h = \frac {5 \sqrt{15}} {6}.$

Answer: $\frac {5 \sqrt{15}} {6}.$

2014 1 Problem 6

Find all pares of real numbers $(x,y)$ satisfying the system of equations $\left\{\begin{array}{l} x^{\frac{3}{2}} + y = 16 ,\\x + y^{\frac{2}{3}} = 8.\end{array}\right.$ Solution

Denote $t= y^{\frac{2}{3}} \implies y = t^{\frac{3}{2}}.$ $\left\{\begin{array}{l} x^{\frac{3}{2}} + t^{\frac{3}{2}} = 16 ,\\x + t = 8.\end{array}\right.$ Denote $u = \frac {x}{4}, v = \frac {t}{4}.$ $\left\{\begin{array}{l} u^{\frac{3}{2}} + v^{\frac{3}{2}} = 2 ,\\u + v = 2.\end{array}\right.$ $u = v = 1, x = 4, y = 8$ is the solution. Let $F(u) = u^{\frac{3}{2}} + (2 - u)^{\frac{3}{2}} \implies F'(u) = 1.5(\sqrt{u} - \sqrt{2 - u}).$ If $u > 1$ then $F'(u) > 0,$ if $u < 1$ then $F'(u) < 0,$ therefore $u = 1$ is the single root.

2014 1 Problem 8

Let $f(x,y) = y + \sqrt{6 - 6x^2 - 14 y^2 - 18 xy}, g(x,y) = y - \sqrt{6 - 6x^2 - 14 y^2 - 18 xy}.$

Find $max_x max_y (f(x,y), g(x,y))$ and $min_x min_y (f(x,y), g(x,y)).$

Solution $6 - 6x^2 - 14 y^2 - 18 xy = 6 - 6 \left(x^2 + 3xy + \frac {9}{4} y^2\right) + \frac {6 \cdot 9}{4} y^2 - 14 y^2 = 6 - 6 \cdot \left(x + \frac {3y}{2} \right)^2 - \frac {1}{2} y^2,$ $6 - 6 \cdot (x + \frac {3y}{2})^2 - \frac {1}{2} y^2 \le 6 - \frac {1}{2} y^2.$ $f(x,y) \ge g(x,y) \implies max_x max_y (f(x,y), g(x,y)) = max_x max_y f(x,y) = max_y (y + \sqrt{6 - \frac {1}{2} y^2}),$ where $y > 0.$ $\frac {u+v+w}{3} \le \sqrt{\frac {u^2 + u^2 + w^2}{3}} \implies$ $y + \sqrt{6 - \frac {1}{2} y^2} = \frac {y}{2}+\frac {y}{2}+\sqrt{6 - \frac {1}{2} y^2} \le \sqrt {3} \cdot \sqrt{6 - \frac {1}{2} y^2 + \frac {1}{4} y^2 + \frac {1}{4} y^2} = \sqrt {3 \cdot 6} = 3 \sqrt{2}.$ $f(-x,-y) = -g(x,y) \implies min_x min_y (f(x,y), g(x,y)) = min_x min_y (g(x,y)) = - max_x max_y f(x,y) = - 3 \sqrt{2}.$

Answer: $3 \sqrt{2}, - 3 \sqrt {2}.$

2015 1 Problem 7

A sphere is inscribed in a regular triangular prism with bases $ABCA'B'C'.$ Find its radius if the distance between straight lines $AE$ and $BD$ is equal to $\sqrt{13},$ where $E$ and $D$ are points lying on $A'B'$ and $B'C'$ , respectively, and $A'E : EB' = B'D : DC' = 1 : 2.$

Solution

The distance from the center of the sphere to the centers of the prism faces is equal to $R,$ so $AA' = 2R, AB = 2 \sqrt{3} R.$

In order to find the distance $PQ$ between the lines $\ell = AD$ and $m = BE$ , one can find the length of two perpendiculars $MM'$ and $DE$ to the line $m$ that are perpendicular to each other. Then $\frac {1}{PQ^2} = \frac{1}{DE^2} + \frac{1}{MM'^2}$ since, when viewed along a straight line $m$ , the segment $PQ$ is the altitude of a right triangle with legs $DE$ and $MM'.$

The plane $\pi = BB'C'$ containe the straight line $m.$ The straight line $\ell$ crossed $\pi$ at the point $M \in BB'.$ $\frac {B'D}{DA'} = \frac {B'M}{AA'} =2 \implies B'M = 4R.$ In a right triangle $\triangle BKM$ $KM = BC = 2 \sqrt{3} R, BM = BB' + B'M = 6R, MM' \perp BE.$ $MM'$ is the height falling on the hypotenuse, $\frac {1}{MM'^2} = \frac{1}{KM^2} + \frac{1}{BM^2}.$

Let $F$ be the projection of $A$ onto plane $\pi \implies F \in BC, BF = FC.$

Therefore $FM$ is the projection of $\ell$ onto plane $\pi, KM = 2 FB \implies m \cap MF$ at the point $E.$ $\frac {B'D}{EB'} = 2 \implies DE \perp \pi \implies DE \perp m.$ $\frac {DE}{AF} = \frac {ME}{MF} = \frac {B'M}{MB} = \frac {2}{3} \implies DE = 2R.$ $\frac {1}{PQ^2} = \frac{1}{DE^2} + \frac{1}{MM'^2} = \frac{1}{DE^2} + \frac{1}{KM^2} + \frac{1}{BM^2} = \frac{1}{(2R)^2} + \frac{1}{(2\sqrt{3}R)^2} + \frac{1}{(6R)^2} = \frac{13}{(6R)^2}.$ $PO = \sqrt{13} = \frac{6R}{\sqrt{13}} \implies R = \frac {13}{6}.$ Answer: $\frac {13}{6}.$

2016 2 Problem 7

Let the base of the regular pyramid with vertex $S$ be the hexagon $ABCDEF$ with side $5.$ The plane $\pi$ is parallel to the edge $AB$ , perpendicular to the plane $SDE$ and intersects the edge $BC$ at point $K,$ so that $\frac {BK}{KC} = \frac {3}{2}.$ The lines along which $\pi$ intersects the $BCS$ plane and the base plane are perpendicular.

Find the area of the triangle cut off by the plane $\pi$ from the face $CDS.$

Solution

Denote $K' = \pi \cap AF, L = \pi \cap SC, L' = \pi \cap SF, M = \pi \cap SD, M' = \pi \cap SE,$

$G,O,H,N$ are the midpoints of $KK',CF,DE, MM',$ respectively.

Plane $SGH$ is the plane symmetry of pyramid, $SGH \perp KK' \implies SGH \perp \pi.$

By condition $GN = \pi \cap SGN \perp SH,$ so exist point $Q = SO \cap GN, Q \in LL'.$

$KL$ is the line along which $\pi$ intersects the $BCS$ plane, $KK'$ is the line along which $\pi$ intersects the base plane, so $\angle LKK' = 90^\circ \implies LK || GN.$

We use the top wiew and get $\frac {SL}{LC} = \frac{KB + CB} {CK}= \frac{3+5} {2} = 4. \frac{GO}{OH} = \frac{CK}{CD} = \frac{2}{5}.$ $AB = 5 \implies OH = \frac {5 \sqrt{3}}{2} \implies GO = \sqrt{3}.$ $Q \in LL' \implies \frac {SQ}{QO} = 4.$ Denote $h = SO, \alpha = \angle SHO = \angle GQO$ and use the side wiew.

$\tan \alpha = \frac {SO}{OH} = \frac {GO}{QH} \implies OH \cdot GO = \frac {15}{2} = SO \cdot QO = \frac {h^2}{5} \implies$ $h = 5 \sqrt {\frac {3}{2}} \implies \tan \alpha = \sqrt{2} \implies \cos \alpha = \frac {1}{\sqrt{3}}.$ Triangle $\triangle OCD$ is the regular triangle with side $5$ , so $[OCD] = \frac {25 \sqrt{3}}{4} \implies [SCD] = \frac{OCD}{\cos \alpha} = \frac {75}{4}.$ $SH = \sqrt {SO^2 + OH^2} = \frac {15}{2}, NH = GH \cos \alpha = \frac {7}{2} \implies \frac {SN}{SH} =\frac{8}{15}= \frac{SM}{SD}.$ $[SML] = [SCD] \cdot \frac {SL}{SC} \cdot \frac {SM}{SD} = \frac {75}{4} \cdot \frac{4}{5} \cdot \frac{8}{15} = 8.$ Answer: 8.

2016 2 Problem 8

Find the smallest value of the expression $f=\sqrt{13 + \log^2_a \cos \frac {x}{a} + \log_a \cos^4 \frac {x}{a}}+\sqrt{97 + \log^2_a \sin \frac {x}{a} - \log_a \sin^8 \frac {x}{a}} + \sqrt{20 + \log^2_a \tan \frac {x}{a} + \log_a \tan^4 \frac {x}{a}}.$ Solution $13 + \log^2_a \cos \frac {x}{a} + \log_a \cos^4 \frac {x}{a} = 9 + \log^2_a \cos \frac {x}{a} + 4\log_a \cos \frac {x}{a} + 4 = 3^2 + (\log_a \cos \frac {x}{a} + 2)^2,$ $97 + \log^2_a \sin \frac {x}{a} - \log_a \sin^8 \frac {x}{a} = 9^2 + (4 - \log_a \sin \frac {x}{a} )^2,$ $20 + \log^2_a \tan \frac {x}{a} + \log_a \tan^4 \frac {x}{a} = 16 + (\log_a \tan \frac {x}{a} + 2)^2 = 4^2 + (\log_a \sin \frac {x}{a} - \log_a \cos \frac {x}{a}+ 2)^2.$ Denote $\vec {AB} = (3, \log_a \cos \frac {x}{a} + 2), \vec {BC} = (9, 4 - \log_a \sin \frac {x}{a}), \vec {CD} = (4, \log_a \sin \frac {x}{a} - \log_a \cos \frac {x}{a}+ 2).$ $f = |\vec {AB}| + |\vec {BC}| + |\vec {CD}|.$ $\vec {AB} + \vec {BC} + \vec {CD} = (3 +9 + 4, \log_a \cos \frac {x}{a} + 2 +4 - \log_a \sin \frac {x}{a} + \log_a \sin \frac {x}{a} - \log_a \cos \frac {x}{a}+ 2) = (16,8).$ The shortest length of a broken line $ABCD$ with fixed ends is equal to the distance between points $A$ and $D,$ which is $|AD|$ and is achieved if points $A,B,C,$ and $D$ are collinear. $|\vec AD| = \sqrt {16^2 + 8^2} = 8 \sqrt{5}, \vec {AB} = (3, 1.5),\vec {BC} = (9, 4.5), \vec {CD} (4,2).$ $\log_a \tan \frac {x}{a} + 2 = 2 \implies \log_a \tan \frac {x}{a} = 0 \implies \tan \frac {x}{a} = 1.$ $\sin \frac {x}{a} >0 \implies \sin \frac {x}{a} = \frac {1}{\sqrt{2}}, \cos \frac {x}{a} = \frac {1}{\sqrt{2}}.$ $4 - \log_a \sin \frac {x}{a} = 4.5 \implies \log_a \frac {1}{\sqrt{2}} = - \frac{1}{2} \implies a = 2.$ $\sin \frac {x}{2} = \cos \frac {x}{2} = \frac {1}{\sqrt{2}} \implies \frac {x}{2} = \frac {\pi}{4} + 2 k \pi \implies x = \frac {\pi}{2} + 4k \pi.$ Answer: $min(f) = 8 \sqrt{5}, a = 2, x = \frac {\pi}{2} + 4k \pi.$

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: $\frac {11}{30}.$

2020 203 problem 6

Let a cube $ABCDA'B'C'D'$ with the base $ABCD$ and side edges $AA', BB', CC', DD', AB = 1$ be given. Find the volume of a polyhedron whose vertices are the midpoints of the edges $AB, AD, AA', CC', C'B', C'D'.$

Solution

Denote the vertices of polyhedron $E, F, G, E', F', G'.$ Triangles $\triangle EFG$ and $\triangle E'F'G'$ are equilateral triangles with sides $\frac {\sqrt{2}}{2}$ and areas $[EFG] = \frac {\sqrt{3}}{8}.$

This triangles lies in parallel planes, which are normal to cube diagonal $AC'.$ The distance $d$ between this planes is $\sqrt{3} - 2 \cdot \frac{\sqrt{3}}{6} = \frac {2}{\sqrt{3}}.$ So the volume of the regular prism with base $\triangle EFG$ and height $d$ is $V_0 = \frac {\sqrt{3}}{8} \cdot \frac {2}{\sqrt{3}} = \frac {1}{4}.$

Let the area $[A(x)]$ be the quadratic function of $x.$ Let $A_1 = A[x_1], A_2 = A[x_2], d = x_2 - x_1,$ $x_0 = \frac{x_1 + x_2}{2}, A_0 = A[x_0] \implies$ $V = \frac{d}{6} \cdot \left(A_1 + A_2 + 4 A_0 \right).$ Suppose, we move point $P$ along axis $AC'$ and cross the solid by plane contains $P$ and normal to axis. Distance from $P$ to each crosspoint this plane with the edge change proportionally position $P$ along axes, so the area is quadratic function from $P$ position. $\frac {OE''}{ME} = \frac {\sqrt{3}}{2} \implies \frac {[E''F''G'']}{[EFG]} = 2 \left (\frac {OE''}{ME} \right)^2 = \frac {3}{2}.$ $V = \frac{d}{6} \cdot ([EFG] + {[E'F'G']} + 4 [E''F''G'']) = d \cdot [EFG] \cdot \frac {4}{3} = \frac {1}{4} \cdot \frac {4}{3} = \frac {1}{3}.$

Answer: $\frac {1}{3}.$

2020 204 problem 6

Let a regular triangular pyramid be given. The circumcenter of the sphere $O$ 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 $12.$

Solution

$OP = ON \implies BP = BN, BS = 2 BP = 2 BN,$ $AB = \sqrt{3}BN, \angle BSN = 30^\circ \implies$ $SN = \frac {3}{2} SO = AB.$ $NM = \frac {BN}{2} = \frac {AB}{2 \sqrt{3}} \implies$ $\tan \angle MSN = \frac {1}{2\sqrt{3}} \implies$ $\sin \angle MSN = \frac {1}{\sqrt{13}} = \frac {ID}{SN - IN}, IN = ID = \frac {AB}{1 + \sqrt{13}}.$ Answer: $\frac {12}{1 + \sqrt{13}}.$

2020 205 problem 6

Let the quadrangular pyramid $ABCDS$ with the base parallelogram $ABCD$ be given.

Point $E \in SB, \frac {SE}{EB} = 2.$ Point $F \in SD, \frac {SF}{FD} = \frac {1}{2}.$

Find the ratio in which the plane $AEF$ divides the volume of the pyramid.

Solution

Let plane $AEF$ cross edge $SC$ at point $G.$ We make the central projection from point $S$ The images of points $A,E,F,G$ are $A,B,D,C,$ respectively. The image of $S$ is the crosspoint of $AC$ and $BD.$ So lines $EF, SO,$ and $AG$ are crossed at point $H.$ $\frac {2 OH}{SH} = \frac {BE}{SE} + \frac {DF}{SF} = \frac {AA}{SA}+ \frac {CG}{SG}.$ $2 + \frac {1}{2} = 0 + \frac {CG}{SG} \implies \frac {CG}{SG} = \frac {5}{2}.$ Let’s compare volumes of some tetrachedrons, denote the volume of $X$ as $[X].$ $\frac {[ABDS]}{[CBDS]} = \frac {[ABD]}{[CBD]} =1.$ $\frac {[AEFS]}{[ABDS]} = \frac {[EFS]}{[BDS]} = \frac {SE \cdot SF}{SB \cdot SD} = \frac{2}{9}.$ $\frac {[GEFS]}{[CBDS]} = \frac {[EFS]}{[BDS]} \cdot \frac {GX}{CO} = \frac{2}{9} \cdot \frac {SG}{SC} = \frac{2}{9} \cdot \frac {2}{7} = \frac{4}{63}.$ $\frac {[AEGFS]}{[ABDS]} = \frac {[AEFS]+[GEFS]}{[ABDS]} = \frac{2}{9} + \frac{4}{63} = \frac{2}{7} \implies \frac {[AEGFS]}{[ABCDS]} = \frac{1}{7}.$ Answer: 1 : 6.

2020 var 206 problem 6

Given a cube $ABCDA'B'C'D'$ with the base $ABCD$ and side edges $AA', BB', CC', DD' =1.$ Find the distance between the line passing through the midpoints of the edges $AB$ and $AA'$ and the line passing through the midpoints of the edges $BB'$ and $B'C'.$

Solution

Let points $E,F,P, G, H, K$ be the midpoints of $AB, AA', A'D', BB', B'C', B'A',$ respectively. We need to prove that planes $GKH$ and $EFP$ are parallel, perpendicular to $B'D.$ Therefore, $B'D = \sqrt{3}.$

Point $O$ is the midpoint $B'D \implies$ $B'O = \frac {\sqrt{3}}{2}, B'H = \frac {1}{2}, GH = \frac {\sqrt{2}}{2},$ $HQ = \frac{GH}{\sqrt{3}}, B'Q = \frac{\sqrt{3}}{6}, OQ = \frac {1}{\sqrt{3}} = IJ.$ For proof we can use one of the following methods:

1. Vectors: $\vec {B'A'} = 2 \vec e_x, \vec {B'B} = 2 \vec e_y, \vec {B'C'} = 2 \vec e_z \implies$ $\vec {B'G} = \vec e_y, \vec {B'H} = \vec e_z, \vec {HG} = \vec e_y - \vec e_z, \vec {B'D} = \vec e_x + \vec e_y + \vec e_z.$ Scalar product $(\vec{B'D} \cdot \vec {HG}) = 0.$ Similarly, $\vec {B'E} = 2\vec e_y + \vec e_x, \vec {B'F} = 2\vec e_x+ \vec e_y, \vec {FE} = \vec e_y - \vec e_x.$

2. $\angle B'OG = \angle B'OE = 90^\circ.$

3. Rotating the cube around its axis $B'D$ we find that the point $G$ move to $H$ , then to $K,$ then to $G.$

Answer: $\frac {1}{\sqrt{3}}$

2021 var 215 problem 7

The sphere touches all edges of the tetrahedron $ABCD.$ It is known that the products of the lengths of crossing edges are equal. It is also known that $AB = 3, BC = 1.$ Find $AC.$

Solution

The tangent segments from the common point to the sphere are equal.

Let us denote the segments from the vertex $A$ to the sphere by $a.$

Similarly, we define $b, c, d.$ $AB = a + b = 3, BC = b + c = 1, a - c = (a+b) - (b+c) = 3 - 1 = 2.$ $AB \cdot CD = AD \cdot BC \implies 3(c+d) = 1(a + d) \implies a = 3c + 2d$ $a = c + 2 \implies c + d = 1 \implies b = d.$ $AD = AB = 3, AD \cdot BC = 3 \cdot 1 = 3 = (a+c)(b+ d) = (3 - b + 1 - b) \cdot 2b.$ If $b = \frac {3}{2}$ then $c < 0.$

If $b = \frac {1}{2} = d = c, a = \frac {5}{2}, AC = 3.$

The tetrahedron $ABCD$ is a regular pyramid with a regular triangle with side $1$ at the base and side edges equal to $3.$

Answer: 3.

2022 var 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 var 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: $\frac {\pi}{4}.$

2022 var 222 problem 6

Find all possible values of the product $xy$ if it is known that $x,y \in \left [ 0, \frac{\pi}{2} \right)$ and it is true $\frac{1 - \sin(x - y)}{1 - \cos(x - y)}= \frac{1 - \sin(x + y)}{1 - \cos(x + y)}.$

Solution

Let $y = 0,$ then for each $x$ equation is true, $xy = 0.$ Let $y > 0.$ $1 - \sin (x - y) - \cos (x + y) + \sin (x – y) \cos (x + y) = 1 - \sin (x + y) - \cos (x - y) + \sin (x + y) \cos (x - y).$ $\sin (x + y) - \sin (x - y) + \cos (x - y) - \cos (x + y) = \sin ((x + y) - (x - y)),$ $2 \cos x \sin y + 2 \sin x \sin y = 2 \cos y \sin y,$ $\cos x + \sin x = \cos y .$ $\cos y < 0, x \in \left [ 0, \frac{\pi}{2} \right) \implies \cos x + \sin x \ge 1,$ no solution.

Answer: $0.$

2022 var 224 problem 6

Find all triples of real numbers $(x,y,z)$ in the interval $\left ( 0; \frac {\pi}{2} \right)$ satisfying the system of equations $\begin{equation} \left\{ \begin{aligned} \sin x &= \sin y - \sin z \cos (x+z) ,\\ \cos x &= \cos z + \cos y \cos (x+y) . \end{aligned} \right.\end{equation}$

Solution

Denote $u = x + z \implies$ $x = u - z, \sin (u - z) = \sin u \cos z - \sin z \cos u = \sin x = \sin y - \sin z \cos (x+z) \implies$ $\sin y = \cos z \sin (x+z).$ Similarly, $\cos z = \sin y \sin (x+y).$

$\begin{equation} \left\{ \begin{aligned} \sin (x+z) = \frac {\sin y}{\cos z},\\ \sin (x+y) = \frac {\cos z}{\sin y}. \end{aligned} \right.\end{equation}$ Therefore $x+y = x+z = \frac{ \pi}{2} \implies y = z,$ $\sin y = \cos z \implies x = y = z = \frac {\pi}{4}.$ Answer: $\left (\frac {\pi}{4},\frac {\pi}{4},\frac {\pi}{4} \right ).$

2023 var 231 problem 6

Let positive numbers $a,b,c$ be such that $\frac {1}{a+1} + \frac {1}{b+1} +\frac {1}{c+1} = 1.$

Find the maximum value of $\frac {a}{2 + a^2} + \frac {b}{2 + b^2} + \frac {c}{2 + c^2}.$

Solution $\frac {a}{a^2 +2} \le \frac {1}{6} +\frac {1}{2(1+a)} \Leftrightarrow 6a + 6a^2 \le a^3 + a^2 +2a + 2 + 3a ^2 + 6 \Leftrightarrow$ $a^3 - 2 a^2 - 4a + 8 \ge 0 \Leftrightarrow (a -2)^2 (a+2) \ge 0.$ Similarly $\frac {b}{b^2 +2} \le \frac {1}{6} +\frac {1}{2(1+b)}, \frac {c}{c^2 +2} \le \frac {1}{6} +\frac {1}{2(1+c)}.$ Adding this equations, we get: $\frac {a}{2 + a^2} + \frac {b}{2 + b^2} + \frac {c}{2 + c^2} \le \frac {1}{2} \left (1 + \frac {1}{a+1} + \frac {1}{b+1} +\frac {1}{c+1} \right ) = 1.$ If $a = b = c = 2$ then $\frac {a}{2 + a^2} + \frac {b}{2 + b^2} + \frac {c}{2 + c^2} = 1.$

Answer: $1.$

Explanation for students

For the function under study $F(x) = \frac {x}{x^2 +2}$ it is required to find the majorizing function $G(x) \ge F(x).$ This function must be a linear combination of the given function $g(x) = \frac {1}{x+1}$ and a constant, $G(x) = k + m \cdot g(x).$

At the supposed extremum point $x_0 = 2$ the functions and their derivatives must coincide $G(x_0) = F(x_0), G'(x_0) = F'(x_0).$ $F(x_0) = \frac {x_0}{x_0^2 +2} = \frac {1}{3} = G(x_0) = k + m \cdot \frac {1}{x_0+1} = k + \frac {m}{3} \implies 1 = 3k + m.$ $F'(x_0) = \frac {2 - x_0^2}{(x_0^2 + 2)^2} = -\frac{1}{18} = G'(x_0) = m \cdot g'(x_0) = - \frac {m}{(1+ x_0)^2} = -\frac{m}{9} \implies m = \frac{1}{2}, k = \frac {1}{6}.$

2023 var 231 EM problem 6

$F(x) = log_{\frac{5}{2}} (2 + \cos x) \cdot log_{\frac{5}{2}} (3 - \cos x).$ Find the maximum value $F_m = max (F(x))$ and all argument values $x_0$ such that $F_m = F(x_0)$ .

Solution $a + t = 2 + \cos x; a - t = 3 - \cos x \implies a = \frac {5}{2}, t = \cos x - \frac {1}{2}.$ $u = \frac {t}{a} = \frac {\cos x - \frac {1}{2}}{a}\implies$ $F(u) = log_a {a(1 + u)} \cdot log_a {a(1 - u)} = (1 + log_a (1 + u)) \cdot (1 + log_a (1 - u)) = 1 + log_a (1- u^2) + log_a (1+ u) \cdot log_a (1 - u) \le 1,$ because $1- u^2 \le 1 \implies log_a (1- u^2) \le 0$ and signs of $log_a (1 + u)$ and $log_a (1 - u)$ are different, so $log_a (1 + u) \cdot log_a (1 - u) \le 0.$ Therefore $F_m = 1, \cos x_0 = \frac {1}{2}, x_0 = \pm \frac{\pi}{3} + 2 k \pi.$

2023 var 232 problem 6

Let positive numbers $a,b,c$ be such that $\left ( a+b+c \right) \cdot \left ( \frac {1}{a} + \frac {1}{b} +\frac {1}{c} \right ) = 10.$ Find the maximum value of $\frac {a+ b}{c}.$

Solution $\left ( a+b+c \right) \cdot \left ( \frac {1}{a} + \frac {1}{b} +\frac {1}{c} \right ) = 10 \implies$ $\frac {b}{a} + \frac {a}{b} + \frac {b}{c} + \frac {a}{c} + \frac {c}{a} + \frac {c}{b} = 7.$

$\frac {a+b}{c} + \frac {c(a+b)}{ab} + \frac {b}{a} +\frac {a}{b} = 7.$ $\frac {a+b}{c} + \frac {c}{a+b}\cdot \frac {(a+b)^2}{ab} + \frac {b}{a} +\frac {a}{b} = 7.$ It is clear that $\frac {b}{a} +\frac {a}{b} \ge 2$ and $\frac {(a+b)^2}{ab} \ge 4,$ Denote $X = \frac {a+b}{c}.$ So $7 = \frac {a+b}{c} + \frac {c}{a+b}\cdot \frac {(a+b)^2}{ab} + \frac {b}{a} +\frac {a}{b} \ge \frac {a+b}{c} + \frac {4c}{a+b}+ 2 = X + \frac {4}{X} + 2,$ $X + \frac {4}{X} \le 5 \implies (X - 4)(X - 1) \le 0 \implies X \le 4.$ If $a = b = 2, c = 1$ then $\frac {a+b}{c} = 4.$

Answer: $4.$

2023 var 233 problem 6

Let positive numbers $a,b,c$ be such that $a^2 + b^2 + c^2 = 1.$

Find the maximum value of $a b + b c \sqrt{3}.$

Solution

Let $\vec X = \{a; c \}, \vec Y = \{1; \sqrt{3} \}$ Then $a + c \sqrt{3} = \vec X \cdot \vec Y \le | \vec X| \cdot |\vec Y| = \sqrt{a^2 + c^2} \cdot \sqrt{1 + 3 } = 2 \sqrt{a^2 + c^2}.$

$2 u v \le u^2 + v^2 \implies 2 b \cdot \sqrt{a^2 + c^2} \le b^2 + (a^2 + c^2) = 1.$

Equality is achieved if $\frac {c}{a} = \frac {\sqrt{3}}{1}, b^2 = a^2 + c^2 \implies a = \frac{1}{2 \sqrt{2}}, b = \frac{1}{ \sqrt{2}}, c = \frac{\sqrt{3}}{2 \sqrt{2}}.$

Answer: $1.$

2024 Problem 18 (EGE)

Find those values of the parameter a for which the system of equations has exactly one solution: $\left\{\begin{array}{l} x |y| = 3 - 2x ,\\a(2y + 1) = 3 - 2x.\end{array}\right.$ Solution

1. Special case $a = 0 \implies x = \frac{3}{2}, y = 0$ exactly one solution.

2. $|y| \ge 0 \implies \frac {3 - 2x}{x} \ge 0 \implies x \in \left [ 0; \frac{3}{2} \right ].$

3. We solve the first equation with respect $y$ and get $y = \pm \left( \frac{3}{x} - 2 \right ).$

This solution is shown in the diagram by red curve.

We solve the second equation with respect $y$ and get $y = \frac {3 - 2x}{2a} - \frac{1}{2} = -\frac {x}{a} + \frac{3 - a}{2a}.$ This solution is shown in the diagram by segments which connect point $\left ( \frac{3}{2}, -\frac{1}{2} \right)$ with axis $x = 0.$

Each solution of the system is shown by the point of crosspoint red curve with segment.

If $a = \frac {1}{3}$ then segment (colored by blue) is tangent to red curve (discriminant is zero), so we have two solutions (1,1) and $\approx(1.4,-0.4).$

If $a \in (0, \frac{1}{3})$ we get three solutions (colored by yellow).

In other cases the system has exactly one solution.

Answer: $(- \infty,0 ]\cup (\frac{1}{3}, \infty ).$

2024 Test problem 7

Find all values of the parameter a for which there is at least one solution to the inequality $\frac {1}{x} + \frac{2}{a} \le \frac {3}{a - x} - \frac{1}{x+a}$ on the interval $x \in [2,3]$

Solution

$\frac {1}{x} + \frac{2}{a} \le \frac {3}{a - x} - \frac{1}{x+a} \leftrightarrow F(x,a) \ge 0,$ where $F(x,a) = (a + 2x) \cdot\left( \frac{2}{a^2-x^2}-\frac{1}{ax} \right) = \frac {2(x+a/2)(x+a_1) (x-a_2)}{ax(x+a)(a - x)},$ where $a_1 = a(\sqrt{2}+1), a_2= a(\sqrt{2}-1).$

The equation $F(x,a) = 0$ has solutions $x= -\frac{a}{2}, x = -a_1,$ and $x = a_2.$ $F(2,a) = \frac {(4+a)(2+a_1) (2-a_2)}{2a(2+a)(a -2)}.$ $F(2,a) \ge 0$ if $a \in [-4, -2) \cup [2-2\sqrt{2},0) \cup (2, 2 \sqrt{2}+2],$ so given inequality has the solution $x=2$ for these $a.$ $F(3,a) = \frac {(6+a)(3+a_1) (3-a_2)}{3a(3+a)(a -3)}.$ $F(3,a) \ge 0$ if $a \in [-6, -3) \cup [3-3\sqrt{2},0) \cup (3, 3 \sqrt{2}+3],$ so given inequality has the solution $x=3$ for these $a.$

$a \in (-\infty, -6), x \in [2,3] \implies F(x,a) < 0,$ no solution of the given inequality.

$F(x,-2) < 0$ no solution of the inequality if $a = -2.$

$a \in (-2, 3(1 - \sqrt{2})).$ If $x \in [2,3] F(x,a) < 0 \implies$ no solution of the inequality.

$a \in (0, 2).$ If $x \in [2,3] F(x,a) < 0 \implies$ no solution of the given inequality.

$a \in (3 + 3\sqrt{2},\infty).$ If $x \in [2,3] F(a,x) < 0 \implies$ no solution of the given inequality.

2024 var 241 Problem 2

The natural numbers $a_1,...a_n$ form a strictly increasing arithmetic progression. Find all possible values of $n$ if it is known that $n$ is odd, $n > 1$ and $a_1 + a_2+...+a_n = 2024.$

Solution

$n$ is odd, so $a_1 + a_2+...+a_n = n a_\frac{n+1}{2} = 2024 = 8 \cdot 11 \cdot 23.$

Let $n = 11 \implies a_6 = 8 \cdot 23 = 184,$ the common difference may be $1,$ increasing arithmetic progression exist.

Let $n = 23 \implies a_{12} = 8 \cdot 11 = 88,$ the common difference may be $1,$ increasing arithmetic progression exist.

Let $n = 11 \cdot 23 = 253 \implies a_{127} = 8 \implies a_{100} < 0$ can not be the natural number.

Answer: $11,23.$

2024 var 242 Problem 7

The base of the pyramid is the trapezoid $ABCD, AD||BC, AD = 2BC.$

A sphere of radius $1$ touches the plane of the base of the pyramid and the planes of its lateral faces $ADS$ and $BCS$ at points $P,Q,$ and $T,$ respectively.

Find the ratio in which the volume of the pyramid is divided by the plane $ADT,$ if the face $ADS$ is perpendicular to the plane $ABD$ and the height of the pyramid is $4.$

Solution

A sketch of the given pyramid is shown in the diagram. The planes $SAD$ and $SBC$ intersect along the straight line $SK||AD||BC,$ that is, the planes $ABCD, SBC, SAD$ form the lateral surface of a prism into which a sphere with center at point $I$ is inscribed.

The plane $\pi$ containing the point $I$ and perpendicular to $AD$ contains points $P, Q,$ and $T.$ Plane $\pi$ intersects parallel lines $AD, BC,$ and $SK$ at points $N,L,$ and $K,$ respectively.

Let $EF$ be the line parallel to $AD, E \in BS, F \in CS.$ The plane $AETFD$ cuts off the pyramid $SAEFD$ with volume $v$ from the pyramid $SABCD$ with volume $V.$

$KN \perp AD$ and equal to the distance from $S$ to $AD, KN = 4, LN \perp AD, LN \perp KN$ and equal to the distance between $BC$ and $AD.$ $V = \frac {1}{3} KN \cdot KL \cdot \frac { AD + BC}{2} = BC \cdot \frac {KN \cdot KL}{2} = BC \cdot [KNL].$ Consider a right triangle $KLN ([KLN]$ is the area of $\triangle KLN)$ into which a circle $PQT$ with radius $r = 1$ is inscribed. $QN = PN = r, KQ = KN - NQ = 3 = KT, x = TL = PL \implies$ $(x+1)^2 + 4^2 = (x+3)^2 \implies x = 2 \implies \frac {EF}{BC} = \frac {KT}{KL} = \frac {3}{5}.$ We are looking for $v.$ Let $h$ be the distance from $S$ to the plane $AETFD.$ $v = \frac {1}{3} h \cdot NT \cdot \frac {AD + EF}{2} = BC \cdot (2 + \frac{3}{5}) \cdot \frac {h \cdot NT}{2} = BC \cdot \frac{13}{15} \cdot [KNT].$ $\frac {[KNT]} {[KNL]} = \frac {KT}{KL} = \frac {3}{5} \implies \frac {v}{V} = \frac{13}{15} \cdot \frac{3}{5} = \frac{13}{25} \implies \frac {v}{V-v} = \frac{13}{12}.$ Answer: $13 : 12.$

2024 var 243 Problem 6

Solve the system of equations in the positive $x, y, z:$ $\left\{\begin{array}{l} (x^2 + xy + y^2) (y^2 + yz + z^2)(z^2 + zx + x^2) = xyz,\\(x^4 + x^2\cdot y^2 + y^4) (y^4 + y^2 \cdot z^2 + z^4)(z^4 + z^2\cdot x^2 + x^4) = (xyz)^3.\end{array}\right.$

Solution (after Natalia Zakharova) $x^4 + x^2\cdot y^2 + y^4 = (x^2 + xy + y^2) \cdot (x^2 - xy + y^2),$ $y^4 + y^2 \cdot z^2 + z^4 = (y^2 + yz + z^2) \cdot (y^2 - yz + z^2),$ $z^4 + z^2\cdot x^2 + x^4 = (x^2 + xz + z^2) \cdot (x^2 - xz + z^2) \implies$ $\frac {(x^4 + x^2\cdot y^2 + y^4) (y^4 + y^2 \cdot z^2 + z^4)(z^4 + z^2\cdot x^2 + x^4)}{(x^2 + xy + y^2) (y^2 + yz + z^2)(z^2 + zx + x^2)} = (x^2 - xy + y^2) (y^2 - yz + z^2)(z^2 - zx + x^2) = (xyz)^2.$ $x^2 - xy + y^2 \ge 2xy - xy = xy, y^2 - yz + z^2 \ge yz, z^2 - zx + x^2 \ge zx \implies$ $(xyz)^2 = (x^2 - xy + y^2) (y^2 - yz + z^2)(z^2 - zx + x^2) \ge xy \cdot yz \cdot zx = (xyz)^2 \implies x = y = z \implies x = \frac{1}{3}.$ Answer: $x = y = z = \frac{1}{3}.$

2024 var 244 Problem 7

Let $ABCDA'B'C'D'$ be the cube, $AB = 1$ . Let $K \in A'B', L \in B'B,$ $M \in BC, N \in CD, Q \in DD', P \in A'D', \angle A'AK = \angle LAK,$ $\angle BAM = \angle MAN, \angle DAQ = \angle PAQ,$ $A'K + LB = BM + ND = DQ + PA' = \frac {5}{4}.$

Find the ratio in which the plane $KMQ$ divides the volume of the cube.

Solution

1. Let $F$ lie on the ray $B'A', A'F = BL.$ $AB = AA', \angle ABL = \angle AA'F \implies \triangle AA'F = \triangle ABL \implies AL = AF.$ $\angle KAF = \angle A'AF + \angle KAA' = \angle BAL + \angle KAA' =$ $= 90^\circ - \angle KAL = 90^\circ - \angle KAA' = \angle AKF \implies$ $AF = KF = KA' + BL = \frac {5}{4} = AL.$ So $BL = \sqrt{AL^2 – AB^2} = \frac {3}{4} \implies KA' = KB' = \frac {1}{2}.$

Similarly, $M$ is the midpoint $BC, Q$ is the midpoint $DD'.$

2. $KQ = KM = QM, AQ = AK = AM = C'M = C'K = C'Q =\frac {\sqrt{5}}{2} \implies$

regular pyramids are equal $AKMQ = C'KMQ.$ So $O$ (midpoint $AC'$ ) lies in plane $KMQ.$

Let $M'$ be the midpoint $A'D' \implies M'$ symmetric to $M$ with respect $O,$ so $M' \in KMQ.$

Similarly $K' \in KMQ, Q' \in KMQ,$ where $K'$ midpoint $CD, Q'$ the midpoint $BB'.$

For each point on the edges of the solid forming a part of the cube cut off by a plane $KMQ$ from the side of vertex $A,$ one can find a point symmetrical relative to the center of the cube $O$ on the edges of the solid forming another part of the cube.

It means that these parts are congruent and the plane $KMQ$ divides the cube in half.

Answer: $1:1.$

2024 var 245 Problem 6

Let $a,b,c,$ and $d$ be the positive real numbers such that $a+b+c+d = 1.$ Find the minimal value of $\frac {a^2}{1-a} + \frac {b^2}{1-b}+ \frac {c^2}{1-c} + \frac {d^2}{1-d}.$

Solution

$\frac {a^2}{1-a} + a = \frac {a}{1-a} = \frac{a}{b+c+d} = \frac{1}{\frac{b}{a}+\frac {c}{a}+\frac{d}{a}} \le \frac{1}{9} \left( \frac{a}{b}+\frac {a}{c}+\frac{a}{d}\right).$ For the last transform we use unequality between the harmonic mean and the arithmetic mean for three numbers. Therefore $\frac {a^2}{1-a} + \frac {b^2}{1-b}+ \frac {c^2}{1-c} + \frac {d^2}{1-d} \le \frac{1}{9} \left( \frac{a}{b}+\frac{b}{a}+\frac {a}{c}+\frac{c}{a}+\frac{a}{d}+\frac{d}{a}+\frac{b}{c}+\frac{c}{b}+\frac{b}{d}+\frac{d}{b}+\frac{c}{d}+\frac{d}{c}\right) - a - b - c - d \le \frac{12}{9} - 1 = \frac{1}{3}.$ Equality we get if $a = b = c = d = \frac {1}{4}.$

Answer: $\frac{1}{3}.$

2024 var 247 Problem 6

Real numbers $a, b,$ and $c$ satisfy the system of equations $\left\{\begin{array}{l} a + b + c = 4,\\a^2 + b^2 + c^2 = 8.\end{array}\right.$ Find the largest possible value of $c.$

Solution

In coordinates $a, b,$ and $c$ the first equation defines the plane $ABC,$ the second - a sphere with the center at the origin. They are shown in the diagram.

The solution of the given system (if it exists) is a circle symmetrical with respect to the plane $a = b.$ This plane intersects the plane of the first equation along the line $CED$ on which the points of maximum (E) and minimum (D) of the values of $c$ are located.

At these points the system takes the form $\left\{\begin{array}{l} 2a + c = 4,\\2a^2 + c^2 = 8.\end{array}\right.$ $(4 - c)^2 = 4a^2 = 2(8 - c^2) \implies c^2 - 8c = - 2 c^2.$ These system has two solutions $c_1 = 0, c_2 = \frac {8}{3},$ so solution of the given system exist.

Answer: $\frac {8}{3}.$

2024 var 247 Problem 7

Let the cube $ABCDA'B'C'D'$ be given. Let points $K \in A'B', L \in BC, M \in CD,$ and $N \in A'D'$ be given, $A'K = BL, A'N = DM.$

Let the plane $A'BD$ cross the plane $ALN$ by the line $\ell,$ and cross the plane $AKM$ by line $\ell'.$

Find the angle between $\ell$ and $\ell'.$

Solution

Denote $F = AL \cap BD, E = AN \cap A'D \implies \ell = EF,$

$E' = AM \cap BD, F' = A'B \cap AK \implies \ell' = E'F'.$ $A'K = BL, \angle BA'B' = \angle KA'F' = \angle CBD = \angle LBF, \angle AKA' = \angle F'KA' =$ $= \angle ALB = \angle FLB \implies \triangle A'KF' = \triangle BLF \implies A'F' = BF.$ Similarly, $A'E = DE'.$ $A'B = BD = DA' \implies A'BD$ is the regular triangle.