drop down menu

Euler Olympiad 2009-22 VIII (Russia) 113p

geometry problems from Euler Olympiads (remote (1st), regional (2nd) and final (3rd) rounds , from Russia) with aops links in the names
a Junior Competition

collected inside aops: here


Final Round (3rd)

2009 Euler Olympiad P3
In the triangle $ ABC $, the sides $ AB $ and $ BC $ are equal. The point $ D $ inside the triangle is such that the angle $ ADC $ is twice as large as the angle $ ABC $. Prove that the double distance from the point $ B $ to the line bisecting the angles externally to the angle $ ADC $ is $ AD + DC $.

2009 Euler Olympiad P6
In the convex quadrilateral $ ABCD $ the relations $ AB = BD $ are satisfied; $ \angle ABD = \angle DBC $. On the diagonal $ BD $ there was a point $ K $ such that $ BK = BC $. Prove that $ \angle KAD = \angle KCD $.

2010 Euler Olympiad P3
In the quadrilateral $ ABCD $, the side $ AB $ is equal to the diagonal $ AC $ and is perpendicular to the side $ AD $, and the diagonal $ AC $ is perpendicular to the side $ CD $. On the side $ AD $, a point $ K $ is taken such that $ AC = AK $. The bisector of the angle $ ADC $ intersects $ BK $ at the point $ M $. Find the angle $ ACM $.

2010 Euler Olympiad P6
In the convex quadrilateral $ ABCD $, the angles $ B $ and $ D $ are equal, $ CD = 4BC $, and the bisector of the angle $ A $ passes through the middle of the side $ CD $. What can the $ AD / AB $ ratio be?

2011 Euler Olympiad P4
Inside the convex quadrilateral $ ABCD $, in which $ AB = CD $, the point $ P $ is chosen in such a way that the sum of the angles $ PBA $ and $ PCD $ is $180$ degrees. Prove that $PB+PC<AD$.

2011 Euler Olympiad P6
The convex pentagon $ABCDE$ is such that $ AB \parallel CD $, $ BC \parallel AD $, $ AC \parallel DE $, $ CE \perp BC $. Prove that $EC$ is the bisector of the angle $BED$.

2012 Euler Olympiad P1
On the side $ BC $ of the triangle $ ABC $ the point $ D $ is taken in such a way that the perpendicular bisector of segment $ AD $ passes through the center of the circle inscribed in the triangle $ ABC $. Prove that this perpendicular passes through a vertice of the triangle $ ABC $.

2012 Euler Olympiad P7
The angles of the triangle $ABC$ satisfy the condition $2 \angle A + \angle B = \angle C$. Inside this triangle, the point $ K $ is chosen on the bisector of the angle $ A $ such that $BK = BC$. Prove that $\angle KBC = 2 \angle KBA$.

2013 Euler Olympiad P3
The diagonals of the convex quadrilateral $ ABCD $ are equal and intersect at the point $ O $. The point $ P $ inside the triangle $ AOD $ is such that $ CD \parallel BP $ and $ AB \parallel CP $. Prove that the point $ P $ lies on the bisector of the angle $ AOD $.

2013 Euler Olympiad P6
In the convex quadrilateral $ ABCD $ in which $ AB = CD $, the points $ K $ and $ M $ are chosen on the sides $ AB $ and $ CD $, respectively. It turned out that $ AM = KC $, $ BM = KD $. Prove that the angle between the lines $ AB $ and $ KM $ is equal to the angle between the lines $ KM $ and $ CD $.

On the side  $AB$ of a  triangle  $ABC$ with an angle of  $100 ^\circ$  at the vertice $C$,  points  $P$ and $Q$ such that  $AP = BC$ and $BQ = AC$ . Let $M$, $N$, $K$ be  midpoints of $AB$, $CP$, $CQ$, respectively. Find the angle $NMK$.

A diagonal of a convex $101$-gon will be called [i]principal [/i] if there are $50$ vertices on one side of it and $49$ vertices on the other. Several principal diagonals have been chosen that have no common ends. Prove that the sum of the lengths of these diagonals is less than the sum of the lengths of the remaining principal diagonals.

In the triangle $ABC$  , side  $AB$ is greater than side $BC$. On the extension of the side $BC$ beyond  the point  $C$,  noted the point $N$  so that  $2BN = AB+BC$. Let  $BS$ be the angle bisector of $ABC$, $M$ be the midpoint of the side $AC$, а $L$  be a point on the side $BS$  such that $ML \parallel AB$ . Prove that $2LN = AC$.

2015 Euler Olympiad P8
Let $CK$  be angle bisector of the triangle $ABC$. On the sides $BC$ and $AC$, are chosen points $L$ and $T$ respectively, such that $CT = BL$ and  $TL = BK$. Prove that the triangle $LTC$ is similar to the original one.

An equilateral triangle $ABC$. is given. Point $D$  is chosen on the extension of the side $AB$ beyond the point $A$, point $E$  on the extension of $BC$ beyond the point $C$, and point $F$ on the extension of $AC$ beyond the point $C$ so that  $CF = AD$ and $AC+EF = DE$. Find the angle $BDE$.

2016 Euler Olympiad P8
A parallelogram $ABCD$. is given. On the sides $AB$ and  $BC$ and the extension of the side  $CD$  beyond the point  $D$ , the points $K$, $L$ and $M$ are chosen, and so that the triangles $KLM$ and  $BCA$  are congruent  (precisely with such a correspondence of the vertices). The segment $KM$ intersects the segment $AD$ at the point $N$. Prove that $LN  \parallel  AB$.

Diagonals of a convex quadrilateral $ABCD$ intersect at a point $E$. It is known that $AB=BC=CD=DE=1$. Prove that $AD<2$

2017 Euler Olympiad P6
In a convex quadrilateral $ABCD$ angles $A$ and  $C$ are equal  $100^\circ$. On the sides $AB$ and $BC$ points $X$ and  $Y$ are selected  so that $AX = CY$ . It turned out that the line $YD$ is parallel to the angle bisector of the angle $ABC$. Find the angle $AXY$.

2018 Euler Olympiad P3
Diagonals of a convex quadrilateral   $ABCD$  are equal and intersect at a point $K$ . Inside the triangles $AKD$ and  $BKC$ select points  $P$ and  $Q$ so that  $\angle KAP = \angle KDP = \angle KBQ = \angle KCQ.$  Prove that line  $PQ$ is parallel to the bisector of the angle $AKD$.

2018 Euler Olympiad P8
Vertex  $F$ of a parallelogram $ACEF$ lies on the side  $BC$ of a parallelogram $ABCD$ . It is known that $AC = AD$ and $AE = 2CD$. Prove that $\angle CDE = \angle BEF.$

2019 Euler Olympiad P4
Given a convex quadrilateral $ ABSC. $ A point $ P $ is chosen on the diagonal $ BC $ so that $ AP = CP > BP. $ Point $ Q $ is symmetric to the point $ P $ with respect to the center of the diagonal $ BC $ ,  $ R $ is symmetric the point $ Q $ with respect to the line $ AC. $ It turned out that $ \angle SAB = \angle QAC $ and $ \angle SBC = \angle BAC. $ Prove that $ SA = SR. $

2019 Euler Olympiad P6
The points $ M $ and $ N $ are the midpoints of the sides $ AB $ and $ BC $ respectively of the triangle $ ABC. $ The point $ D$ is marked on the extension of the segment $ CM $ for the point $ M $.  It turned out that $ BC = BD = 2 $ and $ AN = 3. $ Prove that $ \angle ADC = 90^\circ. $

2020 Euler Olympiad P3
Given is a triangle ABC with 2<B=180°+<A. Point K is in the interior of the triangle, so that <ACK=2<BCK and L is on AB, so that BK=KL. Prove that CK+AL=AC

On every side of convex 100-gon are chosen two points which divide every side into 3 equal parts. We delete all vertices of the 100-gon. Prove that by knowing where the newly added points are, we can restore the 100-gon.

Points $P$ and $Q$ are selected on sides $AB$ and $BC$ of triangle $ABC$ respectively. The segments $CP$ and $AQ$ intersect at point $R$. It turned out that $AR = CR = PR + QR$. Prove that from the segments $AP, CQ$ and $PQ$ you can make a triangle, one of the angles of which is equal to angle $B$.

The diagonals of the trapezoid $ABCD$ ($AD\parallel BC$) meet at point $K$. Inside the triangle $ABK$, there is a point $M$ such that $\angle MBC = \angle MAD$, $\angle MCB = \angle MDA$. Prove that line $MK$ is parallel to the bases of the trapezoid.

Point $D$ is marked on side $BC$ of triangle $ABC$. A point $P$ is chosen on the side $AB$. The segments $PC$ and $AD$ intersect at the point $Q$. Point $R$ is the midpoint of segment $AP$. Prove that there exists a fixed point $X$ through which the line $RQ$ passes for any choice of the point $P$.

In a convex pentagon $ABCDE$, the diagonals $AD$ and $CE$ intersect at the point $X$. It turned out that $ABCX$ is a parallelogram and $BD = CX,$ $BE = AX.$ Prove that $AE = CD.$

Regional Round (2nd)

Point $ K $ is the midpoint of hypotenuse $ AB $ of right-angled isosceles triangle $ ABC $. Points $ L $ and $ M $ are chosen on legs $ BC $ and $ AC $, respectively, such that $ BL = CM $. Prove that $ LMK $ is also a right-angled isosceles triangle.

In the convex quadrilateral $ ABCD $, a point of the diagonal $ AC $ lies on the perpendicular bisectors of the sides $ AB $ and $ CD $, and a point of the diagonal $ BD $ lies on the perpendicular bisectors of the sides $ AD $ and $ BC $. Prove that $ ABCD $ is a rectangle.

On the hypotenuse $ BC $ of the right-angled triangle $ ABC $, the point $ K $ is chosen so that $ AB = AK $. The segment $ AK $ intersects the angle bisector $ CL $ in its midpoint . Find the acute angles of the triangle $ ABC $.

The bisectors of the angles $ A $ and $ C $ of the trapezoid $ ABCD $ intersect at the point $ P $, and the bisectors of the angles $ B $ and $ D $ intersect at the point $ Q $, different from $ P $. Prove that if the segment $ PQ $ is parallel to the base $ AD $, then the trapezoid is isosceles.

You are given a convex quadrilateral $ ABCD $ such that $ AD = AB + CD $. It turned out that the bisector of the angle $ A $ passes through the midpoiint of the side $ BC $. Prove that the bisector of $ D $ also passes through the midpoint of $ BC $.

In triangle $ ABC $ the points $ M $ and $ N $ are the midpoints of the sides $ AC $ and $ AB $, respectively. A point $ P $ is chosen on the median $ BM $ that does not lie on $ CN $. It turned out that $ PC = 2PN $. Prove that $ AP = BC $.

The trapezoid $ ABCD $ with bases $ AD $ and $ BC $ is such that the angle $ ABD $ is right and $ BC + CD = AD $. Find the ratio of the bases $ AD: BC $.

In the convex quadrilateral $ ABCD $, the angles $ ABC $ and $ ADC $ are right. Points $ K $, $ L $, $ M $, $ N $ are taken on the sides $ AB $, $ BC $, $ CD $, $ DA $, respectively, so that $ KLMN $ is a rectangle. Prove that the midpoint of the diagonal $ AC $ is equidistant from the lines $ KL $ and $ MN $.

The point $ M $ is marked on the segment $ AB $. The points $ P $ and $ Q $ are the midpoints of the segments $ AM $ and $ BM$ respectively, the point $ O $ is the midpoint of the segment $ PQ $. Let's choose point $ C $ so that the angle $ ACB $ is right. Let $ MD $ and $ ME $ be the perpendiculars dropped from the point $ M $ onto the lines $ CA $ and $ CB $, and $ F $ the midpoint of the segment $ DE $. Prove that the length of the segment $ OF $ does not depend on the choice of the point $ C $.

Given an acute-angled triangle $ ABC $. The altitude $ AA_1 $ is extended beyond the top $ A $ to the segment $ AA_2 = BC $. The altitude $ CC_1 $ is extended beyond the top $ C $ to the segment $ CC_2 = AB $. Find the angles of the triangle $ A_2BC_2 $.

On the side $ AC $ of the triangle $ ABC $, a point $ D $ is chosen such that $ BD = AC $. The median $ AM $ of this triangle intersects the segment $ BD $ at the point $ K $. It turned out that $ DK = DC $. Prove that $ AM + KM = AB $.

Given a convex pentagon $ ABCDE $, where the line $ BE $ is parallel to the line $ CD $ and the segment $ BE $ is shorter than the segment $ CD $. The points $ F $ and $ G $ are selected inside the pentagon in such a way that $ ABCF $ and $ AGDE $ are parallelograms. Prove that $ CD = BE + FG $.

The perpendicular bisectors to the sides $ AB $ and $ BC $ of the convex quadrilateral $ ABCD $ meet the sides $ CD $ and $ DA $ at the points $ P $ and $ Q $, respectively. It turned out that $ \angle APB = \angle BQC $. Inside the quadrangle, a point $ X $ is selected such that $ QX \parallel AB $ and $ PX \parallel BC $. Prove that the line $ BX $ bisects the diagonal of $ AC $..

In trapezoid $ ABCD $, where $ AD \parallel BC $, the angle $ B $ is equal to the sum of the angles $ A $ and $ D $. On the extension of the segment $ CD $ beyond the vertex $ D $, draw the segment $ DK = BC $. Prove that $ AK = BK $.

In trapezoid $ ABCD $, point $ M $ is the midpoint of the base of $ AD $. It is known that $ \angle ABD = 90 ^\circ $ and $ BC = CD $. On the segment $ BD $, a point $ F $ is chosen such that $ \angle BCF = 90 ^\circ $. Prove that $ MF \perp CD $.

The points $ M $ and $ N $ are the midpoints of the angle bisectors $ AK $ and $ CL $ of the triangle $ ABC $, respectively. Prove that an angle $ ABC $ is right if and only if $ \angle MBN = 45 ^\circ $.
On the lateral sides $ AB $ and $ AC $ of an isosceles triangle $ ABC $, the points $ P $ and $ Q $ are selected, respectively, so that $ PQ \parallel BC $. On the angle bisectors of triangles $ ABC $ and $ APQ $ outgoing from the vertices $ B $ and $ Q $, the points $ X $ and $ Y $, respectively, are chosen so that $ XY \parallel BC $. Prove that $ PX = CY $.

In convex quadrilateral $ ABCD $, the bisector of angle $ B $ passes through the midpoint of side $ AD $, and $ \angle C = \angle A + \angle D $. Find the angle $ ACD $.

Inside the parallelogram $ ABCD $, the point $ E $ is chosen so that $ AE = DE $ and $ \angle ABE = 90 ^ \circ. $ Point $ M $ is the midpoint of the segment $ BC. $ Find the angle $ DME. $

Point $ D $ is selected on the angle bisector $ AL $ of the triangle $ ABC $. It is known that $ \angle BAC = 2 \alpha$, $ \angle ADC = 3 \alpha$, $ \angle ACB = 4 \alpha$. Prove, that $ BC + CD = AB. $

The perimeter of the triangle $ ABC $ is $2$. The point $ P $ is marked on the side $ AC $, and the point $ Q $ is marked on the segment $ CP $ so that $ 2AP = AB $ and $ 2QC = BC. $ Prove that the perimeter of the triangle $ BPQ $ is greater than $1$.

Point $ N $ is the midpoint of side $ BC $ of triangle $ ABC, $ in which $ \angle ACB = 60 ^\circ $. Point $ M $ on side $ AC $ is such that $ AM = BN. $ Point $ K $ is the midpoint of segment $ BM. $ Prove that $ AK = KC. $ ympi

The bisector of the angle $ A $ of the convex quadrilateral $ ABCD $ meets the side $ CD $ at the point $ K. $ It turns out that $ DK = BC $ and $ KC + AB = AD. $ Prove that $ \angle BCD = \angle ADC. $

On the midline of an equilateral triangle $ ABC, $ parallel to the side $ BC, $ point $ D$ is taken. Point $ E $ on the extension of side $ BA $ beyond point $ A $ is such that $ \angle ECA = \angle DCA. $ Point $ F $ on the extension of the side $ CA $ beyond the point $ A $ is such that $ \angle FBA = \angle DBA. $ Prove that the point $ A $ lies on the midline of the triangle $ DEF, $ parallel to the side $ EF. $

$CL$ is the angle bisector of triangle $ABC$. $CLBK$ is a parallelogram. Line $AK$ intersects
segment $CL$ at point $P$. It turned out that point $P$ is equidistant from the diagonals of the
parallelogram $CLBK$. Prove that $AK \ge CL$. 2021 Euler Olympiad Regional 2.2 Point $M$ is the midpoint of side $AC$ of equilateral triangle $ABC$. Points $P$ and $R$ on the
segments $AM$ and $BC$ are respectively chosen so that $AP = BR$. Find the sum of the angles
$ARM, PBM$, and $BMR$.

Angle bisectors $BK$ and $CL$ are drawn in triangle $ABC$. A point $N$ is marked on the segment
$BK$ so that $LN \parallel AC.$ It turned out that $NK = LN$. Find the measure of the angle $ABC$. 2022 Euler Olympiad Regional 1.5
What is the largest $n$ for which there exists a convex $n$-gon whose diagonal lengths are at most two
Is there a triangle in which the lengths of non-coinciding medians and altitudes drawn from one of its
vertices are respectively equal to the lengths of the two sides of this triangle?


Given an isosceles triangle $ ABC $ $ (AC = BC) $. Points $ A_1 $, $ B_1 $ and $ C_1 $ are marked on the sides $ BC $, $ AC $, $ AB $, respectively. It turned out that $ C_1B_1 $ is perpendicular to $ AC $, $ B_1A_1 $ is perpendicular to $ BC $ and $ B_1A_1 = B_1C_1 $. Prove that $ A_1C_1 $ is perpendicular to $ AB $.

The two angle bisectors of the triangle intersect at an angle of $60$ degrees. Prove that one of the angles of this triangle is $60$ degrees.

The length of the rectangle has been decreased by $ 10 \% $ and the width has been decreased by $ 20 \% $. In this case, the perimeter of the rectangle has decreased by $ 12 \% $. By what percentage will the perimeter of the rectangle decrease if its length is reduced by $ 20 \% $, and reduce the width by $ 10 \% $?

Can the distances from a point on the plane to the vertices of a certain square be equal to $1, 1, 2$ and $3$?

The point $ D $ lies on the hypotenuse $ AB $ of the right-angled triangle $ ABC $, but does not coincide with its middle. Prove that there are no equal segments among the segments $ AD $, $ BD $ and $ CD $.

In triangle $ ABC $, the median $ BM $ is half the side of $ AB $ and makes an angle of $40$ degrees with it. Find the angle $ ABC $.

In a right-angled triangle, the altitude dropped on the hypotenuse is four times shorter than the hypotenuse. Find the acute angles of the triangle.

There are four properties of quads:
(1) opposite sides are equal in pairs;
(2) two opposite sides are parallel;
(3) some two adjacent sides are equal;
(4) the diagonals are perpendicular and divided by the point of intersection in the same ratio.
One of these two quadrilaterals has some two of these properties, the other two others.
Prove that one of these two quads is a rhombus.

The points $ E $ and $ F $ are the midpoints of the sides $ BC $ and $ CD $ respectively of the rectangle $ ABCD $. Prove that $ AE <2EF $.

Inside the angle $ AOB $, equal to $ 120 ^\circ $, the rays $ OC $ and $ OD $ are drawn so that each of them is the bisector of one of the angles obtained in the drawing. Find the value of angle $ AOC $. Indicate all possible options.

An arbitrary point $ D $ on the median $ BM $ is marked in the triangle $ ABC $. Then a line parallel to $ AB $ was drawn through $ D $, and a line parallel to $ BM $ through $ C $. These lines intersect at the point $ E $. Prove that $ BE = AD $.

The point $ E $ is marked on the side $ BC $ of the triangle $ ABC $, and the point $ F $ is marked on the angle bisector $ BD $ in such a way that $ EF \parallel AC $ and $ AF = AD $. Prove that $ AB = BE $.

Divide a $ 30 ^ \circ $ right-angled triangle into two smaller triangles so that some median of one of these triangles is parallel to one of the angle bisectors of the second triangle.

Given a pentagon $ ABCDE $ such that $ AB = BC = CD = DE $, $ \angle B = 96 ^ \circ $ $ \angle C = \angle D = 108 ^ \circ $. Find $ \angle E $.

In triangle $ ABC $, the angle $ C $ is three times the angle $ A $, and the side $ AB $ is twice the side $ BC $. Prove that the angle $ ABC $ is $60$ degrees.

The angle bisector $ BL $ is drawn in the triangle $ ABC $, and the point $ K $ is chosen on its extension beyond the point $ L $, for which $ LK = AB $. It turned out that $ AK \parallel BC $. Prove that $ AB> BC $.

Inside the acute angle $ BAC $, we took such a point $ D $ that the angle $ CAD $ is twice the angle $ BAD $. Could point $ D $ be twice as far from line $ AC $ as from line $ AB $?

Can five identical rectangles with a perimeter of $10$ make one rectangle with a perimeter of $22$?

In a triangle $ ABC $ , $ AC = 1 $, $ AB = 2 $, $ O $ is the intersection point of the angle bisectors. A segment passing through point $ O $ parallel to side $ BC $ intersects sides $ AC $ and $ AB $ at points $ K $ and $ M $, respectively. Find the perimeter of the triangle $ AKM $.

In a triangle $ ABC $ $ AB = BC $. Points $ D $, $ E $, and $ F $ are marked on the rays $ CA $, $ AB $ and $ BC $, respectively, so that $ AD = AC $, $ BE = BA $, $ CF = CB $. Find the sum of the angles $ ADB $, $ BEC $ and $ CFA $.

Point $ E $ is the midpoint of the base $ AD $ of trapezoid $ ABCD $. The segments $ BD $ and $ CE $ meet at the point $ F $. It is known that $ AF \perp BD $. Prove that $ BC = FC $.

Point $ D $ lies inside triangle $ ABC $. Could it be that the shortest side of triangle $ BCD $ is $1$, the shortest side of triangle $ ACD $ is $2$, and the shortest side of triangle $ ABD $ is $3$?

A smaller square was cut out of the square, one of the sides of which lies on the side of the original square. The perimeter of the resulting octagon is $ 40 \% $ larger than the perimeter of the original square. How many percent is its area less than the original square?

In parallelogram $ ABCD $ with side $ AB = 1 $, point $ M $ is the midpoint of side $ BC $, and the angle $ AMD $ is $90$ degrees. Find the side $ BC $.

The diagonals $ AD $ and $ BE $ of the convex pentagon $ ABCDE $ meet at the point $ P $. It is known that $ AC = CE = AE $, $ \angle APB = \angle ACE $ and $ AB + BC = CD + DE $. Prove that $ AD = BE $.

The square is cut into rectangles of equal area as shown in the figure. Find the area of the square if the segment $ AB $ is $1$.
Inside the angle $ BAC $, equal to $ 45 ^\circ $, point $ D $ is taken so that each of the angles $ ADB $ and $ ADC $ is $ 45 ^\circ $. Points $ D_1 $ and $ D_2 $ are symmetric to point $ D $ with respect to lines $ AB $ and $ AC $, respectively. Prove that the points $ D_1 $, $ D_2 $, $ B $ and $ C $ are collinear.

Points $ D $ and $ E $ are marked on the side $ AC $ of triangle $ ABC $ with an angle of $120$ degrees at the vertex $ B $ such that $ AD = AB $ and $ CE = CB $. From the point $ D $, the perpendicular $ DF $ is dropped to the line $ BE $. Find the ratio $ BD / DF $.

On the sides $ AB $, $ BC $, $ CD $ and $ DA $ of the quadrilateral $ ABCD $, the points $ K $, $ L $, $ M $, $ N $ are selected, respectively, so that $ AK = AN $, $ BK = BL $, $ CL = CM $, $ DM = DN $ and $ KLMN $ is a rectangle. Prove that $ ABCD $ is a rhombus.

The median $ BM $ is drawn in the triangle $ ABC $. It is known that $ \angle ABM = 40 ^ \circ $ and $ \angle CBM = 70 ^ \circ $. Find the ratio $ AB: BM $.

In the triangle $ ABC $ the angle bisector $ BD $ was drawn, in the triangle $ BDC $ the angle bisector $ DE $, and in the triangle $ DEC $ the angle bisector $ EF $. It turned out that the lines $ BD $ and $ EF $ are parallel. Prove that the angle $ ABC $ is twice the angle $ BAC $.

The bisectors of the angles $ A $ and $ C $ cut the non-isosceles triangle $ ABC $ into a quadrilateral and three triangles, and among these three triangles there are two isosceles. Find the angles of the triangle $ ABC $.

The squares with sides $11, 9, 7$, and $5$ are located approximately as in the picture below. It turned out that the area of the gray parts is twice as large as the area of the black parts. Find the area of the white parts.
In triangle $ ABC $, point $ M $ is the midpoint of $ AC $, in addition, $ BC = 2AC / 3 $ and $ \angle BMC = 2 \angle ABM $. Find the ratio $ AM / AB $ .

Can the median and the angle bisector drawn from the vertex $ A $ of an acute-angled triangle $ ABC $ divide the altitude $ BH $ of this triangle into three equal parts?

In triangle $ ABC $, the angle $ C $ is $2$ times the angle $ B $, $ CD $ is the angle bisector. From the midpoint $ M $ of the side $ BC $ the perpendicular $ MH $ is drawn on the segment $ CD $. There is a point $ K $ on the side $ AB $ such that $ KMH $ is an equilateral triangle. Prove that the points $ M $, $ H $ and $ A $ are collinear.

The diagonals of the parallelogram $ ABCD $ meet at the point $ O $. The point $ P $ such that $ DOCP $ is also a parallelogram ($ CD $ is its diagonal). Let $ Q $ denote the intersection point of $ BP $ and $ AC $, and let $ R $ be the intersection point of $ DQ $ and $ CP $. Prove that $ PC = CR $.

In trapezoid $ ABCD $ $ (AD \parallel BC) $ $ AD = 2 $, $ BC = 1 $, $ \angle ABD = 90 ^ \circ $. Find the side $ CD $.

The teacher drew a rectangle $ ABCD $ on the blackboard. Student Petya divided this rectangle into two rectangles with a straight line parallel to the side $ AB $. It turned out that the areas of these parts are $1: 2$, and the perimeters are $3: 5$ (in the same order). Pupil Vasya divided this rectangle into two parts by a straight line parallel to the $ BC $ side. The area of the new parts is also $1: 2$. What is the ratio of their perimeters ?

Point $ K $ is taken in triangle $ ABC $ on side $ BC $. $ KM $ and $ KP $ are the angle bisectors of triangles $ AKB $ and $ AKC $, respectively. It turned out that the diagonal $ MK $ divides the quadrilateral $ BMPK $ into two equal triangles. Prove that $ M $ is the midpoint of $ AB $.

Petya marks four points on the plane so that all of them cannot be crossed out by two parallel straight lines. Vasya chooses two of the lines passing through pairs of points, measures the angle between them and pays Petya an amount equal to the degree measure of the angle. What is the largest amount Petya can guarantee himself?

The angle bisector $ BD $ was drawn in the triangle $ ABC $, and in the triangles $ ABD $ and $ CBD $ the angle bisectors $ DE $ and $ DF $, respectively. It turned out that $ EF \parallel AC $. Find the angle $DEF $ .

Borya drew nine segments, three of which are equal to three altitudes of the $ ABC $ triangle, three to three angle bisectors, three to three medians. It turned out that for any of the drawn segments, among the other eight, there is an equal to it. Prove that triangle $ ABC $ is isosceles.

The stick is broken into $15$ parts so that no three parts can be used to form a triangle. Prove that among the pieces there is one that is longer than a third of the original stick.

In a convex pentagon $ ABCDE $, $ AB $ is parallel to $ DE $, $ CD = DE $, $ CE $ is perpendicular to $ BC $ and $ AD $. Prove that the line passing through $ A $ parallel to $ CD $, the line passing through $ B $ parallel to $ CE $, and the line passing through $ E $ parallel to $ BC $, intersect at one point.

In trapezoid $ ABCD $, the side of $ AB $ is equal to the diagonal of $ BD. $ The point $ M $ is the midpoint of the diagonal $ AC $. Line $ BM $ intersects $ CD $ at point $ E. $ Prove that $ BE = CE. $

The point $ P $ is located inside the triangle $ ABC $. On the side $ BC $, a point $ H $ is selected that does not coincide with the midpoint of the side. It turned out that the bisector of the angle $ AHP $ is perpendicular to the side $ BC $, the angle $ ABC $ is equal to the angle $ HCP $ and $ BP = AC $. Prove that $ BH = AH $.

In trapezoid $ ABCD $, the base $ AD $ is greater than the lateral side $ CD. $ The bisector of the angle $ D $ meets the side $ AB $ at the point $ K. $ Prove that $ AK> KB. $

Inside the trapezoid $ ABCD $ $ (BC \parallel AD), $ where $ AD = 2BC, $ is the point $ F, $ for which $ AB = FB. $ The point $ M $ is the midpoint of the segment $ FD. $ Prove that $ CM \perp FA. $

The points $ D $ and $ E $ lie on the extensions of the sides $ AB $ and $ BC $ of an acute-angled triangle $ ABC $ beyond the points $ B $ and $ C $, respectively. The points $ M $ and $ N $ are the midpoints of the segments $ AE $ and $ DC. $ Prove that $ MN> AD / 2. $

In an acute-angled triangle $ ABC $, the angle at the vertex $ A $ is $45$ degrees. Prove that the perimeter of this triangle is less than twice the sum of its altitudes drawn from the vertices $ B $ and $ C $.

In triangle $ABC$ ($\angle C = 90^o$) on leg $BC$, points $K$ and $L$ are marked such that $\angle CAK = \angle KAL = \angle LAB$. A point $M$ is marked on the hypotenuse $AB$ such that $ML = KL$. Prove that the perpendicular from the point $C$ to the line $AK$ does not bisect the segment $ML$.

$AH$ is the altitude of an isosceles triangle $ABC$ ($AB = BC$). $HK$ is the altitude of the triangle $AHB$. It turned out that $4HK = AB$. What could be the measure of the angle $ABC$?

In trapezoid $ABCD$, the bisector of angle $B$ intersects the base $AD$ at point $L$. Point $M$ is the midpoint of side $CD$. A straight line parallel to $BM$ and passing through $L$ intersects side $AB$ at point $K$. It turned out that the angle $BLM$ is a straight line. Find the ratio $BK / KA$ .

Given a triangle $ABC$ in which $AB = BC$. On the side $BC$ there is a point $D$ such that $CD = AC$. The point $E$ on the ray $DA$ is such that $DE = AC$. Which segment is longer , $EC$ or $AC?$

$ABCD$ is a convex quadrilateral, where $AB = 7,$ $BC = 4,$ $AD = DC,$ $\angle ABD = \angle DBC.$ The point $E$ on the segment $AB$ is such that $\angle DEB = 90^\circ.$ Find the length of the segment $AE.$

source: matol.kz/nodes/74 (official site)

No comments:

Post a Comment

Subscribe to: Posts (Atom)