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Question Number 17524 Answers: 1 Comments: 0

The circle ω touches the circle Ω internally at P. The centre O of Ω is outside ω. Let XY be a diameter of Ω which is also tangent to ω. Assume PY > PX. Let PY intersect ω at Z. If YZ = 2PZ, what is the magnitude of ∠PYX in degrees?

$$\mathrm{The}\:\mathrm{circle}\:\omega\:\mathrm{touches}\:\mathrm{the}\:\mathrm{circle}\:\Omega \\ $$$$\mathrm{internally}\:\mathrm{at}\:{P}.\:\mathrm{The}\:\mathrm{centre}\:{O}\:\mathrm{of}\:\Omega\:\mathrm{is} \\ $$$$\mathrm{outside}\:\omega.\:\mathrm{Let}\:{XY}\:\mathrm{be}\:\mathrm{a}\:\mathrm{diameter}\:\mathrm{of}\:\Omega \\ $$$$\mathrm{which}\:\mathrm{is}\:\mathrm{also}\:\mathrm{tangent}\:\mathrm{to}\:\omega.\:\mathrm{Assume} \\ $$$${PY}\:>\:{PX}.\:\mathrm{Let}\:{PY}\:\mathrm{intersect}\:\omega\:\mathrm{at}\:{Z}.\:\mathrm{If} \\ $$$${YZ}\:=\:\mathrm{2}{PZ},\:\mathrm{what}\:\mathrm{is}\:\mathrm{the}\:\mathrm{magnitude}\:\mathrm{of} \\ $$$$\angle{PYX}\:\mathrm{in}\:\mathrm{degrees}? \\ $$

Question Number 17520 Answers: 1 Comments: 1

Find the coordinate of the point in RΛ3 which is the reflection the point (1,2,3) with respect to plane X+Y+Z=1 .

$$\mathrm{Find}\:\mathrm{the}\:\mathrm{coordinate}\:\mathrm{of}\:\mathrm{the}\:\mathrm{point}\:\mathrm{in} \\ $$$$\mathrm{R}\Lambda\mathrm{3}\:\mathrm{which}\:\mathrm{is}\:\mathrm{the}\:\mathrm{reflection}\:\mathrm{the}\:\mathrm{point} \\ $$$$\left(\mathrm{1},\mathrm{2},\mathrm{3}\right)\:\mathrm{with}\:\mathrm{respect}\:\mathrm{to}\:\mathrm{plane}\: \\ $$$$\mathrm{X}+\mathrm{Y}+\mathrm{Z}=\mathrm{1}\:. \\ $$

Question Number 17449 Answers: 1 Comments: 4

Between 2:00 and 2:15, what time is it exactly when the hour, minute, and second′s hand of a clock occupy the same angular position.

$$\mathrm{Between}\:\mathrm{2}:\mathrm{00}\:\mathrm{and}\:\mathrm{2}:\mathrm{15},\:\mathrm{what}\:\mathrm{time} \\ $$$$\mathrm{is}\:\mathrm{it}\:\mathrm{exactly}\:\mathrm{when}\:\mathrm{the}\:\mathrm{hour}, \\ $$$$\mathrm{minute},\:\mathrm{and}\:\mathrm{second}'\mathrm{s}\:\mathrm{hand}\:\mathrm{of}\:\mathrm{a} \\ $$$$\mathrm{clock}\:\mathrm{occupy}\:\mathrm{the}\:\mathrm{same}\:\mathrm{angular} \\ $$$$\mathrm{position}. \\ $$$$ \\ $$

Question Number 17440 Answers: 1 Comments: 0

3x−4y=0,4y−5z=0,5z−3x=0 then x,y,z is AP,GP,HP,AGP??????

$$\mathrm{3}{x}−\mathrm{4}{y}=\mathrm{0},\mathrm{4}{y}−\mathrm{5}{z}=\mathrm{0},\mathrm{5}{z}−\mathrm{3}{x}=\mathrm{0} \\ $$$${then}\:{x},{y},{z}\:{is}\:{AP},{GP},{HP},{AGP}?????? \\ $$

Question Number 17645 Answers: 2 Comments: 1

Suppose that the point M lying in the interior of the parallelogram ABCD, two parallels to AB and AD are drawn, intersecting the sides of ABCD at the points P, Q, R, S (See Figure). Prove that M lies on the diagonal AC if and only if [MRDS] = [MPBQ].

$$\mathrm{Suppose}\:\mathrm{that}\:\mathrm{the}\:\mathrm{point}\:{M}\:\mathrm{lying}\:\mathrm{in}\:\mathrm{the} \\ $$$$\mathrm{interior}\:\mathrm{of}\:\mathrm{the}\:\mathrm{parallelogram}\:{ABCD}, \\ $$$$\mathrm{two}\:\mathrm{parallels}\:\mathrm{to}\:{AB}\:\mathrm{and}\:{AD}\:\mathrm{are}\:\mathrm{drawn}, \\ $$$$\mathrm{intersecting}\:\mathrm{the}\:\mathrm{sides}\:\mathrm{of}\:{ABCD}\:\mathrm{at}\:\mathrm{the} \\ $$$$\mathrm{points}\:{P},\:{Q},\:{R},\:{S}\:\left(\mathrm{See}\:\mathrm{Figure}\right).\:\mathrm{Prove} \\ $$$$\mathrm{that}\:{M}\:\mathrm{lies}\:\mathrm{on}\:\mathrm{the}\:\mathrm{diagonal}\:{AC}\:\mathrm{if}\:\mathrm{and} \\ $$$$\mathrm{only}\:\mathrm{if}\:\left[{MRDS}\right]\:=\:\left[{MPBQ}\right]. \\ $$

Question Number 17373 Answers: 2 Comments: 0

Find the point in interior of a convex quadrilateral such that the sum of its distances to the 4 vertices is minimal. Find the point in interior of a convex quadrilateral such that the sum of its distances to the 4 sides is minimal.

$$\mathrm{Find}\:\mathrm{the}\:\mathrm{point}\:\mathrm{in}\:\mathrm{interior}\:\mathrm{of}\:\mathrm{a}\:\mathrm{convex} \\ $$$$\mathrm{quadrilateral}\:\mathrm{such}\:\mathrm{that}\:\mathrm{the}\:\mathrm{sum}\:\mathrm{of}\:\mathrm{its} \\ $$$$\mathrm{distances}\:\mathrm{to}\:\mathrm{the}\:\mathrm{4}\:\mathrm{vertices}\:\mathrm{is}\:\mathrm{minimal}. \\ $$$$ \\ $$$$\mathrm{Find}\:\mathrm{the}\:\mathrm{point}\:\mathrm{in}\:\mathrm{interior}\:\mathrm{of}\:\mathrm{a}\:\mathrm{convex} \\ $$$$\mathrm{quadrilateral}\:\mathrm{such}\:\mathrm{that}\:\mathrm{the}\:\mathrm{sum}\:\mathrm{of}\:\mathrm{its} \\ $$$$\mathrm{distances}\:\mathrm{to}\:\mathrm{the}\:\mathrm{4}\:\mathrm{sides}\:\mathrm{is}\:\mathrm{minimal}. \\ $$

Question Number 17273 Answers: 1 Comments: 2

The intersection of the ABC triangle median is at G point. The corner of the BGC is 90°. If the AG cut length is 12 cm, locate the BC side.

Question Number 17177 Answers: 1 Comments: 0

Question Number 17158 Answers: 0 Comments: 4

Please solve Q. 16069. Ask from me the solution if needed and please explain it.

$$\mathrm{Please}\:\mathrm{solve}\:\mathrm{Q}.\:\mathrm{16069}.\:\mathrm{Ask}\:\mathrm{from}\:\mathrm{me}\:\mathrm{the} \\ $$$$\mathrm{solution}\:\mathrm{if}\:\mathrm{needed}\:\mathrm{and}\:\mathrm{please}\:\mathrm{explain}\:\mathrm{it}. \\ $$

Question Number 16980 Answers: 0 Comments: 1

To Q16066: I have posted my solution there. Those who are intetested in this interesting question please have a critical view at it. Maybe there are alternative solutions which are easier and more direct and straight on.

$$\mathrm{To}\:\mathrm{Q16066}: \\ $$$$\mathrm{I}\:\mathrm{have}\:\mathrm{posted}\:\mathrm{my}\:\mathrm{solution}\:\mathrm{there}. \\ $$$$\mathrm{Those}\:\mathrm{who}\:\mathrm{are}\:\mathrm{intetested}\:\mathrm{in}\:\mathrm{this}\:\mathrm{interesting} \\ $$$$\mathrm{question}\:\mathrm{please}\:\mathrm{have}\:\mathrm{a}\:\mathrm{critical}\:\mathrm{view}\:\mathrm{at} \\ $$$$\mathrm{it}.\:\mathrm{Maybe}\:\mathrm{there}\:\mathrm{are}\:\mathrm{alternative}\:\mathrm{solutions} \\ $$$$\mathrm{which}\:\mathrm{are}\:\mathrm{easier}\:\mathrm{and}\:\mathrm{more}\:\mathrm{direct}\:\mathrm{and} \\ $$$$\mathrm{straight}\:\mathrm{on}. \\ $$

Question Number 16958 Answers: 0 Comments: 0

Let ABCD be a quadrilateral with an inscribed circle. Prove that the circles inscribed in triangles ABC and ADC are tangent to each other.

$$\mathrm{Let}\:{ABCD}\:\mathrm{be}\:\mathrm{a}\:\mathrm{quadrilateral}\:\mathrm{with}\:\mathrm{an} \\ $$$$\mathrm{inscribed}\:\mathrm{circle}.\:\mathrm{Prove}\:\mathrm{that}\:\mathrm{the}\:\mathrm{circles} \\ $$$$\mathrm{inscribed}\:\mathrm{in}\:\mathrm{triangles}\:{ABC}\:\mathrm{and}\:{ADC} \\ $$$$\mathrm{are}\:\mathrm{tangent}\:\mathrm{to}\:\mathrm{each}\:\mathrm{other}. \\ $$

Question Number 16951 Answers: 0 Comments: 0

Let M be a point in interior of ΔABC. Three lines are drawn through M, parallel to triangle′s sides, thereby producing three trapezoids. Suppose a diagonal is drawn in each trapezoid in such a way that the diagonals have no common endpoints. These three diagonals divide ABC into seven parts, four of them being triangles. Prove that the area of one of the four triangles equals the sum of the areas of the other three.

$$\mathrm{Let}\:{M}\:\mathrm{be}\:\mathrm{a}\:\mathrm{point}\:\mathrm{in}\:\mathrm{interior}\:\mathrm{of}\:\Delta{ABC}. \\ $$$$\mathrm{Three}\:\mathrm{lines}\:\mathrm{are}\:\mathrm{drawn}\:\mathrm{through}\:{M}, \\ $$$$\mathrm{parallel}\:\mathrm{to}\:\mathrm{triangle}'\mathrm{s}\:\mathrm{sides},\:\mathrm{thereby} \\ $$$$\mathrm{producing}\:\mathrm{three}\:\mathrm{trapezoids}.\:\mathrm{Suppose}\:\mathrm{a} \\ $$$$\mathrm{diagonal}\:\mathrm{is}\:\mathrm{drawn}\:\mathrm{in}\:\mathrm{each}\:\mathrm{trapezoid}\:\mathrm{in} \\ $$$$\mathrm{such}\:\mathrm{a}\:\mathrm{way}\:\mathrm{that}\:\mathrm{the}\:\mathrm{diagonals}\:\mathrm{have}\:\mathrm{no} \\ $$$$\mathrm{common}\:\mathrm{endpoints}.\:\mathrm{These}\:\mathrm{three} \\ $$$$\mathrm{diagonals}\:\mathrm{divide}\:{ABC}\:\mathrm{into}\:\mathrm{seven} \\ $$$$\mathrm{parts},\:\mathrm{four}\:\mathrm{of}\:\mathrm{them}\:\mathrm{being}\:\mathrm{triangles}. \\ $$$$\mathrm{Prove}\:\mathrm{that}\:\mathrm{the}\:\mathrm{area}\:\mathrm{of}\:\mathrm{one}\:\mathrm{of}\:\mathrm{the}\:\mathrm{four} \\ $$$$\mathrm{triangles}\:\mathrm{equals}\:\mathrm{the}\:\mathrm{sum}\:\mathrm{of}\:\mathrm{the}\:\mathrm{areas} \\ $$$$\mathrm{of}\:\mathrm{the}\:\mathrm{other}\:\mathrm{three}. \\ $$

Question Number 16947 Answers: 0 Comments: 0

Through the vertices of the smaller base AB of the trapezoid ABCD two parallel lines are drawn, intersecting the segment CD. These lines and the trapezoid′s diagonals divide it into seven triangles and a pentagon. Show that the area of the pentagon equals the sum of the areas of the three triangles that share a common side with the trapezoid.

Question Number 16946 Answers: 0 Comments: 0

Consider the quadrilateral ABCD. The points M, N, P and Q are the midpoints of the sides AB, BC, CD and DA. Let X = AP ∩ BQ, Y = BQ ∩ CM, Q = CM ∩ DN and T= DN ∩ AP. Prove that [XYZT] = [AQX] + [BMY] + [CNZ] + [DPT].

$$\mathrm{Consider}\:\mathrm{the}\:\mathrm{quadrilateral}\:{ABCD}. \\ $$$$\mathrm{The}\:\mathrm{points}\:{M},\:{N},\:{P}\:\mathrm{and}\:{Q}\:\mathrm{are}\:\mathrm{the} \\ $$$$\mathrm{midpoints}\:\mathrm{of}\:\mathrm{the}\:\mathrm{sides}\:{AB},\:{BC},\:{CD} \\ $$$$\mathrm{and}\:{DA}. \\ $$$$\mathrm{Let}\:{X}\:=\:{AP}\:\cap\:{BQ},\:{Y}\:=\:{BQ}\:\cap\:{CM}, \\ $$$${Q}\:=\:{CM}\:\cap\:{DN}\:\mathrm{and}\:{T}=\:{DN}\:\cap\:{AP}. \\ $$$$\mathrm{Prove}\:\mathrm{that}\:\left[{XYZT}\right]\:=\:\left[{AQX}\right]\:+\:\left[{BMY}\right] \\ $$$$+\:\left[{CNZ}\right]\:+\:\left[{DPT}\right]. \\ $$

Question Number 16940 Answers: 1 Comments: 1

Six points A, B, C, D, E, and F are placed on a square rigid, as shown. How many triangles that are not right-angled can be drawn by using 3 of these 6 points as vertices?

$$\mathrm{Six}\:\mathrm{points}\:\mathrm{A},\:\mathrm{B},\:\mathrm{C},\:\mathrm{D},\:\mathrm{E},\:\mathrm{and}\:\mathrm{F}\:\mathrm{are} \\ $$$$\mathrm{placed}\:\mathrm{on}\:\mathrm{a}\:\mathrm{square}\:\mathrm{rigid},\:\mathrm{as}\:\mathrm{shown}. \\ $$$$\mathrm{How}\:\mathrm{many}\:\mathrm{triangles}\:\mathrm{that}\:\mathrm{are}\:\boldsymbol{\mathrm{not}} \\ $$$$\mathrm{right}-\mathrm{angled}\:\mathrm{can}\:\mathrm{be}\:\mathrm{drawn}\:\mathrm{by}\:\mathrm{using}\:\mathrm{3} \\ $$$$\mathrm{of}\:\mathrm{these}\:\mathrm{6}\:\mathrm{points}\:\mathrm{as}\:\mathrm{vertices}? \\ $$

Question Number 16882 Answers: 0 Comments: 2

^• In plane parallel lines are the lines which don′t meet each other. ^• What is the condition in space that two lines be parallel?

$$\:^{\bullet} \mathrm{In}\:\mathrm{plane}\:\mathrm{parallel}\:\mathrm{lines}\:\mathrm{are}\:\mathrm{the}\:\mathrm{lines} \\ $$$$\mathrm{which}\:\mathrm{don}'\mathrm{t}\:\mathrm{meet}\:\mathrm{each}\:\mathrm{other}. \\ $$$$ \\ $$$$\:^{\bullet} \mathrm{What}\:\mathrm{is}\:\mathrm{the}\:\mathrm{condition}\:\mathrm{in}\:\mathrm{space} \\ $$$$\mathrm{that}\:\mathrm{two}\:\mathrm{lines}\:\mathrm{be}\:\mathrm{parallel}? \\ $$

Question Number 16879 Answers: 0 Comments: 0

Let ABCD be a parallelogram. The points M, N and P are chosen on the segments BD, BC and CD, respectively, so that CNMP is a parallelogram. Let E = AN ∩ BD and F = AP ∩ BD. Prove that [AEF] = [DFP] + [BEN].

$$\mathrm{Let}\:{ABCD}\:\mathrm{be}\:\mathrm{a}\:\mathrm{parallelogram}.\:\mathrm{The} \\ $$$$\mathrm{points}\:{M},\:{N}\:\mathrm{and}\:{P}\:\mathrm{are}\:\mathrm{chosen}\:\mathrm{on}\:\mathrm{the} \\ $$$$\mathrm{segments}\:{BD},\:{BC}\:\mathrm{and}\:{CD}, \\ $$$$\mathrm{respectively},\:\mathrm{so}\:\mathrm{that}\:{CNMP}\:\mathrm{is}\:\mathrm{a} \\ $$$$\mathrm{parallelogram}.\:\mathrm{Let}\:{E}\:=\:{AN}\:\cap\:{BD}\:\mathrm{and} \\ $$$${F}\:=\:{AP}\:\cap\:{BD}.\:\mathrm{Prove}\:\mathrm{that} \\ $$$$\left[{AEF}\right]\:=\:\left[{DFP}\right]\:+\:\left[{BEN}\right]. \\ $$

Question Number 16878 Answers: 0 Comments: 2

Let P be a point on the circumcircle of the equilateral triangle ABC. Prove that the projections of any point Q onto the lines PA, PB and PC are the vertices of an equilateral triangle.

$$\mathrm{Let}\:{P}\:\mathrm{be}\:\mathrm{a}\:\mathrm{point}\:\mathrm{on}\:\mathrm{the}\:\mathrm{circumcircle}\:\mathrm{of} \\ $$$$\mathrm{the}\:\mathrm{equilateral}\:\mathrm{triangle}\:{ABC}.\:\mathrm{Prove} \\ $$$$\mathrm{that}\:\mathrm{the}\:\mathrm{projections}\:\mathrm{of}\:\mathrm{any}\:\mathrm{point}\:{Q} \\ $$$$\mathrm{onto}\:\mathrm{the}\:\mathrm{lines}\:{PA},\:{PB}\:\mathrm{and}\:{PC}\:\mathrm{are}\:\mathrm{the} \\ $$$$\mathrm{vertices}\:\mathrm{of}\:\mathrm{an}\:\mathrm{equilateral}\:\mathrm{triangle}. \\ $$

Question Number 16877 Answers: 0 Comments: 0

From a point on the circumcircle of an equilateral triangle ABC parallels to the sides BC, CA and AB are drawn, intersecting the sides CA, AB and BC at the points M, N, P, respectively. Prove that the points M, N and P are collinear.

$$\mathrm{From}\:\mathrm{a}\:\mathrm{point}\:\mathrm{on}\:\mathrm{the}\:\mathrm{circumcircle}\:\mathrm{of}\:\mathrm{an} \\ $$$$\mathrm{equilateral}\:\mathrm{triangle}\:{ABC}\:\mathrm{parallels}\:\mathrm{to} \\ $$$$\mathrm{the}\:\mathrm{sides}\:{BC},\:{CA}\:\mathrm{and}\:{AB}\:\mathrm{are}\:\mathrm{drawn}, \\ $$$$\mathrm{intersecting}\:\mathrm{the}\:\mathrm{sides}\:{CA},\:{AB}\:\mathrm{and}\:{BC} \\ $$$$\mathrm{at}\:\mathrm{the}\:\mathrm{points}\:{M},\:{N},\:{P},\:\mathrm{respectively}. \\ $$$$\mathrm{Prove}\:\mathrm{that}\:\mathrm{the}\:\mathrm{points}\:{M},\:{N}\:\mathrm{and}\:{P}\:\mathrm{are} \\ $$$$\mathrm{collinear}. \\ $$

Question Number 16875 Answers: 0 Comments: 0

Let P_1 , P_2 , ..., P_n be a convex polygon with the following property : for any two vertices P_i and P_j , there exists a vertex P_k such that the segment P_i P_j is seen from P_k under an angle of 60°. Prove that the polygon is an equilateral triangle.

$$\mathrm{Let}\:{P}_{\mathrm{1}} ,\:{P}_{\mathrm{2}} ,\:...,\:{P}_{{n}} \:\mathrm{be}\:\mathrm{a}\:\mathrm{convex}\:\mathrm{polygon} \\ $$$$\mathrm{with}\:\mathrm{the}\:\mathrm{following}\:\mathrm{property}\::\:\mathrm{for}\:\mathrm{any} \\ $$$$\mathrm{two}\:\mathrm{vertices}\:{P}_{{i}} \:\mathrm{and}\:{P}_{{j}} ,\:\mathrm{there}\:\mathrm{exists}\:\mathrm{a} \\ $$$$\mathrm{vertex}\:{P}_{{k}} \:\mathrm{such}\:\mathrm{that}\:\mathrm{the}\:\mathrm{segment}\:{P}_{{i}} {P}_{{j}} \\ $$$$\mathrm{is}\:\mathrm{seen}\:\mathrm{from}\:{P}_{{k}} \:\mathrm{under}\:\mathrm{an}\:\mathrm{angle}\:\mathrm{of}\:\mathrm{60}°. \\ $$$$\mathrm{Prove}\:\mathrm{that}\:\mathrm{the}\:\mathrm{polygon}\:\mathrm{is}\:\mathrm{an} \\ $$$$\mathrm{equilateral}\:\mathrm{triangle}. \\ $$

Question Number 16874 Answers: 0 Comments: 0

Let ABC be an acute triangle. The interior bisectors of the angles ∠B and ∠C meet the opposite sides at the points L and M, respectively. Prove that there exists a point K in the interior of the side BC such that ΔKLM is equilateral if and only if ∠A = 60°.

$$\mathrm{Let}\:{ABC}\:\mathrm{be}\:\mathrm{an}\:\mathrm{acute}\:\mathrm{triangle}.\:\mathrm{The} \\ $$$$\mathrm{interior}\:\mathrm{bisectors}\:\mathrm{of}\:\mathrm{the}\:\mathrm{angles}\:\angle{B}\:\mathrm{and} \\ $$$$\angle{C}\:\mathrm{meet}\:\mathrm{the}\:\mathrm{opposite}\:\mathrm{sides}\:\mathrm{at}\:\mathrm{the} \\ $$$$\mathrm{points}\:{L}\:\mathrm{and}\:{M},\:\mathrm{respectively}.\:\mathrm{Prove} \\ $$$$\mathrm{that}\:\mathrm{there}\:\mathrm{exists}\:\mathrm{a}\:\mathrm{point}\:{K}\:\mathrm{in}\:\mathrm{the} \\ $$$$\mathrm{interior}\:\mathrm{of}\:\mathrm{the}\:\mathrm{side}\:{BC}\:\mathrm{such}\:\mathrm{that} \\ $$$$\Delta{KLM}\:\mathrm{is}\:\mathrm{equilateral}\:\mathrm{if}\:\mathrm{and}\:\mathrm{only}\:\mathrm{if} \\ $$$$\angle{A}\:=\:\mathrm{60}°. \\ $$

Question Number 16911 Answers: 0 Comments: 1

Please solve Q. 16066 or please tell me whether to post its solution or not. I don′t understood the solution.

$$\mathrm{Please}\:\mathrm{solve}\:\mathrm{Q}.\:\mathrm{16066}\:\mathrm{or}\:\mathrm{please}\:\mathrm{tell}\:\mathrm{me} \\ $$$$\mathrm{whether}\:\mathrm{to}\:\mathrm{post}\:\mathrm{its}\:\mathrm{solution}\:\mathrm{or}\:\mathrm{not}.\:\mathrm{I} \\ $$$$\mathrm{don}'\mathrm{t}\:\mathrm{understood}\:\mathrm{the}\:\mathrm{solution}. \\ $$

Question Number 16873 Answers: 0 Comments: 0

Let I be the incenter of ΔABC. It is known that for every point M ∈ (AB), one can find the points N ∈ (BC) and P ∈ (AC) such that I is the centroid of ΔMNP. Prove that ABC is an equilateral triangle.

$$\mathrm{Let}\:{I}\:\mathrm{be}\:\mathrm{the}\:\mathrm{incenter}\:\mathrm{of}\:\Delta{ABC}.\:\mathrm{It}\:\mathrm{is} \\ $$$$\mathrm{known}\:\mathrm{that}\:\mathrm{for}\:\mathrm{every}\:\mathrm{point}\:{M}\:\in\:\left({AB}\right), \\ $$$$\mathrm{one}\:\mathrm{can}\:\mathrm{find}\:\mathrm{the}\:\mathrm{points}\:{N}\:\in\:\left({BC}\right)\:\mathrm{and} \\ $$$${P}\:\in\:\left({AC}\right)\:\mathrm{such}\:\mathrm{that}\:{I}\:\mathrm{is}\:\mathrm{the}\:\mathrm{centroid}\:\mathrm{of} \\ $$$$\Delta{MNP}.\:\mathrm{Prove}\:\mathrm{that}\:{ABC}\:\mathrm{is}\:\mathrm{an} \\ $$$$\mathrm{equilateral}\:\mathrm{triangle}. \\ $$

Question Number 16803 Answers: 1 Comments: 1

Question Number 16785 Answers: 2 Comments: 2

Question Number 16771 Answers: 1 Comments: 1

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