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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

Question Number 16756 Answers: 2 Comments: 1

Alternate vertices of a regular hexagon are joined as shown. What fraction of the total area of the hexagon is shaded? (Justify your answer.)

$$\mathrm{Alternate}\:\mathrm{vertices}\:\mathrm{of}\:\mathrm{a}\:\mathrm{regular}\:\mathrm{hexagon} \\ $$$$\mathrm{are}\:\mathrm{joined}\:\mathrm{as}\:\mathrm{shown}.\:\mathrm{What}\:\mathrm{fraction}\:\mathrm{of} \\ $$$$\mathrm{the}\:\mathrm{total}\:\mathrm{area}\:\mathrm{of}\:\mathrm{the}\:\mathrm{hexagon}\:\mathrm{is} \\ $$$$\mathrm{shaded}?\:\left(\mathrm{Justify}\:\mathrm{your}\:\mathrm{answer}.\right) \\ $$

Question Number 16740 Answers: 1 Comments: 3

The maximum value of cos^2 (cos (33π + θ)) + sin^2 (sin (45π + θ)) is (1) 1 + sin^2 1 (2) 2 (3) 1 + cos^2 1 (4) cos^2 2

$$\mathrm{The}\:\mathrm{maximum}\:\mathrm{value}\:\mathrm{of} \\ $$$$\mathrm{cos}^{\mathrm{2}} \:\left(\mathrm{cos}\:\left(\mathrm{33}\pi\:+\:\theta\right)\right)\:+\:\mathrm{sin}^{\mathrm{2}} \:\left(\mathrm{sin}\:\left(\mathrm{45}\pi\:+\:\theta\right)\right) \\ $$$$\mathrm{is} \\ $$$$\left(\mathrm{1}\right)\:\mathrm{1}\:+\:\mathrm{sin}^{\mathrm{2}} \mathrm{1} \\ $$$$\left(\mathrm{2}\right)\:\mathrm{2} \\ $$$$\left(\mathrm{3}\right)\:\mathrm{1}\:+\:\mathrm{cos}^{\mathrm{2}} \mathrm{1} \\ $$$$\left(\mathrm{4}\right)\:\mathrm{cos}^{\mathrm{2}} \mathrm{2} \\ $$

Question Number 16739 Answers: 0 Comments: 0

Let M be a point in the interior of the equilateral triangle ABC and let A′, B′ and C′ be its projections onto the sides BC, CA and AB, respectively. Prove that the sum of lengths of the inradii of triangles MAC′, MBA′ and MCB′ equals the sum of lengths of the inradii of trianges MAB′, MBC′ and MCA′.

$$\mathrm{Let}\:{M}\:\mathrm{be}\:\mathrm{a}\:\mathrm{point}\:\mathrm{in}\:\mathrm{the}\:\mathrm{interior}\:\mathrm{of}\:\mathrm{the} \\ $$$$\mathrm{equilateral}\:\mathrm{triangle}\:{ABC}\:\mathrm{and}\:\mathrm{let}\:{A}', \\ $$$${B}'\:\mathrm{and}\:{C}'\:\mathrm{be}\:\mathrm{its}\:\mathrm{projections}\:\mathrm{onto}\:\mathrm{the} \\ $$$$\mathrm{sides}\:{BC},\:{CA}\:\mathrm{and}\:{AB},\:\mathrm{respectively}. \\ $$$$\mathrm{Prove}\:\mathrm{that}\:\mathrm{the}\:\mathrm{sum}\:\mathrm{of}\:\mathrm{lengths}\:\mathrm{of}\:\mathrm{the} \\ $$$$\mathrm{inradii}\:\mathrm{of}\:\mathrm{triangles}\:{MAC}',\:{MBA}'\:\mathrm{and} \\ $$$${MCB}'\:\mathrm{equals}\:\mathrm{the}\:\mathrm{sum}\:\mathrm{of}\:\mathrm{lengths}\:\mathrm{of}\:\mathrm{the} \\ $$$$\mathrm{inradii}\:\mathrm{of}\:\mathrm{trianges}\:{MAB}',\:{MBC}'\:\mathrm{and} \\ $$$${MCA}'. \\ $$

Question Number 16738 Answers: 0 Comments: 0

Prove that the segments joining the midpoints of the opposite sides of an equiangular hexagon are concurrent.

$$\mathrm{Prove}\:\mathrm{that}\:\mathrm{the}\:\mathrm{segments}\:\mathrm{joining}\:\mathrm{the} \\ $$$$\mathrm{midpoints}\:\mathrm{of}\:\mathrm{the}\:\mathrm{opposite}\:\mathrm{sides}\:\mathrm{of}\:\mathrm{an} \\ $$$$\mathrm{equiangular}\:\mathrm{hexagon}\:\mathrm{are}\:\mathrm{concurrent}. \\ $$

Question Number 16737 Answers: 0 Comments: 0

A convex hexagon is given in which any two opposite sides have the following property: the distance between their midpoints is ((√3)/2) times the sum of their lengths. Prove that the hexagon is equiangular.

$$\mathrm{A}\:\mathrm{convex}\:\mathrm{hexagon}\:\mathrm{is}\:\mathrm{given}\:\mathrm{in}\:\mathrm{which} \\ $$$$\mathrm{any}\:\mathrm{two}\:\mathrm{opposite}\:\mathrm{sides}\:\mathrm{have}\:\mathrm{the} \\ $$$$\mathrm{following}\:\mathrm{property}:\:\mathrm{the}\:\mathrm{distance} \\ $$$$\mathrm{between}\:\mathrm{their}\:\mathrm{midpoints}\:\mathrm{is}\:\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}\:\mathrm{times}\:\mathrm{the} \\ $$$$\mathrm{sum}\:\mathrm{of}\:\mathrm{their}\:\mathrm{lengths}.\:\mathrm{Prove}\:\mathrm{that}\:\mathrm{the} \\ $$$$\mathrm{hexagon}\:\mathrm{is}\:\mathrm{equiangular}. \\ $$

Question Number 16736 Answers: 0 Comments: 0

The side lengths of an equiangular octagon are rational numbers. Prove that the octagon has a symmetry center.

$$\mathrm{The}\:\mathrm{side}\:\mathrm{lengths}\:\mathrm{of}\:\mathrm{an}\:\mathrm{equiangular} \\ $$$$\mathrm{octagon}\:\mathrm{are}\:\mathrm{rational}\:\mathrm{numbers}.\:\mathrm{Prove} \\ $$$$\mathrm{that}\:\mathrm{the}\:\mathrm{octagon}\:\mathrm{has}\:\mathrm{a}\:\mathrm{symmetry} \\ $$$$\mathrm{center}. \\ $$

Question Number 16735 Answers: 0 Comments: 0

Let a_1 , a_2 , ..., a_n be the side lengths of an equiangular polygon. Prove that if a_1 ≥ a_2 ≥ ... ≥ a_n , then the polygon is regular.

$$\mathrm{Let}\:{a}_{\mathrm{1}} ,\:{a}_{\mathrm{2}} ,\:...,\:{a}_{{n}} \:\mathrm{be}\:\mathrm{the}\:\mathrm{side}\:\mathrm{lengths}\:\mathrm{of}\: \\ $$$$\mathrm{an}\:\mathrm{equiangular}\:\mathrm{polygon}.\:\mathrm{Prove}\:\mathrm{that}\:\mathrm{if} \\ $$$${a}_{\mathrm{1}} \:\geqslant\:{a}_{\mathrm{2}} \:\geqslant\:...\:\geqslant\:{a}_{{n}} ,\:\mathrm{then}\:\mathrm{the}\:\mathrm{polygon}\:\mathrm{is} \\ $$$$\mathrm{regular}. \\ $$

Question Number 16734 Answers: 0 Comments: 0

An equiangular polygon with an odd number of sides is inscribed in a circle. Prove that the polygon is regular.

$$\mathrm{An}\:\mathrm{equiangular}\:\mathrm{polygon}\:\mathrm{with}\:\mathrm{an}\:\mathrm{odd} \\ $$$$\mathrm{number}\:\mathrm{of}\:\mathrm{sides}\:\mathrm{is}\:\mathrm{inscribed}\:\mathrm{in}\:\mathrm{a}\:\mathrm{circle}. \\ $$$$\mathrm{Prove}\:\mathrm{that}\:\mathrm{the}\:\mathrm{polygon}\:\mathrm{is}\:\mathrm{regular}. \\ $$

Question Number 16754 Answers: 2 Comments: 1

Question Number 16665 Answers: 0 Comments: 1

Question Number 16641 Answers: 0 Comments: 0

Prove that p is a prime number if and only if every equiangular polygon with p sides of rational lengths is regular.

$$\mathrm{Prove}\:\mathrm{that}\:{p}\:\mathrm{is}\:\mathrm{a}\:\mathrm{prime}\:\mathrm{number}\:\mathrm{if}\:\mathrm{and} \\ $$$$\mathrm{only}\:\mathrm{if}\:\mathrm{every}\:\mathrm{equiangular}\:\mathrm{polygon}\:\mathrm{with} \\ $$$${p}\:\mathrm{sides}\:\mathrm{of}\:\mathrm{rational}\:\mathrm{lengths}\:\mathrm{is}\:\mathrm{regular}. \\ $$

Question Number 16595 Answers: 1 Comments: 4

Question Number 16592 Answers: 1 Comments: 1

please what does the question mean by the overlapping portion of A and B.

$$\mathrm{please}\:\mathrm{what}\:\mathrm{does}\:\mathrm{the}\:\mathrm{question}\:\mathrm{mean} \\ $$$$\mathrm{by}\:\mathrm{the}\:\mathrm{overlapping}\:\mathrm{portion}\:\mathrm{of}\:\mathrm{A}\: \\ $$$$\mathrm{and}\:\mathrm{B}. \\ $$

Question Number 16579 Answers: 0 Comments: 3

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