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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 16944 Answers: 0 Comments: 2

Find the number of digits in the number 2^(2005) × 5^(2000) when written in full.

$$\mathrm{Find}\:\mathrm{the}\:\mathrm{number}\:\mathrm{of}\:\mathrm{digits}\:\mathrm{in}\:\mathrm{the} \\ $$$$\mathrm{number}\:\mathrm{2}^{\mathrm{2005}} \:×\:\mathrm{5}^{\mathrm{2000}} \:\mathrm{when}\:\mathrm{written}\:\mathrm{in} \\ $$$$\mathrm{full}. \\ $$

Question Number 16969 Answers: 0 Comments: 0

Question Number 16942 Answers: 1 Comments: 0

A distance of 200 km is to be covered by car in less than 10 hours. Yash does it in two parts. He first drives for 150 km at an average speed of 36 km/hr, without stopping. After taking rest for 30 minutes, he starts again and covers the remaining distance non-stop. His average for the entire journey (including the period of rest) exceeds that for the second part by 5 km/hr. Find the speed at which he covers the second part.

$$\mathrm{A}\:\mathrm{distance}\:\mathrm{of}\:\mathrm{200}\:\mathrm{km}\:\mathrm{is}\:\mathrm{to}\:\mathrm{be}\:\mathrm{covered}\:\mathrm{by} \\ $$$$\mathrm{car}\:\mathrm{in}\:\mathrm{less}\:\mathrm{than}\:\mathrm{10}\:\mathrm{hours}.\:\mathrm{Yash}\:\mathrm{does}\:\mathrm{it} \\ $$$$\mathrm{in}\:\mathrm{two}\:\mathrm{parts}.\:\mathrm{He}\:\mathrm{first}\:\mathrm{drives}\:\mathrm{for}\:\mathrm{150}\:\mathrm{km} \\ $$$$\mathrm{at}\:\mathrm{an}\:\mathrm{average}\:\mathrm{speed}\:\mathrm{of}\:\mathrm{36}\:\mathrm{km}/\mathrm{hr}, \\ $$$$\mathrm{without}\:\mathrm{stopping}.\:\mathrm{After}\:\mathrm{taking}\:\mathrm{rest}\:\mathrm{for} \\ $$$$\mathrm{30}\:\mathrm{minutes},\:\mathrm{he}\:\mathrm{starts}\:\mathrm{again}\:\mathrm{and}\:\mathrm{covers} \\ $$$$\mathrm{the}\:\mathrm{remaining}\:\mathrm{distance}\:\mathrm{non}-\mathrm{stop}.\:\mathrm{His} \\ $$$$\mathrm{average}\:\mathrm{for}\:\mathrm{the}\:\mathrm{entire}\:\mathrm{journey} \\ $$$$\left(\mathrm{including}\:\mathrm{the}\:\mathrm{period}\:\mathrm{of}\:\mathrm{rest}\right)\:\mathrm{exceeds} \\ $$$$\mathrm{that}\:\mathrm{for}\:\mathrm{the}\:\mathrm{second}\:\mathrm{part}\:\mathrm{by}\:\mathrm{5}\:\mathrm{km}/\mathrm{hr}. \\ $$$$\mathrm{Find}\:\mathrm{the}\:\mathrm{speed}\:\mathrm{at}\:\mathrm{which}\:\mathrm{he}\:\mathrm{covers}\:\mathrm{the} \\ $$$$\mathrm{second}\:\mathrm{part}. \\ $$

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 16938 Answers: 1 Comments: 2

The Object shown in the diagram is made by gluing together the adjacent faces of six wooden cubes, each having edges of length 2 cm. Find the total surface area of the object in square centimetres.

$$\mathrm{The}\:\mathrm{Object}\:\mathrm{shown}\:\mathrm{in}\:\mathrm{the}\:\mathrm{diagram}\:\mathrm{is} \\ $$$$\mathrm{made}\:\mathrm{by}\:\mathrm{gluing}\:\mathrm{together}\:\mathrm{the}\:\mathrm{adjacent} \\ $$$$\mathrm{faces}\:\mathrm{of}\:\mathrm{six}\:\mathrm{wooden}\:\mathrm{cubes},\:\mathrm{each}\:\mathrm{having} \\ $$$$\mathrm{edges}\:\mathrm{of}\:\mathrm{length}\:\mathrm{2}\:\mathrm{cm}.\:\mathrm{Find}\:\mathrm{the}\:\mathrm{total} \\ $$$$\mathrm{surface}\:\mathrm{area}\:\mathrm{of}\:\mathrm{the}\:\mathrm{object}\:\mathrm{in}\:\mathrm{square} \\ $$$$\mathrm{centimetres}. \\ $$

Question Number 16936 Answers: 0 Comments: 5

In the diagram, it is possible to travel only along an edge in the direction indicated by the arrow. How many different routes from A to B are there in all?

$$\mathrm{In}\:\mathrm{the}\:\mathrm{diagram},\:\mathrm{it}\:\mathrm{is}\:\mathrm{possible}\:\mathrm{to}\:\mathrm{travel} \\ $$$$\mathrm{only}\:\mathrm{along}\:\mathrm{an}\:\mathrm{edge}\:\mathrm{in}\:\mathrm{the}\:\mathrm{direction} \\ $$$$\mathrm{indicated}\:\mathrm{by}\:\mathrm{the}\:\mathrm{arrow}.\:\mathrm{How}\:\mathrm{many} \\ $$$$\mathrm{different}\:\mathrm{routes}\:\mathrm{from}\:\mathrm{A}\:\mathrm{to}\:\mathrm{B}\:\mathrm{are}\:\mathrm{there} \\ $$$$\mathrm{in}\:\mathrm{all}? \\ $$

Question Number 16935 Answers: 1 Comments: 0

A two-digit number has the property that the square of its tens digit plus ten times its units digit is equal to the square of its units digit plus ten times its tens digit. Find all two digit numbers which have this property, and are prime numbers.

$$\mathrm{A}\:\mathrm{two}-\mathrm{digit}\:\mathrm{number}\:\mathrm{has}\:\mathrm{the}\:\mathrm{property} \\ $$$$\mathrm{that}\:\mathrm{the}\:\mathrm{square}\:\mathrm{of}\:\mathrm{its}\:\mathrm{tens}\:\mathrm{digit}\:\mathrm{plus} \\ $$$$\mathrm{ten}\:\mathrm{times}\:\mathrm{its}\:\mathrm{units}\:\mathrm{digit}\:\mathrm{is}\:\mathrm{equal}\:\mathrm{to}\:\mathrm{the} \\ $$$$\mathrm{square}\:\mathrm{of}\:\mathrm{its}\:\mathrm{units}\:\mathrm{digit}\:\mathrm{plus}\:\mathrm{ten}\:\mathrm{times} \\ $$$$\mathrm{its}\:\mathrm{tens}\:\mathrm{digit}.\:\mathrm{Find}\:\mathrm{all}\:\mathrm{two}\:\mathrm{digit} \\ $$$$\mathrm{numbers}\:\mathrm{which}\:\mathrm{have}\:\mathrm{this}\:\mathrm{property},\:\mathrm{and} \\ $$$$\mathrm{are}\:\mathrm{prime}\:\mathrm{numbers}. \\ $$

Question Number 16919 Answers: 1 Comments: 0

ax2+bx+c=0

$${ax}\mathrm{2}+{bx}+{c}=\mathrm{0} \\ $$

Question Number 16918 Answers: 1 Comments: 0

2x2+9x=10

$$\mathrm{2}{x}\mathrm{2}+\mathrm{9}{x}=\mathrm{10} \\ $$

Question Number 16917 Answers: 1 Comments: 0

If x + iy = (1/(a + ib)) prove that : (x^2 + y^2 )(a^2 + b^2 ) = 1

$$\mathrm{If}\:\:\:\mathrm{x}\:+\:\mathrm{iy}\:=\:\frac{\mathrm{1}}{\mathrm{a}\:+\:\mathrm{ib}} \\ $$$$\mathrm{prove}\:\mathrm{that}\::\:\:\:\left(\mathrm{x}^{\mathrm{2}} \:+\:\mathrm{y}^{\mathrm{2}} \right)\left(\mathrm{a}^{\mathrm{2}} \:+\:\mathrm{b}^{\mathrm{2}} \right)\:=\:\mathrm{1} \\ $$

Question Number 16909 Answers: 1 Comments: 0

(√x^x^6 ) = 729 , Find x

$$\sqrt{\mathrm{x}^{\mathrm{x}^{\mathrm{6}} } }\:=\:\mathrm{729}\:,\:\:\:\:\mathrm{Find}\:\mathrm{x} \\ $$

Question Number 16906 Answers: 1 Comments: 0

x^x = 100, find x.

$$\mathrm{x}^{\mathrm{x}} \:=\:\mathrm{100},\:\mathrm{find}\:\mathrm{x}. \\ $$

Question Number 16904 Answers: 0 Comments: 1

Proove that Σ_(n=1) ^∞ (m^2 /(xm))=Σ_(n=1) ^∞ ((xm)/x^2 )

$${Proove}\:{that} \\ $$$$\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{{m}^{\mathrm{2}} }{{xm}}=\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{{xm}}{{x}^{\mathrm{2}} } \\ $$

Question Number 16893 Answers: 0 Comments: 4

if y=u^n show that d^n y/dx^n =n!

$$\mathrm{if}\:\mathrm{y}=\mathrm{u}^{\mathrm{n}} \:\:\:\mathrm{show}\:\mathrm{that}\:\mathrm{d}^{\mathrm{n}} \mathrm{y}/\mathrm{dx}^{\mathrm{n}} =\mathrm{n}! \\ $$

Question Number 16892 Answers: 0 Comments: 0

∫_( 0) ^( ∞) ((ln(x))/(x^2 + 2x + 4)) dx

$$\int_{\:\mathrm{0}} ^{\:\infty} \:\frac{\mathrm{ln}\left(\mathrm{x}\right)}{\mathrm{x}^{\mathrm{2}} \:+\:\mathrm{2x}\:+\:\mathrm{4}}\:\mathrm{dx} \\ $$

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 16881 Answers: 2 Comments: 0

The quadratic polynomials p(x) = a(x − 3)^2 + bx + 1 and q(x) = 2x^2 + c(x − 2) + 13 are equal for all values of x. Find the values of a, b, and c.

$$\mathrm{The}\:\mathrm{quadratic}\:\mathrm{polynomials} \\ $$$${p}\left({x}\right)\:=\:{a}\left({x}\:−\:\mathrm{3}\right)^{\mathrm{2}} \:+\:{bx}\:+\:\mathrm{1}\:\mathrm{and} \\ $$$${q}\left({x}\right)\:=\:\mathrm{2}{x}^{\mathrm{2}} \:+\:{c}\left({x}\:−\:\mathrm{2}\right)\:+\:\mathrm{13}\:\mathrm{are}\:\mathrm{equal} \\ $$$$\mathrm{for}\:\mathrm{all}\:\mathrm{values}\:\mathrm{of}\:{x}.\:\mathrm{Find}\:\mathrm{the}\:\mathrm{values}\:\mathrm{of}\:{a}, \\ $$$${b},\:\mathrm{and}\:{c}. \\ $$

Question Number 16880 Answers: 0 Comments: 0

Find the number of positive integers less than or equal to 300 that are multiples of 3 or 5, but are not multiples of 10 or 15.

$$\mathrm{Find}\:\mathrm{the}\:\mathrm{number}\:\mathrm{of}\:\mathrm{positive}\:\mathrm{integers} \\ $$$$\mathrm{less}\:\mathrm{than}\:\mathrm{or}\:\mathrm{equal}\:\mathrm{to}\:\mathrm{300}\:\mathrm{that}\:\mathrm{are} \\ $$$$\mathrm{multiples}\:\mathrm{of}\:\mathrm{3}\:\mathrm{or}\:\mathrm{5},\:\mathrm{but}\:\mathrm{are}\:\mathrm{not} \\ $$$$\mathrm{multiples}\:\mathrm{of}\:\mathrm{10}\:\mathrm{or}\:\mathrm{15}. \\ $$

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 16876 Answers: 1 Comments: 0

In how many ways can a family of 5 brothers be seated round a table if (i) 2 brothers must seat next to each other. (ii) 2 brothers must not seat together.

$$\mathrm{In}\:\mathrm{how}\:\mathrm{many}\:\mathrm{ways}\:\mathrm{can}\:\mathrm{a}\:\mathrm{family}\:\mathrm{of}\:\mathrm{5}\:\mathrm{brothers}\:\mathrm{be}\:\mathrm{seated}\:\mathrm{round}\:\mathrm{a}\:\mathrm{table} \\ $$$$\mathrm{if}\:\left(\mathrm{i}\right)\:\mathrm{2}\:\mathrm{brothers}\:\mathrm{must}\:\mathrm{seat}\:\mathrm{next}\:\mathrm{to}\:\mathrm{each}\:\mathrm{other}. \\ $$$$\left(\mathrm{ii}\right)\:\mathrm{2}\:\mathrm{brothers}\:\mathrm{must}\:\mathrm{not}\:\mathrm{seat}\:\mathrm{together}. \\ $$

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.

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

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