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Question Number 14905 Answers: 0 Comments: 5

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Question Number 14810 Answers: 0 Comments: 0

7 real numbers are given in the interval (1, 13). Prove that atleast 3 of them are the lengths of a triangle′s sides.

$$\mathrm{7}\:\mathrm{real}\:\mathrm{numbers}\:\mathrm{are}\:\mathrm{given}\:\mathrm{in}\:\mathrm{the}\:\mathrm{interval} \\ $$$$\left(\mathrm{1},\:\mathrm{13}\right).\:\mathrm{Prove}\:\mathrm{that}\:\mathrm{atleast}\:\mathrm{3}\:\mathrm{of}\:\mathrm{them} \\ $$$$\mathrm{are}\:\mathrm{the}\:\mathrm{lengths}\:\mathrm{of}\:\mathrm{a}\:\mathrm{triangle}'\mathrm{s}\:\mathrm{sides}. \\ $$

Question Number 14809 Answers: 1 Comments: 2

Let ABC be an acute triangle. Find the positions of the points M, N, P on the sides BC, CA, AB, respectively, such that the perimeter of the triangle MNP is minimal.

$$\mathrm{Let}\:{ABC}\:\mathrm{be}\:\mathrm{an}\:\mathrm{acute}\:\mathrm{triangle}.\:\mathrm{Find} \\ $$$$\mathrm{the}\:\mathrm{positions}\:\mathrm{of}\:\mathrm{the}\:\mathrm{points}\:{M},\:{N},\:{P}\:\mathrm{on} \\ $$$$\mathrm{the}\:\mathrm{sides}\:{BC},\:{CA},\:{AB},\:\mathrm{respectively}, \\ $$$$\mathrm{such}\:\mathrm{that}\:\mathrm{the}\:\mathrm{perimeter}\:\mathrm{of}\:\mathrm{the}\:\mathrm{triangle} \\ $$$${MNP}\:\mathrm{is}\:\mathrm{minimal}. \\ $$

Question Number 14807 Answers: 0 Comments: 0

Prove that the medians of a given triangle can form a triangle.

$$\mathrm{Prove}\:\mathrm{that}\:\mathrm{the}\:\mathrm{medians}\:\mathrm{of}\:\mathrm{a}\:\mathrm{given} \\ $$$$\mathrm{triangle}\:\mathrm{can}\:\mathrm{form}\:\mathrm{a}\:\mathrm{triangle}. \\ $$

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Question Number 14630 Answers: 0 Comments: 2

solve the eqn dr/dθ=[r(a^2 −r^2 )/a^2 +r^2 ]cotθ hint. let a^2 +r^2 =a^2 −r^2 +2r^2 .

$$\mathrm{solve}\:\mathrm{the}\:\mathrm{eqn} \\ $$$$\mathrm{dr}/\mathrm{d}\theta=\left[\mathrm{r}\left(\mathrm{a}^{\mathrm{2}} −\mathrm{r}^{\mathrm{2}} \right)/\mathrm{a}^{\mathrm{2}} +\mathrm{r}^{\mathrm{2}} \right]\mathrm{cot}\theta \\ $$$$\mathrm{hint}.\:\mathrm{let}\:\mathrm{a}^{\mathrm{2}} +\mathrm{r}^{\mathrm{2}} =\mathrm{a}^{\mathrm{2}} −\mathrm{r}^{\mathrm{2}} +\mathrm{2r}^{\mathrm{2}} . \\ $$

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Related to Q#14157 a^2 +b^2 −ab=α^2 b^2 +c^2 −bc=β^2 c^2 +d^2 −cd=γ^2 d^2 +e^2 −de=δ^2 e^2 +a^2 −ea=ξ^2

Question Number 14364 Answers: 0 Comments: 9

Modification of Q#14157 x^2 +y^2 −xy=a^2 y^2 +z^2 −yz=b^2 z^2 +x^2 −zx=c^2 Pl discuss also geometrical/ trigonometrical aspects.

$$\mathrm{Modification}\:\mathrm{of}\:\mathrm{Q}#\mathrm{14157} \\ $$$$\mathrm{x}^{\mathrm{2}} +\mathrm{y}^{\mathrm{2}} −\mathrm{xy}=\mathrm{a}^{\mathrm{2}} \\ $$$$\mathrm{y}^{\mathrm{2}} +\mathrm{z}^{\mathrm{2}} −\mathrm{yz}=\mathrm{b}^{\mathrm{2}} \\ $$$$\mathrm{z}^{\mathrm{2}} +\mathrm{x}^{\mathrm{2}} −\mathrm{zx}=\mathrm{c}^{\mathrm{2}} \\ $$$$\mathrm{Pl}\:\mathrm{discuss}\:\mathrm{also}\:\mathrm{geometrical}/ \\ $$$$\mathrm{trigonometrical}\:\mathrm{aspects}. \\ $$

Question Number 14145 Answers: 1 Comments: 1

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Question Number 14071 Answers: 2 Comments: 1

Solve the Partial fraction ((3x^4 − 9x^3 + 16x^2 + 9x + 13)/((x − 1)^2 (x^2 + 2x − 2)^2 ))

$$\mathrm{Solve}\:\mathrm{the}\:\mathrm{Partial}\:\mathrm{fraction}\: \\ $$$$\frac{\mathrm{3x}^{\mathrm{4}} \:−\:\mathrm{9x}^{\mathrm{3}} \:+\:\mathrm{16x}^{\mathrm{2}} \:+\:\mathrm{9x}\:+\:\mathrm{13}}{\left(\mathrm{x}\:−\:\mathrm{1}\right)^{\mathrm{2}} \left(\mathrm{x}^{\mathrm{2}} \:+\:\mathrm{2x}\:−\:\mathrm{2}\right)^{\mathrm{2}} } \\ $$

Question Number 13738 Answers: 1 Comments: 1

P,Q,R,S are four locations on the same horizontal plane.Q is on a bearing of 041° from P and the distance is 40km. S is 28km from R on a bearing 074°, R is directly due north of P and the distance between Q and R is 38km. (a)the bearing of R from Q (b)the distance between Q and S (c)the distance between P and R

$$\mathrm{P},\mathrm{Q},\mathrm{R},\mathrm{S}\:\mathrm{are}\:\mathrm{four}\:\mathrm{locations}\:\mathrm{on}\:\mathrm{the} \\ $$$$\mathrm{same}\:\mathrm{horizontal}\:\mathrm{plane}.\mathrm{Q}\:\mathrm{is}\:\mathrm{on}\:\mathrm{a}\: \\ $$$$\mathrm{bearing}\:\mathrm{of}\:\mathrm{041}°\:\mathrm{from}\:\mathrm{P}\:\mathrm{and}\:\mathrm{the} \\ $$$$\mathrm{distance}\:\mathrm{is}\:\mathrm{40km}. \\ $$$$\mathrm{S}\:\mathrm{is}\:\mathrm{28km}\:\mathrm{from}\:\mathrm{R}\:\mathrm{on}\:\mathrm{a}\:\mathrm{bearing}\:\mathrm{074}°, \\ $$$$\mathrm{R}\:\mathrm{is}\:\mathrm{directly}\:\mathrm{due}\:\mathrm{north}\:\mathrm{of}\:\mathrm{P}\:\mathrm{and} \\ $$$$\mathrm{the}\:\mathrm{distance}\:\mathrm{between}\:\mathrm{Q}\:\mathrm{and}\:\mathrm{R}\:\mathrm{is} \\ $$$$\mathrm{38km}. \\ $$$$\left(\mathrm{a}\right)\mathrm{the}\:\mathrm{bearing}\:\mathrm{of}\:\mathrm{R}\:\mathrm{from}\:\mathrm{Q} \\ $$$$\left(\mathrm{b}\right)\mathrm{the}\:\mathrm{distance}\:\mathrm{between}\:\mathrm{Q}\:\mathrm{and}\:\mathrm{S} \\ $$$$\left(\mathrm{c}\right)\mathrm{the}\:\mathrm{distance}\:\mathrm{between}\:\mathrm{P}\:\mathrm{and}\:\mathrm{R} \\ $$

Question Number 13735 Answers: 2 Comments: 6

Question Number 13501 Answers: 1 Comments: 0

Find x 10_C_x = 5_C_2

$$\mathrm{Find}\:\mathrm{x} \\ $$$$\mathrm{10}_{\mathrm{C}_{\mathrm{x}} } \:=\:\mathrm{5}_{\mathrm{C}_{\mathrm{2}} } \\ $$

Question Number 13548 Answers: 1 Comments: 11

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Question Number 13434 Answers: 1 Comments: 1

ΔABC∼ΔDEF ΔABC=25 ΔDEF=35 ((ΣMadian^2 of ΔABC)/(ΣMadian^2 of ΔDEF))=?

$$\Delta{ABC}\sim\Delta{DEF}\: \\ $$$$\Delta{ABC}=\mathrm{25} \\ $$$$\Delta{DEF}=\mathrm{35} \\ $$$$\frac{\Sigma{Madian}^{\mathrm{2}} {of}\:\Delta{ABC}}{\Sigma{Madian}^{\mathrm{2}} \:{of}\:\Delta{DEF}}=? \\ $$

Question Number 13305 Answers: 1 Comments: 0

Does sin A=sin B imply sin (A/2)=sin(B/2)? For example A=π and B=−π?

$$\mathrm{Does} \\ $$$$\mathrm{sin}\:\mathrm{A}=\mathrm{sin}\:\mathrm{B} \\ $$$$\mathrm{imply}\:\mathrm{sin}\:\frac{\mathrm{A}}{\mathrm{2}}=\mathrm{sin}\frac{\mathrm{B}}{\mathrm{2}}? \\ $$$$\mathrm{For}\:\mathrm{example}\:\mathrm{A}=\pi\:\mathrm{and}\:\mathrm{B}=−\pi? \\ $$

Question Number 13888 Answers: 1 Comments: 3

Prove that if A′, B′ and C′ are the midpoints of the sides BC, CA and AB, respectively, then AA′ + BB′ + CC′ < AB + BC + CA

$$\mathrm{Prove}\:\mathrm{that}\:\mathrm{if}\:{A}',\:{B}'\:\mathrm{and}\:{C}'\:\mathrm{are}\:\mathrm{the} \\ $$$$\mathrm{midpoints}\:\mathrm{of}\:\mathrm{the}\:\mathrm{sides}\:{BC},\:{CA}\:\mathrm{and}\:{AB}, \\ $$$$\mathrm{respectively},\:\mathrm{then} \\ $$$${AA}'\:+\:{BB}'\:+\:{CC}'\:<\:{AB}\:+\:{BC}\:+\:{CA} \\ $$

Question Number 13236 Answers: 0 Comments: 16

For positive a,b,c such that a b c=1 show that a^(b+c) b^(c+a) c^(a+b) ≤1 solution: a^(b+c) b^(c+a) c^(a+b) =(a^b a^c b^c b^a c^a c^b ) =(b×c)^a (a×c)^b (a×b)^c =(a^0 ×b×c)^a (a×b^0 ×c)^b (a×b×c^0 )^c ≤(a×b×c)^(a ) (a×b^ × c)^b (a×b×c)^c ≤(1)^a (1)^b (1)^c ;since a b c=1 ≤1

$${For}\:{positive}\:\:{a},{b},{c}\:\:{such}\:{that}\:\:{a}\:{b}\:{c}=\mathrm{1} \\ $$$${show}\:{that}\:\:{a}^{{b}+{c}} \:\:{b}^{{c}+{a}} \:\:{c}^{{a}+{b}} \:\leqslant\mathrm{1} \\ $$$${solution}: \\ $$$${a}^{{b}+{c}} \:\:{b}^{{c}+{a}} \:\:{c}^{{a}+{b}} \:=\left({a}^{{b}} {a}^{{c}} \:{b}^{{c}} {b}^{{a}} \:\:{c}^{{a}} {c}^{{b}} \right) \\ $$$$\:\:\:\:\:\:=\left({b}×{c}\right)^{{a}} \:\left({a}×{c}\right)^{{b}} \:\left({a}×{b}\right)^{{c}} \\ $$$$\:\:\:\:\:\:\:=\left({a}^{\mathrm{0}} ×{b}×{c}\right)^{{a}} \:\left({a}×{b}^{\mathrm{0}} ×{c}\right)^{{b}} \left({a}×{b}×{c}^{\mathrm{0}} \right)^{{c}} \\ $$$$\:\:\:\:\:\:\:\leqslant\left({a}×{b}×{c}\right)^{{a}\:\:\:\:} \left({a}×{b}^{} ×\:{c}\right)^{{b}} \:\left({a}×{b}×{c}\right)^{{c}} \\ $$$$\:\:\:\:\:\:\:\leqslant\left(\mathrm{1}\right)^{{a}} \:\left(\mathrm{1}\right)^{{b}} \:\left(\mathrm{1}\right)^{{c}} \:\:\:\:;{since}\:\:{a}\:{b}\:{c}=\mathrm{1} \\ $$$$\:\:\:\:\:\:\:\:\leqslant\mathrm{1} \\ $$

Question Number 13228 Answers: 0 Comments: 5

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