A triangle ABC has the following properties BC=1, AB=AC and that the angle bisector from vertex B is also a median. Find all possible triangle(s) with its/their side−lengths and angles.


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Question Number 112318 by Aina Samuel Temidayo last updated on 07/Sep/20

$A triangle ABC has the following$ $properties BC = 1, AB = AC and that$ $the angle bisector from vertex B is$ $also a median . Find all possible$ $triangle (s) with its / their$ $side - lengths and angles .$
Commented by Aina Samuel Temidayo last updated on 07/Sep/20

$Solution please ?$
Commented by mr W last updated on 07/Sep/20

$t h e o n l y p o s s i b i l i t y :$ $e q u i l a t e r a l t r i a n g l e$ $A B = B C = C A = 1$
	Answered by mr W last updated on 07/Sep/20

	Commented by mr W last updated on 07/Sep/20

	$\frac{C D}{B C} = \frac{\sin β}{\sin ∠ B D C}$ $\frac{D A}{B A} = \frac{\sin β}{\sin ∠ B D A} = \frac{\sin β}{\sin ∠ B D C} = \frac{C D}{B C}$ $D A = C D a s g i v e n$ $\Rightarrow B C = B A$ $\Rightarrow B C = B A = C A$ $\Rightarrow Δ A B C i s e q u i l a t e r a l$
	Commented by Aina Samuel Temidayo last updated on 07/Sep/20

	$Thanks .$
	Commented by Aina Samuel Temidayo last updated on 07/Sep/20

	Commented by Aina Samuel Temidayo last updated on 07/Sep/20

	$Can you also help with this please ?$
	Commented by 1549442205PVT last updated on 08/Sep/20

	$The length of the median from the$ $vertex A equal to m_{a} = \frac{1}{2} \sqrt{2 (b^{2} + c^{2}) - a^{2}}$ $The length of the bisector of the angle$ $\hat{ABC} equal to l_{b} = \sqrt{ac - \frac{b^{2} ac}{{(a + c)}^{2}}}$ $The length of the altitude from the$ $vertex C equal to h_{C} = \frac{2}{c} \sqrt{p (p - a) (p - b) (p - c)}$ $From the hypothesis we get$ $m_{a} = l_{b} = h_{c} \Leftrightarrow \frac{1}{2} \sqrt{2 (b^{2} + c^{2}) - a^{2}} =$ $\sqrt{ac} (\sqrt{1 - \frac{b^{2}}{{(a + c)}^{2}}}) = \frac{2}{c} \sqrt{p (p - a) (p - b) (p - c)}$
	Commented by Aina Samuel Temidayo last updated on 08/Sep/20

	$Is it complete ?$
	Commented by 1549442205PVT last updated on 11/Sep/20

	$You only need give c one some value$ $e . x, c = 6 and solve equation to find a, b$ $It can have some of triangles$ $satisfying that conditions$
	Commented by Aina Samuel Temidayo last updated on 10/Sep/20

	$I {don}^{'} t understand .$
	Answered by 1549442205PVT last updated on 07/Sep/20

	$From the hypothesis AB = AC we infer$ $the triangle ABC is isosceles at the$ $vertex A . From the hypothesis the bisector$ $of the angle \hat{B} is also the median we$ $infer the triangle ABC is isosceles at$ $the vertex B \Rightarrow BA = BC . Therefore we$ $have AB = AC = BC = 1$ $(In course of secondary school level$ $proved the result that : if one triangle$ $has the bisector belong to any side$ $that is also the median (or altitude)$ $then that triangle is isosceles)$
	Commented by Aina Samuel Temidayo last updated on 07/Sep/20

	$Thanks .$
	Commented by Aina Samuel Temidayo last updated on 07/Sep/20

	Commented by Aina Samuel Temidayo last updated on 07/Sep/20

	$Please help @ 1549442205 PVT, @ mr$ $W$
	Commented by 1549442205PVT last updated on 08/Sep/20
	$2)Given x,y,z∈N;yz+x is prime, (yz+x)∣(zx+y),(yz+x)∣(xy+z)(∗) We need find all possible values of P=(((xy+z)((zx+y))/((yz+x)^2 )) i)Case z=0 then P=((xy^2 )/x^2 )=(y^2 /x)=m^2 where y=mx, x∈P(prime) ii)Case y=0 then P=((xz^2 )/x^2 )=(z^2 /x)=n^2 where z=nx,x∈P(prime) iii)Case y,z≠0 From the hypothesis we have zx+y=m(yz+x),xy+z=n(yz+x) where m,n∈N^∗ .Adding two equalities we get x(y+z)+y+z=(m+n)(yz+x) ⇔(x+1)(y+z)=(m+n)(yz+x) Since yz+x is prime we infer that (yz+x)∣(x+1) or (yz+x)∣(y+z) a) If (yz+x)∣(x+1)then yz+x≤x+1 ⇒yz≤1⇒y=z=1(since y,z∈N,≠0) Then P=(((x+1)^2 )/((x+1)^2 ))=1(Here we see that has infinite x so that x+1 is prime) b)If (yz+x)∣(y+z)then yz+x≤y+z ⇔(y−1)(z−1)≤1−x(1) Since y−1≥0,z−1≥0,1−x≤0 ,we see that the equality (1)ocurrs if and only if x=y=z=1.Then P=((2.2)/2^2 )=1 Combining all cases we have P=m^2 for z=0,y=mx,x∈P P=n^(2 ) for y=0,z=nx,x∈P or P=1 when y=z=1, x∈P or x=y=z=1 3)Find ∣{n∈N∣n≤100,4!∣2^n −n^3 }∣ We need find n∈N∣n≤100 and A=2^n −n^3 ⋮4!=24 If n is odd then 2^n −n^3 is odd number ,so A isn′t divisible by 24.Hence,n must be an even number ,infer n=2k⇒A=2^(2k) −(2k)^3 =4^k −8k^3 (1) We need find k so that A⋮24 We need must have 0≤A≤100,so k=5 because when k=4 we get A=4^4 −8.4^3 <0.When k=6 we get A=4^6 −8.6^3 =2368 when k=5 we get A=4^5 −8.5^3 = 1024−8.125=24⋮4! Thus,has unique n=10 satisfying the condition of our problem:n∈N, n≤100,2^n −n^3 ⋮4!.Therefore ∣{n∈N∣n≤100,2^n −n^3 ⋮4!}∣=1$
	$2) Given x, y, z \in N; yz + x is prime,$ $(yz + x) ∣ (zx + y), (yz + x) ∣ (xy + z) ()$ $We need find all possible values of$ $P = \frac{(xy + z) ((zx + y)}{{(yz + x)}^{2}}$ $i) Case z = 0 then P = \frac{{xy}^{2}}{x^{2}} = \frac{y^{2}}{x} = m^{2}$ $where y = mx, x \in P (prime)$ $ii) Case y = 0 then P = \frac{{xz}^{2}}{x^{2}} = \frac{z^{2}}{x} = n^{2}$ $where z = nx, x \in P (prime)$ $iii) Case y, z \neq 0$ $From the hypothesis we have$ $zx + y = m (yz + x), xy + z = n (yz + x)$ $where m, n \in N^{} . Adding two equalities$ $we get x (y + z) + y + z = (m + n) (yz + x)$ $\Leftrightarrow (x + 1) (y + z) = (m + n) (yz + x)$ $Since yz + x is prime we infer that$ $(yz + x) ∣ (x + 1) or (yz + x) ∣ (y + z)$ $a) If (yz + x) ∣ (x + 1) then yz + x ⩽ x + 1$ $\Rightarrow yz ⩽ 1 \Rightarrow y = z = 1 (since y, z \in N, \neq 0)$ $Then P = \frac{{(x + 1)}^{2}}{{(x + 1)}^{2}} = 1 (Here we see that$ $has infinite x so that x + 1 is prime)$ $b) If (yz + x) ∣ (y + z) then yz + x ⩽ y + z$ $\Leftrightarrow (y - 1) (z - 1) ⩽ 1 - x (1)$ $Since y - 1 ⩾ 0, z - 1 ⩾ 0, 1 - x ⩽ 0, we see$ $that the equality (1) ocurrs if and only$ $if x = y = z = 1 . Then$ $P = \frac{2.2}{2^{2}} = 1$ $Combining all cases we have$ $P = m^{2} for z = 0, y = mx, x \in P$ $P = n^{2} for y = 0, z = nx, x \in P$ $or P = 1 when y = z = 1, x \in P$ $or x = y = z = 1$ $3) Find ∣ {n \in N ∣ n ⩽ 100, 4! ∣ 2^{n} - n^{3}} ∣$ $We need find n \in N ∣ n ⩽ 100 and$ $A = 2^{n} - n^{3} ⋮ 4! = 24$ $If n is odd then 2^{n} - n^{3} is odd number$ $, so A {isn}^{'} t divisible by 24 . Hence, n$ $must be an even number, infer$ $n = 2 k \Rightarrow A = 2^{2 k} - {(2 k)}^{3} = 4^{k} - {8 k}^{3} (1)$ $We need find k so that A ⋮ 24$ $We need must have 0 ⩽ A ⩽ 100, so$ $k = 5 because when k = 4 we get$ $A = 4^{4} - 8 . 4^{3} < 0 . When k = 6 we get$ $A = 4^{6} - 8 . 6^{3} = 2368$ $when k = 5 we get A = 4^{5} - 8 . 5^{3} =$ $1024 - 8.125 = 24 ⋮ 4!$ $Thus, has unique n = 10 satisfying$ $the condition of our problem : n \in N,$ $n ⩽ 100, 2^{n} - n^{3} ⋮ 4! . Therefore$ $∣ {n \in N ∣ n ⩽ 100, 2^{n} - n^{3} ⋮ 4!} ∣= 1$
	Commented by Aina Samuel Temidayo last updated on 07/Sep/20

	$Thanks . Please do the rest .$
	Commented by Aina Samuel Temidayo last updated on 07/Sep/20

	$Why did you change the solution for$ $No . 2 number theory ? {What}^{'} s wrong$ $with the first ? We {weren}^{'} t asked to$ $find values for x, y and z .$
	Commented by Aina Samuel Temidayo last updated on 07/Sep/20

	$x, y, z are natural numbers . They {can}^{'} t$ $be 0 . I think your first solution is$ $correct . Please send it again .$
	Commented by 1549442205PVT last updated on 07/Sep/20
	$Ok,excuse me .I mistake.Need note the condition (∗).Set of natural numbers by new definition include zero N={0,1,2,.....},N^∗ ={1,2,3,...}$
	$Ok, excuse me . I mistake . Need note$ $the condition () . Set of natural$ $numbers by new definition include$ $zero N = {0, 1, 2, . . . . .}, N^{} = {1, 2, 3, . . .}$
	Commented by Aina Samuel Temidayo last updated on 07/Sep/20

	$But you are going against the$ $question .$
	Commented by 1549442205PVT last updated on 08/Sep/20

	$I answered correctly requirements of$ $question$
	Commented by Aina Samuel Temidayo last updated on 08/Sep/20

	$Answer to no . 3 is 17$
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