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.

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?

Solution please ?

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

the only possibility: equilateral triangle AB=BC=CA=1

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

((CD)/(BC))=((sin β)/(sin ∠BDC)) ((DA)/(BA))=((sin β)/(sin ∠BDA))=((sin β)/(sin ∠BDC))=((CD)/(BC)) DA=CD as given ⇒BC=BA ⇒BC=BA=CA ⇒ΔABC is equilateral

\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.

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?

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 = (1/2)(√(2(b^2 +c^2 )−a^2 )) The length of the bisector of the angle ABC^(�) equal to l_b =(√(ac−((b^2 ac)/((a+c)^2 )))) The length of the altitude from the vertex C equal to h_C =(2/c)(√(p(p−a)(p−b)(p−c))) From the hypothesis we get m_a =l_b =h_c ⇔(1/2)(√(2(b^2 +c^2 )−a^2 )) = (√(ac))((√(1−(b^2 /((a+c)^2 )))) )=(2/c)(√(p(p−a)(p−b)(p−c)))

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?

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

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.

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 B^(�) is also the median we infer the triangle ABC is isosceles at the vertex B⇒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)

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.

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 @1549442205PVT, @mr W

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.

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.

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.

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, \dots . .}, N^{*} = {1, 2, 3, \dots}

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

But you are going against the question.

But you are going against the

question .

Commented by 1549442205PVT last updated on 08/Sep/20

I answered correctly requirements of question

I answered correctly requirements of

question

Commented by Aina Samuel Temidayo last updated on 08/Sep/20

Answer to no. 3 is 17

Answer to no . 3 is 17

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