Question-216188

Question Number 216188 by BaliramKumar last updated on 29/Jan/25

Answered by som(math1967) last updated on 29/Jan/25

A C : B C = 5 : 12

i f A C = 5 c m B C = 12 c m

t h e n A C + B C = 17 = A B

n o t p o s s i b l e

A C = 10 c m

B C = 24 c m

t h e n A C + A B = 10 + 17 = 27 > B C

s o p o s s i b l e p e r i m e t e r

10 + 17 + 24 = 51 c m

Answered by Rasheed.Sindhi last updated on 29/Jan/25

A B = A D + B D = 5 + 12 = 17

A C : B C = 5 : 12

l e t A C = 5 m & B C = 12 m; m \in N

\underset{―}{A C + B C > {A B}_{}}

5 m + 12 m > 17 \Rightarrow 17 m > 17 \Rightarrow m > 1

\underset{―}{A C + A B > {B C}_{}}

5 m + 17 > 12 m \Rightarrow 7 m < 17 \Rightarrow m < 3

∴ m = 2

∴ A C = 10 & B C = 24

P e r i m e t e r = A B + B C + A C

= 17 + 24 + 10 = 51 c m

Answered by dionigi last updated on 31/Jan/25

\frac{A C}{B C} = \frac{A D}{B D} = \frac{5}{12}

\frac{A C}{A D} = \frac{B C}{B D} = x (x ⩾ 1)

\begin{matrix} (A C, B C) \in N^{2} \\ ∄ m, 5 \mod m = 12 \mod m = 0 \end{matrix}} \Rightarrow x \in N

p = A C + A D + B D + B C

p = (A D + B D) (1 + x)

p = 17 (1 + x)

p = 17 n (n = x + 1 ⩾ 2, n \in N)

n = 2 \Rightarrow p = 34

n = 3 \Rightarrow p = 51

n = 4 \Rightarrow p = 68

Commented by Rasheed.Sindhi last updated on 31/Jan/25

F o r p = 34, 68 w h a t a r e t h e v a l u e s

o f t h e s i d e s o f t h e t r i a n g l e ?

Commented by dionigi last updated on 01/Feb/25

f o r p = 34

∠ C A B = ∠ A B C = 0; ∠ A C B = π

A C = 5; B C = 12; A B = 17

∠ C A B + ∠ A B C + ∠ A C B = π

A C + B C + A B = 34

s e t Δ t h e l i n e b i s e c t o r o f ∠ A C B

D = C \Rightarrow D \in Δ

a n d \frac{A C}{B C} = \frac{5}{12}

b u t p = 68 i s n o t a s o l u t i o n

\frac{A C}{B C} = \frac{5}{12} i s n o t a s u f f i c i e n t p a t t e r n

B C ⩽ A B + A C i s m a n d a t o r y a l s o

Commented by som(math1967) last updated on 01/Feb/25

B C < A B + A C, n o t ⩽

o n l y s o l u t i o n i s 51

Commented by Rasheed.Sindhi last updated on 01/Feb/25

B u t \frac{A C}{B C} = \frac{5}{22} \neq \frac{5}{12}

Commented by dionigi last updated on 01/Feb/25

B C = A B + A C r e m a i n s c o r r e c t

T h i s m e a n s t h a t C, A, D, B a r e a l i g n e d

o n t h e s a m e l i n e Δ

A C = 5, B C = 22 a n d A B = 17

A C, B C, a n d A B a r e a l l i n t e g e r s .

T h e 3 a n g l e s o f t h e t r i a n g l e a r e

∠ A C B = ∠ A B C = 0 a n d ∠ C A B = π

L i n e Δ i s b i s e c t o r o f ∠ A C B a n d D \in Δ

A l l t h e p a t t e r n s o f t h e i n i t i a l p r o b l e m

a r e t r u e, b u t p e r i m e t e r p = 44 w a s

n o t p r o p o s e d i n t h e c h o i c e .

Commented by dionigi last updated on 01/Feb/25

R i g h t!

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