Strange triangle I

Geometry Level 4

In △ A B C \triangle ABC with A B = c AB=c , A C = b AC=b and B C = a BC=a , where ∠ A = 5 0 ∘ \angle A=50^{\circ} and ∠ B = 10 0 ∘ \angle B=100^{\circ} , which of the following expressions hold?

1. a 3 + c 3 = 3 a c 2 2. a 3 + c 3 = 3 a 2 c 3. 3 a 3 + c 3 = 3 a c 2 4. 3 a 3 + c 3 = 3 a 2 c 5. 3 a 3 + b 3 = 3 a b 2 6. 3 b c 2 + 3 c 3 = b 3 1. \quad a^3+c^3=3ac^2\\ 2. \quad a^3+c^3=3a^2c\\ 3. \quad 3a^3+c^3=3ac^2\\ 4. \quad 3a^3+c^3=3a^2c\\ 5. \quad 3a^3+b^3=3ab^2\\ 6. \quad 3bc^2+\sqrt{3}c^3=b^3

Only 6 1 and 6 Only 4 Only 1 2 and 5 3 and 5

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Alan Enrique Ontiveros Salazar by Alan Enrique Ontiveros S… , 22

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2 solutions

First calculate ∠ C \angle C , which is simply ∠ C = 18 0 ∘ − ∠ A − ∠ B = 3 0 ∘ \angle C=180^{\circ}-\angle A-\angle B=30^{\circ}

By Law of Sines on △ A B C \triangle ABC we get:

a sin ⁡ 5 0 ∘ = c sin ⁡ 3 0 ∘ sin ⁡ 5 0 ∘ = a 2 c \dfrac{a}{\sin 50^{\circ}}=\dfrac{c}{\sin 30^{\circ}} \\ \sin 50^{\circ}=\dfrac{a}{2c}

Now, in the identity of triple angle sin ⁡ 3 θ = 3 sin ⁡ θ − 4 sin ⁡ 3 θ \sin 3\theta=3\sin \theta-4\sin^3 \theta , put θ = 5 0 ∘ \theta=50^{\circ} and rearrange:

sin ⁡ 15 0 ∘ = 3 sin ⁡ 5 0 ∘ − 4 sin ⁡ 3 5 0 ∘ 1 2 = 3 sin ⁡ 5 0 ∘ − 4 sin ⁡ 3 5 0 ∘ 1 = 6 sin ⁡ 5 0 ∘ − 8 sin ⁡ 3 5 0 ∘ 1 = 6 ( a 2 c ) − 8 ( a 2 c ) 3 1 = 3 a c − a 3 c 3 c 3 = 3 a c 2 − a 3 a 3 + c 3 = 3 a c 2 \sin 150^{\circ}=3\sin 50^{\circ}-4\sin^3 50^{\circ} \\ \dfrac{1}{2}=3\sin 50^{\circ}-4\sin^3 50^{\circ} \\ 1=6\sin 50^{\circ}-8\sin^3 50^{\circ} \\ 1=6\left(\dfrac{a}{2c}\right)-8\left(\dfrac{a}{2c}\right)^3 \\ 1=\dfrac{3a}{c}-\dfrac{a^3}{c^3} \\ c^3=3ac^2-a^3 \\ \boxed{a^3+c^3=3ac^2}

Similarly, by Law of Sines we get:

b sin ⁡ 10 0 ∘ = c sin ⁡ 3 0 ∘ sin ⁡ 8 0 ∘ = b 2 c \dfrac{b}{\sin 100^{\circ}}=\dfrac{c}{\sin 30^{\circ}} \\ \sin 80^{\circ}=\dfrac{b}{2c}

Apply again the identity of triple angle, this time with θ = 80 ° \theta=80° :

sin ⁡ 24 0 ∘ = 3 sin ⁡ 8 0 ∘ − 4 sin ⁡ 3 8 0 ∘ − 3 2 = 3 sin ⁡ 8 0 ∘ − 4 sin ⁡ 3 8 0 ∘ − 3 = 6 sin ⁡ 8 0 ∘ − 8 sin ⁡ 3 8 0 ∘ − 3 = 6 ( b 2 c ) − 8 ( b 2 c ) 3 − 3 = 3 b c − b 3 c 3 − 3 c 3 = 3 b c 2 − b 3 3 b c 2 + 3 c 3 = b 3 \sin 240^{\circ}=3\sin 80^{\circ}-4\sin^3 80^{\circ} \\ -\dfrac{\sqrt{3}}{2}=3\sin 80^{\circ}-4\sin^3 80^{\circ} \\ -\sqrt{3}=6\sin 80^{\circ}-8\sin^3 80^{\circ} \\ -\sqrt{3}=6\left(\dfrac{b}{2c}\right)-8\left(\dfrac{b}{2c}\right)^3 \\ -\sqrt{3}=\dfrac{3b}{c}-\dfrac{b^3}{c^3} \\ -\sqrt{3}c^3=3bc^2-b^3 \\\boxed{3bc^2+\sqrt{3}c^3=b^3}

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Out of the box way of looking at the problem. Though since Sin Law is used, the solution is straight forward without going for expanding Sin cube..

Niranjan Khanderia - 3 years ago

S i n L a w : − a S i n ( 50 ) = b S i n ( 100 ) = c S i n ( 30 ) . ∴ c = a 2 ∗ S i n ( 50 ) . . . . . . . . . ( A ) D i v i d i n g [ 1. ] b y a 3 a n d s u b s t i t u t i n g f r o m ( A ) , w e g e t L H S = R H S . S o [ 1. ] i s a s o l u t i o n . A g a i n b = 2 ∗ c ∗ S i n 100........ ( B ) D i v i d i n g [ 6. ] b y c 3 a n d s u b s t i t u t i n g f r o m ( B ) , w e g e t L H S = R H S . S o [ 6. ] i s a s o l u t i o n . S o 1 a n d 6 a r e t h e s o l u t i o n s . Sin~Law:- ~~~\dfrac a {Sin(50)}=\dfrac b {Sin(100)}=\dfrac c {Sin(30)}.\\ \therefore~c=\dfrac a {2*Sin(50)}.........(A)\\ Dividing~[ 1. ]~ by~a^3~and~substituting~from~(A),~we~get~LHS=RHS.~~~So~[ 1. ]~is~a~solution.\\ Again~b=2*c*Sin100........(B)\\ Dividing~[ 6. ]~ by~c^3~and~substituting~from~(B),~we~get~LHS=RHS.~~~So~[ 6. ]~is~a~solution.\\ So~~~ 1~~and~~6~~~are~~~the~~ solutions.

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