Find the shape of △ A B C \triangle ABC

Geometry Level 2

In △ A B C \triangle ABC , the opposite sides of ∠ A \angle A , ∠ B \angle B , and ∠ C \angle C are a a , b b , and c c respectively.

If c − a cos ⁡ B = ( 2 a − b ) cos ⁡ A c-a\cos B= (2a-b) \cos A , what is the shape of △ A B C \triangle ABC ?

Isosceles triangle or right triangle Isosceles triangle Right triangle Isosceles and right triangle

Alice Smith by Alice Smith , 20, China

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2 solutions

Applying cosine rule to △ A B C \triangle {ABC} we get c − a × c 2 + a 2 − b 2 2 c a = ( 2 a − b ) × b 2 + c 2 − a 2 2 b c ⟹ b ( c 2 − a 2 + b 2 ) = ( 2 a − b ) ( b 2 + c 2 − a 2 ) ⟹ a 2 = b 2 + c 2 c-a\times \dfrac{c^2+a^2-b^2}{2ca}=(2a-b)\times \dfrac{b^2+c^2-a^2}{2bc}\implies b(c^2-a^2+b^2)=(2a-b)(b^2+c^2-a^2)\implies a^2=b^2+c^2 or b = 2 a − b ⟹ b = a b=2a-b\implies b=a . Therefore △ A B C \triangle {ABC} is either isosceles triangle or right triangle .

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Something is amiss with your algebra. Consider a 30-60-90 triangle with a = 1 a = 1 , c = 3 c = \sqrt{3} , and b = 2 b = 2 . This satisfies the original equation as cos ⁡ B = 0 \cos B = 0 and ( 2 a − b ) = 0 (2a - b) = 0 . But b ( c 2 − a 2 + b 2 ) ≠ 0 b(c^2 - a^2 + b^2) \neq 0 .

Your ⟹ \implies symbol is doing a lot of work and appears to have covered up an error. The right triangle I've found has b b as the length of the hypotenuse, not a a .

Richard Desper - 1 year, 3 months ago

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The 30-60-90 triangle with a = 1 a = 1 , c = 3 c = \sqrt{3} , and b = 2 b = 2 does not satisfy the original equation, because then

c − a cos ⁡ B = ( 2 a − b ) cos ⁡ A ⇒ 3 − 1 ⋅ 0 = ( 2 ⋅ 1 − 2 ) 3 2 ⇒ 3 = 0 c - a \cos B = (2a - b) \cos A \Rightarrow \sqrt{3} - 1 \cdot 0 = (2 \cdot 1 - 2) \frac{\sqrt{3}}{2} \Rightarrow \sqrt{3} = 0

@Alak Bhattacharya 's algebra is correct because b ( c 2 − a 2 + b 2 ) = ( 2 a − b ) ( b 2 + c 2 − a 2 ) ⇒ 2 ( a − b ) ( a 2 − b 2 − c 2 ) = 0 b(c^2 - a^2 + b^2) = (2a - b)(b^2 + c^2 - a^2) \Rightarrow 2(a - b)(a^2 - b^2 - c^2) = 0 . Therefore, either a = b a = b (isosceles) or a 2 = b 2 + c 2 a^2 = b^2 + c^2 (right triangle).

David Vreken - 1 year, 3 months ago

By sine rule , we have a sin ⁡ A = b sin ⁡ B = c sin ⁡ C = k \dfrac a{\sin A} = \dfrac b{\sin B} = \dfrac c{\sin C} = k , ⟹ a = k sin ⁡ A \implies a = k\sin A , b = k sin ⁡ B b = k \sin B , and c = k sin ⁡ C c = k \sin C . Then,

c − a cos ⁡ B = ( 2 a − b ) cos ⁡ A k sin ⁡ C − k sin ⁡ A cos ⁡ B = 2 k sin ⁡ A cos ⁡ A − k sin ⁡ B cos ⁡ A Divide both sides by k sin ⁡ ( 18 0 ∘ − A − B ) − sin ⁡ A cos ⁡ B = 2 sin ⁡ A cos ⁡ A − sin ⁡ B cos ⁡ A Note that sin ⁡ ( 18 0 ∘ − θ ) = sin ⁡ θ sin ⁡ ( A + B ) − sin ⁡ A cos ⁡ B = 2 sin ⁡ A cos ⁡ A − sin ⁡ B cos ⁡ A sin ⁡ A cos ⁡ B + sin ⁡ B cos ⁡ A − sin ⁡ A cos ⁡ B = 2 sin ⁡ A cos ⁡ A − sin ⁡ B cos ⁡ A 2 sin ⁡ B cos ⁡ A = 2 sin ⁡ A cos ⁡ A cos ⁡ A ( sin ⁡ B − sin ⁡ A ) = 0 \begin{aligned} c - a\cos B & = (2a - b) \cos A \\ k\sin C - k \sin A \cos B & = 2k\sin A \cos A - k \sin B \cos A & \small \blue{\text{Divide both sides by }k} \\ \sin (180^\circ - A-B) - \sin A \cos B & = 2\sin A \cos A - \sin B \cos A & \small \blue{\text{Note that }\sin (180^\circ - \theta) = \sin \theta} \\ \sin (A+B) - \sin A \cos B & = 2\sin A \cos A - \sin B \cos A \\ \sin A\cos B + \sin B \cos A - \sin A \cos B & = 2\sin A \cos A - \sin B \cos A \\ 2 \sin B \cos A & = 2\sin A \cos A \\ \cos A (\sin B - \sin A) & = 0 \end{aligned}

⟹ { cos ⁡ A = 0 ⟹ A = 9 0 ∘ sin ⁡ A − sin ⁡ B ⟹ A = B \implies \begin{cases} \cos A = 0 & \implies A = 90^\circ \\ \sin A - \sin B & \implies A = B \end{cases} . Therefore △ A B C \triangle ABC can be either an isosceles triangle or right triangle .

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