What a triangle is that?

Geometry Level 4

Given a triangle A B C ABC with ∡ A C B = 100 \measuredangle ACB=100 and ∡ B A C = 30 \measuredangle BAC=30 , point M M lies on the side A C AC ,such that C M = C B CM=CB . If the ratio B M : A C = p : q BM:AC=p:q , where p p and q q are co-prime positive integers, compute p + q p+q .


The answer is 2.

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Kristian Vasilev by Kristian Vasilev , 24, Bulgaria

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2 solutions

Let C M = C B = a CM=CB=a , then by Sine Rule, we have:

q sin ⁡ 5 0 ∘ = a sin ⁡ 3 0 ∘ ⇒ q = 2 a sin ⁡ 5 0 ∘ \quad \dfrac {q}{\sin{50^\circ}} = \dfrac {a}{\sin{30^\circ}}\quad \Rightarrow q = 2a\sin{50^\circ}

By Cosine Rule:

p 2 = a 2 + a 2 − 2 a 2 cos ⁡ 10 0 ∘ = 2 a 2 ( 1 − cos ⁡ 10 0 ∘ ) \quad p^2 = a^2+a^2-2a^2\cos{100^\circ}= 2a^2(1-\cos{100^\circ})

= 2 a 2 ( 1 − ( 1 − 2 sin ⁡ 2 5 0 ∘ ) = 4 a 2 sin ⁡ 2 5 0 ∘ = q 2 \quad \quad = 2a^2( 1 - (1-2\sin^2 {50^\circ}) = 4a^2\sin^2 {50^\circ} = q^2

⇒ q = p ⇒ q : p = 1 : 1 ⇒ q + p = 2 \quad \Rightarrow q=p \quad \Rightarrow q:p = 1:1 \quad \Rightarrow q+p = \boxed {2}

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nice solution, but please try without trigonometry

Kristian Vasilev - 6 years, 6 months ago

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Ur free to do that :)

Chirayu Bhardwaj - 5 years, 3 months ago
Ahmad Saad
Nov 3, 2015

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