Irreducible Ratio

Geometry Level pending

Let A B C ABC be a triangle and let a a , b b and c c be the lengths of the sides opposite to the vertices A A , B B and C C , respectively. If a − 4 b + c = 0 , 2 a + b − 2 c = 0 , a - 4b + c = 0, 2a + b - 2c = 0 , then sin ⁡ A : sin ⁡ B : sin ⁡ C \sin A : \sin B : \sin C can be expressed as an irreducible ratio p : q : r p:q:r , where p p , q q and r r are positive integers. What is p + q + r ? p+q+r?

22 22 20 20 19 19 21 21

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Mahindra Jain by Mahindra Jain

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1 solution

Tom Engelsman
Nov 15, 2020

Solving the above system of equations yields a = 7 9 k , b = 4 9 k , c = k a = \frac{7}{9}k, b = \frac{4}{9}k, c = k for k ∈ R + . k \in \mathbb{R^{+}}. Applying the Law of Sines gives us:

b a ⋅ sin ⁡ A = sin ⁡ B ⇒ 4 7 ⋅ sin ⁡ A = sin ⁡ B ; \frac{b}{a} \cdot \sin A=\sin B \Rightarrow \frac{4}{7} \cdot \sin A = \sin B;

c a ⋅ sin ⁡ A = sin ⁡ C ⇒ 9 7 ⋅ sin ⁡ A = sin ⁡ C \frac{c}{a} \cdot \sin A = \sin C \Rightarrow \frac{9}{7} \cdot \sin A = \sin C .

Finally, we have the ratio sin ⁡ A : sin ⁡ B : sin ⁡ C = sin ⁡ A : 4 7 ⋅ sin ⁡ A : 9 7 ⋅ sin ⁡ A = ( sin ⁡ A 7 ) ⋅ [ 7 : 4 : 9 ] ⇒ 7 : 4 : 9 = p : q : r . \sin A : \sin B: \sin C = \sin A: \frac{4}{7} \cdot \sin A: \frac{9}{7} \cdot \sin A = (\frac{\sin A}{7}) \cdot [7:4:9] \Rightarrow 7:4:9 = p:q:r. Hence, p + q + r = 7 + 4 + 9 = 20 . p+q+r = 7+4+9 = \boxed{20}.

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