It's not Pythagoras

Geometry Level 4

Let a a , b b , and c c be the three sides of a triangle, and let α \alpha , β \beta , and γ \gamma be the angles opposite them respectively. If a 2 + b 2 = 1989 c 2 a^2+b^2=1989c^2 , find

tan ⁡ α tan ⁡ β tan ⁡ α tan ⁡ γ + tan ⁡ β tan ⁡ γ \large \frac{\color{#3D99F6}{\tan \alpha \ \tan \beta}}{\color{magenta}{\tan \alpha \ \tan \gamma}+\color{#69047E}{\tan \beta \ \tan \gamma}}


The answer is 994.

Skye Rzym by SKYE RZYM , 20, Indonesia

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2 solutions

Relevant wiki: Sine Rule (Law of Sines)

By cosine rule , we have:

c 2 = a 2 + b 2 − 2 a b cos ⁡ γ Given that a 2 + b 2 = 1989 c 2 c 2 = 1989 c 2 − 2 a b cos ⁡ γ 2 a b cos ⁡ γ = 1988 c 2 a b cos ⁡ γ = 994 c 2 Sine rule: a sin ⁡ α = b sin ⁡ β = c sin ⁡ γ = k k 2 sin ⁡ α sin ⁡ β cos ⁡ γ = 994 k 2 sin ⁡ 2 γ Dividing both sides by k 2 cos ⁡ γ sin ⁡ α sin ⁡ β = 994 sin ⁡ γ tan ⁡ γ As sin ⁡ γ = sin ⁡ ( π − α − β ) = sin ⁡ ( α + β ) sin ⁡ α sin ⁡ β = 994 sin ⁡ ( α + β ) tan ⁡ γ sin ⁡ α sin ⁡ β = 994 ( sin ⁡ α cos ⁡ β + sin ⁡ β cos ⁡ β ) tan ⁡ γ Dividing both sides by cos ⁡ α cos ⁡ β tan ⁡ α tan ⁡ β = 994 ( tan ⁡ α + tan ⁡ β ) tan ⁡ γ \begin{aligned} c^2 & = {\color{#3D99F6}a^2 + b^2} - 2ab \cos \gamma & \small \color{#3D99F6} \text{Given that } a^2 + b^2 = 1989c^2 \\ c^2 & = 1989c^2 - 2ab \cos \gamma \\ 2ab \cos \gamma & = 1988c^2 \\ ab \cos \gamma & = 994c^2 & \small \color{#3D99F6} \text{Sine rule: } \frac a {\sin \alpha} = \frac b{\sin \beta} = \frac c{\sin \gamma} = k \\ k^2 \sin \alpha \sin \beta \cos \gamma & = 994k^2 \sin^2 \gamma & \small \color{#3D99F6} \text{Dividing both sides by }k^2 \cos \gamma \\ \sin \alpha \sin \beta & = 994 {\color{#3D99F6}\sin \gamma} \tan \gamma & \small \color{#3D99F6} \text{As }\sin \gamma = \sin (\pi - \alpha - \beta) = \sin (\alpha + \beta) \\ \sin \alpha \sin \beta & = 994 {\color{#3D99F6}\sin (\alpha + \beta)} \tan \gamma \\ \sin \alpha \sin \beta & = 994 (\sin \alpha \cos \beta + \sin \beta \cos \beta) \tan \gamma & \small \color{#3D99F6} \text{Dividing both sides by }\cos \alpha \cos \beta \\ \tan \alpha \tan \beta & = 994 (\tan \alpha + \tan \beta) \tan \gamma \end{aligned}

⟹ tan ⁡ α tan ⁡ β tan ⁡ α tan ⁡ γ + tan ⁡ β tan ⁡ γ = 994 \begin{aligned} \implies \frac {\tan \alpha \tan \beta}{\tan \alpha \tan \gamma + \tan \beta \tan \gamma} & = \boxed{994} \end{aligned}

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Nivedit Jain
Mar 1, 2017

\frac{a}{sina})=2R \(\frac{abc}{4R}=Area of triangle so from 1st and 2nd we get SinA=\(\frac{2 times Area)}{bc} similary for all angles can be calculated using sine rule now we now cosine rule so cosA can be calculated fromthere So tanA 4 t i m e s A R e a t w i c e b c \frac{4 times ARea}{twice bc} similarly all tans put all the values and we get above expssion is (a a + b b - c c)/(c c)=994 from given equation

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