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P P , Q Q and R R are sides of a triangle such that

P 2 + Q 2 + R 2 = 114 P^2+Q^2+R^2=114 P 4 + Q 4 + R 4 = 5346 P^4+Q^4+R^4=5346

What is the area of the triangle?


The answer is 12.

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2 solutions

Pi Han Goh
Dec 31, 2013

By Heron's formula ,

Area = 1 4 ( P 2 + Q 2 + R 2 ) 2 2 ( P 4 + Q 4 + R 4 ) = 12 \text{Area} = \frac {1}{4} \sqrt{ \left (P^2 + Q^2 + R^2 \right )^2 - 2\left (P^4 + Q^4 + R^4 \right ) } = \boxed{12}

p = P + Q + R 2 p = \frac{P+Q+R}{2} S = p ( p P ) ( p Q ) ( p R ) S = \sqrt{p(p-P)(p-Q)(p-R)} S 2 = ( P + Q + R 2 ) ( P + Q R 2 ) ( P Q + R 2 ) ( P + Q + R 2 ) S^2 = \left (\frac{P+Q+R}{2} \right )\left (\frac{P+Q-R}{2} \right )\left (\frac{P-Q+R}{2} \right )\left (\frac{-P+Q+R}{2} \right ) 16 S 2 = ( P + Q + R ) ( P + Q R ) ( P Q + R ) ( P + Q + R ) 16S^2 = (P+Q+R)(P+Q-R)(P-Q+R)(-P+Q+R) 16 S 2 = 2 ( P 2 Q 2 + P 2 R 2 + Q 2 R 2 ) ( P 4 + Q 4 + R 4 ) 16S^2 = 2(P^2Q^2 + P^2R^2 + Q^2R^2) - (P^4+Q^4+R^4) 16 S 2 = ( P 2 + Q 2 + R 2 ) 2 2 ( P 4 + Q 4 + R 4 ) 16S^2 = (P^2 + Q^2 + R^2)^2 - 2(P^4 + Q^4 + R^4) 16 S 2 = 11 4 2 2 5346 16S^2 = 114^2 - 2 \cdot 5346 16 S 2 = 2304 16S^2 = 2304 S = 12. \boxed{S = 12.}

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