Turning translation

Geometry Level 3

There is a rod which length is 1 moving at this situation:

  1. The rod will always hinged on point O O and not going anyway.
  2. The angles α \alpha , β \beta , and the lengths a a , b b will always follow the rule: a β = b α a\beta=b\alpha .

Then find the area which this rod made.

Detail:

  1. The area is said made from entire rod not only the part of length b b .
  2. The range of α \alpha is from 0 to π 2 \frac{\pi}{2} .
π \pi π 4 \frac{\pi}{4} π 3 \frac{\pi}{3} π 6 \frac{\pi}{6}

Kelvin Hong by Kelvin Hong , 21, Malaysia

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1 solution

Kelvin Hong
Sep 21, 2017

From the rule, we can get that

( 1 − b ) β = b ( π 2 − β ) (1-b)\beta = b(\frac{\pi}{2}-\beta)

β − b β = b ( π 2 − β ) \beta -b\beta = b(\frac{\pi}{2}-\beta)

β = b π 2 \beta = b\frac{\pi}{2}

b = 2 β π b=\frac{2\beta}{\pi}

Let's find the Area made by b b ,

A r e a = 1 2 ∫ 0 π 2 b 2 d β Area = \frac{1}{2} \int_0^{\frac{\pi}{2}} b^2d\beta

Then do some calculate, get A r e a = π 12 Area = \frac{\pi}{12}

Then do the same on a a and α \alpha , we can see that it is same as b b and β \beta .

So the total area is twice the b b has made,

T o t a l A r e a = π 6 Total Area = \boxed{\frac{\pi}{6}}

2 Helpful 0 Interesting 0 Brilliant 0 Confused

This should be a calculus problem lol. Anyway nice problem, enjoyed solving it.

donglin loo - 2 years, 11 months ago

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You're right, it should be, hahaha

Kelvin Hong - 2 years, 11 months ago

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