Geometry Arc Circle Area – PUTNAM Problem A2 1998 – Trigonometry Identity Proof

Geometry Arc Circle Area

I Found this Interesting Math Olympiad PUTNAM Problem and Its Solution (not mine) whilst browsing.

 

 

 

 

 

 

 

 

 

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A Comment for edification

 

A Very Clever Solution you have there.

 

Comment: Why Multiply by 2?

 

(i) Twice the First Picture in Addend

 

Note that

 

“First Picture in Addend” = (Area of Triangle) = (half of area of Top Left Rectangle)

 

Thus doubling it,

 

“Twice the First Picture in Addend” = Twice (half of area of Top Left Rectangle) = (Area of Top Left Rectangle)

 

(ii) Twice the Second Picture in Addend

 

Note that

 

“Second Picture in Addend” = (Area of Triangle) = (half of area of Bottom Right Rectangle)

Thus doubling it,

 

“Twice the Second Picture in Addend” = Twice (half of area of Bottom Right Rectangle) = (Area of Bottom Right Rectangle)

 

(iii) Twice the Third Picture in Addend

 

Note that

 

“Third Picture in Addend” = (Area of part-Arc)

 

Thus doubling it,

 

“Twice the Third Picture in Addend” = Twice (Area of part-Arc) = (Area of part-Arc COUNTED TWICE, one for its calculation in A and the other for its calculation in B.)

 

Yes only this area is counted “twice” in the Overall Sum because it occurs in both “A and B respectively” .

 

(iv) Total Shaded Area

 

(Total Shaded Area) or (A + B.)

 

= 2 x [ (half of area of Top Left Rectangle) + (half of area of Bottom Right Rectangle) + (Area of part-Arc)]

 

= 2 x [ Area of Arc with angle “theta” & radius “1”]

 

= 2 x [ (1/2) x r^2 x (theta) ], radius given as 1

 

= 2 x [ (1/2) x 1^2 x (theta) ]

 

= 2 x (1/2) x (theta)

 

= (theta)

 

= s

 

Depending only on the arc length s, when radius is given as claimed.

 

QED

 

Note:

 

Using (theta) in Radians,

 

r = radius = 1

 

(theta) = angle in Radians

 

s = arc length

 

Then,

 

s = r (theta)

 

s = (1) (theta)

 

s = (theta) which was substituted in the last line above.

 

 

P/S: A Visual Geometrical Proof  for the infamous elementary trigonometric identities

 

 

 

 

 

 

 

 

How to understand the proof above?

It’s difficult to explain here. Here are some useful pointers:

Example: Look at shaded triangle in Second Picture above.

 

Sin (a + b)

 

= (Vertical dotted line)/ (Hypotenuse)

 

= (Cos a Sin b + Sin a Cos b) / 1

 

= (Cos a Sin b + Sin a Cos b)

 

Can you see it?

 

Note:

 

(Vertical dotted line) = (Cos a Sin b + Sin a Cos b)

 

By diagram construction.

 

Example: In the right triangle with angle “a” in the Top Right Corner,

 

It has (hypotenuse) = Sin b

 

and an ‘angle a’.

 

So, its

 

Sin a = (‘horizontal length’)/(hypotenuse)

 

Sin a = (‘horizontal length’)/(Sin b)

 

So, rewriting this we obtain:

 

(‘horizontal length’) = (Sin a) x (Sin b) as shown in Pic etc.

 

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