I ♡ \heartsuit You! Father's Day Variation

Calculus Level 4

For Father's Day, Little Albert wants to show just how much he loves his dad by making a card for him with a big, colorful heart on the front. Remembering his previous mistake on Mother's Day, he tries extra carefully to draw a heart without a loop.

Unfortunately, Little Albert got a little carried away. Not only did he still draw a loop inside the heart, but he also drew one around it! The polar equation for the curve he drew is r = sin ⁡ θ 5 . r = \sin \dfrac{\theta}{5}. Little Albert decides to cut out the heart shape in the middle and use that to make a card for his dad. This shape is shaded in red in the diagram.

The area of this heart-shaped region is equal to π a − b c d − f , \frac{\pi}{a} - \frac{b}{c} \sqrt{d - \sqrt{f}}, where a , b , c , d , f a, b, c, d, f are all positive integers such that gcd ⁡ ( b , c ) = 1 \gcd(b, c) = 1 and f f is as small as possible.

Find the value of a + b + c + d + f . a + b + c + d + f.


The answer is 40.

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Steven Yuan by Steven Yuan , 20, USA

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1 solution

Steven Yuan
Jun 18, 2017

Let's focus on one portion of the shaded region, which is the portion of r = sin ⁡ θ 5 r = \sin \dfrac{\theta}{5} in the interval π 2 ≤ θ ≤ 3 π 2 . \dfrac{\pi}{2} \leq \theta \leq \dfrac{3\pi}{2}. Since the graph of r = sin ⁡ θ 5 r = \sin \dfrac{\theta}{5} is symmetric about the y-axis, twice the area of this portion will equal the area of the heart. Let A A be the area of the heart Little Albert cuts out. We have

A = 2 ⋅ 1 2 ∫ π 2 3 π 2 sin ⁡ 2 θ 5 d θ = ∫ π 2 3 π 2 1 − cos ⁡ 2 θ 5 2 d θ = 1 2 [ θ − 5 2 sin ⁡ 2 θ 5 ] π 2 3 π 2 = 1 2 [ ( 3 π 2 − 5 2 sin ⁡ 3 π 5 ) − ( π 2 − 5 2 sin ⁡ π 5 ) ] = π 2 − 5 4 ( sin ⁡ 3 π 5 − sin ⁡ π 5 ) \begin{aligned} A &= 2 \cdot \dfrac{1}{2} \int_{\frac{\pi}{2}}^{\frac{3\pi}{2}} \sin^2 \dfrac{\theta}{5} d \theta \\ &= \int_{\frac{\pi}{2}}^{\frac{3\pi}{2}} \dfrac{1 - \cos \frac{2\theta}{5}}{2} d \theta \\ &= \dfrac{1}{2} \left [ \theta - \dfrac{5}{2} \sin \dfrac{2\theta}{5} \right ]_{\frac{\pi}{2}}^{\frac{3\pi}{2}} \\ &= \dfrac{1}{2} \left [ \left (\dfrac{3\pi}{2} - \dfrac{5}{2} \sin \dfrac{3\pi}{5} \right ) - \left (\dfrac{\pi}{2} - \dfrac{5}{2} \sin \dfrac{\pi}{5} \right ) \right ] \\ &= \dfrac{\pi}{2} - \dfrac{5}{4} \left ( \sin \dfrac{3\pi}{5} - \sin \dfrac{\pi}{5} \right ) \\ \end{aligned}

To simplify this into the form we desire, first we have sin ⁡ 3 π 5 = sin ⁡ 2 π 5 . \sin \dfrac{3\pi}{5} = \sin \dfrac{2\pi}{5}. Then, since

cos ⁡ π 5 = 5 + 1 4 cos ⁡ 2 π 5 = 5 − 1 4 , \begin{aligned} \cos \dfrac{\pi}{5} &= \dfrac{\sqrt{5} + 1}{4} \\ \cos \dfrac{2\pi}{5} &= \dfrac{\sqrt{5} - 1}{4}, \end{aligned}

we get sin ⁡ π 5 = 1 − cos ⁡ 2 π 5 = 5 − 5 8 \sin \dfrac{\pi}{5} = \sqrt{1 - \cos^2 \dfrac{\pi}{5}} = \sqrt{\dfrac{5 - \sqrt{5}}{8}} and sin ⁡ 2 π 5 = 1 − cos ⁡ 2 2 π 5 = 5 + 5 8 . \sin \dfrac{2\pi}{5} = \sqrt{1 - \cos^2 \dfrac{2\pi}{5}} = \sqrt{\dfrac{5 + \sqrt{5}}{8}}. So,

sin ⁡ 3 π 5 − sin ⁡ π 5 = sin ⁡ 2 π 5 − sin ⁡ π 5 = 5 + 5 8 − 5 − 5 8 . \sin \dfrac{3\pi}{5} - \sin \dfrac{\pi}{5} = \sin \dfrac{2\pi}{5} - \sin \dfrac{\pi}{5} = \sqrt{\dfrac{5 + \sqrt{5}}{8}} - \sqrt{\dfrac{5 - \sqrt{5}}{8}}.

Let x = 5 + 5 8 − 5 − 5 8 . x = \sqrt{\dfrac{5 + \sqrt{5}}{8}} - \sqrt{\dfrac{5 - \sqrt{5}}{8}}. Then,

x = 5 + 5 8 − 5 − 5 8 x 2 = 1 8 ( ( 5 + 5 ) + ( 5 − 5 ) − 2 ( 5 + 5 ) ( 5 − 5 ) ) x 2 = 1 8 ( 10 − 2 20 ) x 2 = 1 4 ( 5 − 20 ) x = 1 2 5 − 20 . \begin{aligned} x &= \sqrt{\dfrac{5 + \sqrt{5}}{8}} - \sqrt{\dfrac{5 - \sqrt{5}}{8}} \\ x^2 &= \dfrac{1}{8} \left ( \left (5 + \sqrt{5} \right) + \left (5 - \sqrt{5} \right ) - 2 \sqrt{\left (5 + \sqrt{5} \right ) \left (5 - \sqrt{5} \right)} \right ) \\ x^2 &= \dfrac{1}{8} \left (10 - 2 \sqrt{20} \right ) \\ x^2 &= \dfrac{1}{4} \left (5 - \sqrt{20} \right ) \\ x &= \dfrac{1}{2} \sqrt{5 - \sqrt{20}}. \\ \end{aligned}

Therefore, sin ⁡ 3 π 5 − sin ⁡ π 5 = x = 1 2 5 − 20 . \sin \dfrac{3\pi}{5} - \sin \dfrac{\pi}{5} = x = \dfrac{1}{2} \sqrt{5 - \sqrt{20}}.

Finally,

A = π 2 − 5 4 ( sin ⁡ 3 π 5 − sin ⁡ π 5 ) = π 2 − 5 4 ( 1 2 5 − 20 ) = π 2 − 5 8 5 − 20 , \begin{aligned} A &= \dfrac{\pi}{2} - \dfrac{5}{4} \left ( \sin \dfrac{3\pi}{5} - \sin \dfrac{\pi}{5} \right ) \\ &= \dfrac{\pi}{2} - \dfrac{5}{4} \left ( \dfrac{1}{2} \sqrt{5 - \sqrt{20}} \right ) \\ &= \dfrac{\pi}{2} - \dfrac{5}{8} \sqrt{5 - \sqrt{20}}, \end{aligned}

and a + b + c + d + f = 2 + 5 + 8 + 5 + 20 = 40 . a + b + c + d + f = 2 + 5 + 8 + 5 + 20 = \boxed{40}.

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