Dynamic Geometry: P67

Geometry Level pending

The diagram shows a black semicircle with radius 1 1 . The yellow and the cyan semicircle are tangent to each other and internally tangent to the black semicircle. They are growing and shrinking freely so that the sum of their radius is always equal to 1 1 . We inscribe a green circle so that it's tangent to all three semicircles. The intersection points between the green circle and the two moving semicircles trace a locus (purple curve). The area bounded by the purple curve and the diameter of the black semicircle can be expressed as π p − q \dfrac{\pi }{p}-q , where p p and q q are positive integers. Find p − q p-q .


The answer is 1.

Valentin Duringer by Valentin Duringer , 30, Belgium

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1 solution

David Vreken
Mar 17, 2021

Label the diagram as follows, and let θ = ∠ H B E \theta = \angle HBE , r r be the radius of the yellow semicircle, and R R be the radius of the green circle:

Since the black semicircle has a radius of 1 1 , A F = 2 AF = 2 , so the radius of the cyan semicircle is 1 2 C F = 1 2 ( A F − A C ) = 1 2 ( 2 − 2 r ) = 1 − r \frac{1}{2}CF = \frac{1}{2}(AF - AC) = \frac{1}{2}(2 - 2r) = 1 - r .

That means B D = A D − A B = 1 − r BD = AD - AB = 1 - r , B H = B I + I H = r + R BH = BI + IH = r + R , H E = H J + J E = R + 1 − r HE = HJ + JE = R + 1 - r , B E = A F − A B − E F = 2 − r − ( 1 − r ) = 1 BE = AF - AB - EF = 2 - r - (1 - r) = 1 , and H D = G D − G H = 1 − R HD = GD - GH = 1 - R .

By the law of cosines on △ H B D \triangle HBD , cos ⁡ θ = B D 2 + B H 2 − H D 2 2 ⋅ B D ⋅ B H = ( r + R ) 2 + ( r + R ) 2 − ( 1 − R ) 2 2 ⋅ ( 1 − r ) ⋅ ( r + R ) \cos \theta = \cfrac{BD^2 + BH^2 - HD^2}{2 \cdot BD \cdot BH} = \cfrac{(r + R)^2 + (r + R)^2 - (1 - R)^2}{2 \cdot (1 - r) \cdot (r + R)} .

By the law of cosines on △ H B E \triangle HBE , cos ⁡ θ = B E 2 + B H 2 − H E 2 2 ⋅ B E ⋅ B H = 1 2 + ( r + R ) 2 − ( R + 1 − r ) 2 2 ⋅ 1 ⋅ ( r + R ) \cos \theta = \cfrac{BE^2 + BH^2 - HE^2}{2 \cdot BE \cdot BH} = \cfrac{1^2 + (r + R)^2 - (R + 1 - r)^2}{2 \cdot 1 \cdot (r + R)}

So cos ⁡ θ = ( r + R ) 2 + ( r + R ) 2 − ( 1 − R ) 2 2 ⋅ ( 1 − r ) ⋅ ( r + R ) = 1 2 + ( r + R ) 2 − ( R + 1 − r ) 2 2 ⋅ 1 ⋅ ( r + R ) \cos \theta = \cfrac{(r + R)^2 + (r + R)^2 - (1 - R)^2}{2 \cdot (1 - r) \cdot (r + R)} = \cfrac{1^2 + (r + R)^2 - (R + 1 - r)^2}{2 \cdot 1 \cdot (r + R)} , which rearranges to R = r ( 1 − r ) r 2 − r + 1 R = \cfrac{r(1 - r)}{r^2 - r + 1} .

Substituting R = r ( 1 − r ) r 2 − r + 1 R = \cfrac{r(1 - r)}{r^2 - r + 1} into cos ⁡ θ = ( r + R ) 2 + ( r + R ) 2 − ( 1 − R ) 2 2 ⋅ ( 1 − r ) ⋅ ( r + R ) \cos \theta = \cfrac{(r + R)^2 + (r + R)^2 - (1 - R)^2}{2 \cdot (1 - r) \cdot (r + R)} simplifies to cos ⁡ θ = ( 2 − r ) r r 2 − 2 r + 2 \cos \theta = \cfrac{(2 - r)r}{r^2 - 2r + 2} .

If D D is placed at the origin, then I I has an x x -coordinate of x = r − 1 + r cos ⁡ θ x = r - 1 + r \cos \theta and a y y -coordinate of y = r sin ⁡ θ = r 1 − cos ⁡ 2 θ y = r \sin \theta = r \sqrt{1 - \cos^2 \theta} .

Substituting cos ⁡ θ = ( 2 − r ) r r 2 − 2 r + 2 \cos \theta = \cfrac{(2 - r)r}{r^2 - 2r + 2} and combining gives x 2 + ( y + 1 ) 2 = 2 x^2 + (y + 1)^2 = 2 , so the purple locus is a fourth of a circle with a center of ( 0 , − 1 ) (0, -1) and a radius of 2 \sqrt{2} .

The area bounded by the purple curve and the diameter of the black semicircle is then 1 4 π ( 2 ) 2 − 1 2 ( 2 ) 2 = π 2 − 1 \cfrac{1}{4}\pi(\sqrt{2})^2 - \cfrac{1}{2}(\sqrt{2})^2 = \cfrac{\pi}{2} - 1 , so p = 2 p = 2 , q = 1 q = 1 , and p − q = 1 p - q = \boxed{1} .

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Thank you for posting David !

Valentin Duringer - 2 months, 3 weeks ago

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