Dynamic Geometry: P23

Geometry Level 4

The diagram shows a unit square and a unit quadrant. The green circle is any circle which is always tangent to a side of the square and the quadrant. The centers of all such green circles trace a locus (blue curves).

The area bounded by the blue curves and the quadrant can be expressed as:

a b c d π e \frac {a\sqrt b-c}d - \frac \pi e

where a a , b b , c c , d d , and e e are positive integers, and b b and d d are primes. Find a + b + c + d + e + π \sqrt{a+b+c+d+e+\lceil \pi \rceil } .


The answer is 7.

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1 solution

Let the bottom-left vertex of the square be the origin ( 0 , 0 ) (0,0) of the x y xy -plane. Consider an arbitrary point P ( x , y ) P(x,y) on the top blue curve, where the radius of the circle is r r . Then we note that x 2 + y 2 = ( 1 + r ) 2 x^2+y^2 = (1+r)^2 . But we note that r = 1 y r = 1-y . Then we have:

x 2 + y 2 = ( 1 + 1 y ) 2 = y 2 4 y + 4 x 2 = 4 4 y y = 1 x 2 4 \begin{aligned} x^2 + y^2 & = (1+1-y)^2 = y^2 - 4y + 4 \\ x^2 & = 4-4y \\ \implies y & = 1 - \frac {x^2}4 \end{aligned}

By symmetry, the right blue curve is given by x = 1 y 2 4 x = 1 - \dfrac {y^2}4 . The two curves meet at y = x y=x or

x = 1 x 2 4 x 2 + 4 x 4 = 0 x 2 + 4 x + 4 = 8 ( x + 2 ) 2 = 8 x = 2 ( 2 1 ) \begin{aligned} x & = 1 - \frac {x^2}4 \\ x^2 + 4x - 4 & = 0 \\ x^2 + 4x + 4 & = 8 \\ (x+2)^2 & = 8 \\ \implies x & = 2(\sqrt 2 - 1) \end{aligned}

The area bounded by the blue curves and the red quadrant A A is equal to the area bounded by the blue curves and the axes minus the area of the quadrant. The area of bounded by the blue curves and the axes is 2 A 1 + A 2 2A_1 + A_2 . [ U p d a t e d : ] \rm \red{[Updated: \ ]} Since y = 1 x 2 4 y = 1 - \dfrac {x^2}4 is a parabola, A 1 = 2 3 ( 1 2 ( 2 1 ) ) ( 2 ( 2 1 ) ) = 4 ( 5 2 7 ) 3 A_1 = \dfrac 23(1- 2(\sqrt 2-1))(2(\sqrt2 -1)) = \dfrac {4(5\sqrt 2-7)}3 . A 2 = ( 2 ( 2 1 ) ) 2 = 12 8 2 A_2 = (2(\sqrt 2-1))^2 = 12-8\sqrt 2 . Therefore A = 2 A 1 + A 2 π 4 = 16 2 20 3 π 4 A = 2A_1 + A_2 - \dfrac \pi 4= \dfrac {16\sqrt 2 -20}3 - \dfrac \pi 4 . And the required answer is 7 \boxed 7 .

Thank you for posting !

Valentin Duringer - 4 months ago

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You are welcome. I have edited the problem statement. Since there is only one square and one quadrant you don't need mention their color. Because nobody will be confused. Instead of of coprime and square-free we can fix the answer values using primes.

Chew-Seong Cheong - 4 months ago

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Understood thank you !

Valentin Duringer - 4 months ago

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@Valentin Duringer In geometry we talk about locus rather than moving. It is not a Mechanical Engineering problem. A moving object does not change in size. That was why I mentioned about variable circle. I have changed the problem statement again. Check if you understand,

Chew-Seong Cheong - 4 months ago

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@Chew-Seong Cheong I did understand.

Valentin Duringer - 4 months ago

You wrote: "By symmetry the right blue curve is given by" x 2 = 1 y 2 4 x^2=1-\frac{y^2}{4} The square on x x shouldn't be here (a typo, I guess). Nice solution nevertheless.

Veselin Dimov - 4 months ago

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Thanks, I have amended it.

Chew-Seong Cheong - 4 months ago

I did it the same way but I was lazy to do the integration, @Valentin Duringer I like your problems , please keep posting more

OM Krish - 4 months ago

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Hi, im happy you like them. I have made 100 of them so I ll be posting more everyday !

Valentin Duringer - 4 months ago

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Thanks a lot !!

OM Krish - 4 months ago

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@Om Krish @OM Krish I'm posting right, enjoy and share ! Thanks !

Valentin Duringer - 4 months ago

Actually integration is unnecessary for parabola. I have updated the solution above.

Chew-Seong Cheong - 4 months ago

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Sorry for the late reply I had actually changed accounts so I didn't see this , I like your solution sir

OM Krish - 3 months, 3 weeks ago

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