A Red Circle

Geometry Level pending

A B C D ABCD is the unit square. A quadrant is centered at B B . A D AD and A B AB are semicircle diameters. The red circle touches all three curves. If the radius of the red circle is expressed as p q \frac{p}{q} , where p p and q q are positive coprime integers, submit p + q p+q .


The answer is 37.

Fletcher Mattox by Fletcher Mattox , 72, USA

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2 solutions

Let A A be the origin ( 0 , 0 ) (0,0) of the x y xy -plane, r r be the radius and O ( x , y ) O(x,y) be the center of the red circle, and M ( 0 , 1 2 ) M \left(0, \frac 12 \right) and N ( 1 2 , 0 ) N \left(\frac 12, 0 \right) be the midpoints of D A DA and A B AB respectively. By Pythagorean theorem :

{ O M 2 : x 2 + ( y − 1 2 ) 2 = ( 1 2 − r ) 2 . . . ( 1 ) O N 2 : ( x − 1 2 ) 2 + y 2 = ( 1 2 + r ) 2 . . . ( 2 ) O B 2 : ( x − 1 ) 2 + y 2 = ( 1 − r ) 2 . . . ( 3 ) \begin{cases} OM^2: & x^2 + \left(y - \dfrac 12\right)^2 = \left(\dfrac 12 - r\right)^2 & ...(1) \\ ON^2: & \left(x - \dfrac 12\right)^2 + y^2 = \left(\dfrac 12 + r\right)^2 & ...(2) \\ OB^2: & (x-1)^2 + y^2 = (1-r)^2 & ...(3) \end{cases}

( 2 ) − ( 3 ) : x − 3 4 = 3 r − 3 4 ⟹ x = 3 r \begin{aligned} (2)-(3): \quad x - \frac 34 & = 3r - \frac 34 \\ \implies x & = 3r \end{aligned}

( 1 ) − ( 2 ) : x − 1 4 − y + 1 4 = − 2 r x − y = − 2 r ⟹ y = x + 2 r = 5 r \begin{aligned} (1)-(2): \quad x - \frac 14 - y + \frac 14 & = - 2r \\ x - y & = -2r \\ \implies y & = x + 2r = 5r \end{aligned}

( 3 ) : ( 3 r − 1 ) 2 + 25 r 2 = ( 1 − r ) 2 9 r 2 − 6 r + 1 + 25 r 2 = r 2 − 2 r + 1 33 r 2 − 4 r = 0 Since r > 0 ⟹ r = 4 33 \begin{aligned} (3): \quad (3r-1)^2 + 25r^2 & = (1-r)^2 \\ 9r^2 - 6r + 1 + 25r^2 & = r^2 - 2r + 1 \\ 33r^2 - 4r & = 0 & \small \blue{\text{Since }r >0} \\ \implies r & = \frac 4{33} \end{aligned}

Therefore p + q = 4 + 33 = 37 p+q = 4+33 = \boxed{37} .

2 Helpful 1 Interesting 2 Brilliant 0 Confused
Saya Suka
Jan 3, 2021

Let the littlest circle radius be 1 and square's side be 2r. We'll have a quadrilateral with sides r, r, r-1 and r+1 at the square's lower left corner with 2 perpendicular half sides, radii difference of internally kissing circles (littlest and the left semicircle) and sum of radii of externally kissing circles (littlest and the lower semicircle). Additionally, we have a conjoined obtuse scalene triangle with base r, shared side r+1 and longest side 2r-1 from radii difference of internally kissing circles (littlest and the quarter).
Also let's have the triangle height as y and the other leg of right triangle with hypothenuse r+1 as x.

(r + 1)² = x² + y² .......(1)
(2r - 1)² = (r + x)² + y² .......(2)
(r - 1)² = (r - x)² + (y - r)² .......(3)
Manipulating (1) & (3), we get
r + 2 = x + y .......(4)
Manipulating (1) & (4), we get
2r + 3 = 2xy .......(5)
Manipulating (1) & (2), we get
r - 3 = x .......(6)
From (4) & (6), we get y = 5
Putting x and y into (5), we get
r = 33/8
square's side = 2r = 33/4
If it was a unit square, the littlest radius would be 1/(33/4) = 4/33
Answer = p + q = 4 + 33 = 37


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