Circumcircle to four unit circles

Geometry Level pending

As shown in the figure, you are given a unit circle C 1 C_1 centered at A ( 2.5 , 0 ) A (2.5, 0) , and unit circle C 2 C_2 centered at B ( 1.25 , 1 ) B (1.25, -1) . You also have circles C 3 C_3 and C 4 C_4 which are reflections of C 1 C_1 and C 2 C_2 about the y y -axis. Find the radius R R of the circumscribed circle that is tangent to all four circles and has them inside it, and submit 1000 R \lfloor 1000 R \rfloor


The answer is 4106.

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2 solutions

As the system is symmetrical about the y y -axis, we can expect that the center of the circle tangent to the four circles lies on the y y -axis. Let the center be P ( 0 , y 0 ) P(0,y_0) . Since the P P is equidistance from the circumferences of the four circles, it must also be equidistance from the four centers. As the system is symmetrical about the y y -axis, we need only to consider one side and I am choosing C 1 C_1 and C 2 C_2 and we have:

( 5 2 ) 2 + y 0 2 = ( 5 4 ) 2 + ( y 0 + 1 ) 2 25 4 = 25 16 + 2 y 0 + 1 y 0 = 59 32 \begin{aligned} \left(\frac 52 \right)^2 + y_0^2 & = \left(\frac 54 \right)^2 + (y_0+1)^2 \\ \frac {25}4 & = \frac {25}{16} + 2y_0 + 1 \\ \implies y_0 & = \frac {59}{32} \end{aligned}

Then the radius of the circle tangent internally to the four circles is R = A P + 1 = ( 5 2 ) 2 + ( 59 32 ) 2 + 1 4.10635060199 1000 R = 4106 R = AP+1 =\sqrt{\left(\frac 52 \right)^2+\left(\frac {59}{32} \right)^2} + 1 \approx 4.10635060199 \implies \lfloor 1000R \rfloor = \boxed{4106} .

Nice solution! A slight typo: instead of 59 16 \dfrac{59}{16} it is 59 32 \dfrac{59}{32} (in two occasions).

Thanos Petropoulos - 5 months, 1 week ago

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Thanks a lot. I have changed it.

Chew-Seong Cheong - 5 months, 1 week ago
Saya Suka
Jan 3, 2021

Equate the locus equations of C1 & C2 (both being unit circles, therefore the same radii) to get their division line of
y = (-5/4)x + 59/32
Knowing that the other two are originals' reflection by y-axis, their (C3 & C4's pair) division line is
y = (5/4)x + 59/32
So our big circle is centred at (0,59/32).
The distance between this center and point A is √[(5/2)² + (59/32)²] = 3.106. I thought that R should be both 3.106 ± 1, but the answer is 1000(3.106 + 1) = 4106



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