GMAT Club World Cup 2022 (DAY 9): Two circles are inscribed in a recta

Hello everyone,

The correct answer is B.
The detailed explanation of the solution is in the pic. attached

Refering to the picture, 3R (in red) = 3; R=1

Also, 4-2R=2.

Distance between the center will be 1ˆ2+2ˆ2= Dˆ2,

So D = \sqrt{5}

Option B is the correct answer

if you take 1 circle it become inradius of pyth. 3,4,5 so r=3+4-5/2=1 and 2r=2
answer is A

Let the circle on the left be Circle A and the circle on the right be Circle B.

The Perpendicular of both Right angled triangles (together forming a rectangle) = 4, the base = 3. Therefore, the hypotenuse = 5.

With this, we can find the inradius (r) of both the triangles, i.e., (P + B - H)/2
r = 1

If you draw a line from the centre of circle A to the perimeter of Circle B, it gives us a right angled triangle with Perpendicular as 2, base as 1. And the hypotenuse of this triangle is the distance between the two centres, i.e., Sqrt(4 + 1) = rt(5).

Hence, option B is the answer.

In the figure below we extend the circle's radius to meet the sides and construct a right angled triangle as shown.



In \(\triangle ABC\) = \(\sqrt{(3-2r)^2 + (4-2r)^2}\) ----------- (1)

Now lets construct three triangles as shown below -



PM = 5 (3 - 4 -5 triplet)

Area of \(\triangle MPN\) = Area( \(\triangle MOP\) ) + Area( \(\triangle PON\) ) + Area( \(\triangle MON\) )

\(\frac{(4*3)}{2}\) = \(\frac{1}{2} * 4 * r +\frac{1}{2} * 3 * r + \frac{1}{2} * 5 * r\) = 6r

r = 1 -------------------- (2)

Substituting 2 in 1

AB =\( \sqrt{1^2 + 2^2}\) = \(\sqrt{5}\)

IMO B

Answer is B.

the length has to be smaller than 3...

Correct answer : choice B

Let r be the radius of the circles
Drop two perpendiculars each at each of the centers of the circles .

Connect the perpendicular from circle1 to circle 2 to form a right angled triangle with the line joining the two centers as the hypotenous

let one side of this newly formed triangle = a
let other side of this newly formed triangle = b
now we have r + a + r = 3
=> a = 3 - 2r

simlarly
we have r + b + 4 = 4
=> b = 4 - 2r

there fore hypotenuse = Root of ( (4-2r) whole square + (3-2r) whole square) ----> main equation to find the distance between the centers

need to find r
area of half of the rectangle = 1/2 * 3 * 4 = 6
area of half of the rectangle is also equal to = 3r/2 + 4r/2 + 5r/2
=> 6r
there fore 6r = 6
=> r = 1

substituing r in the main equaiton
root of ( (4-2) whole square + (3-2) whole square
root(4+1)
=> root 5
=> choice B

the diagonal length is 5
so option C,D,E are ruled out
between A&B. with A as distance, the two identical circles may not touch the two sides and the diagonal
Option B is correct

Hope this helps!!

Posted from my mobile device

Given: Two circles are inscribed in a rectangle, as shown above.

Asked: What is the distance between the centers of these circles?

For circle with centre C1: -
Let us assume centre = (x,y)
Radius r = x = 3-y = |4y - 3x|/5 = (4y-3x)/5 : since y>x
4(3-x) - 3x = 5x
12 - 7x = 5x
x = 1
y = 2

For circle with centre C2: -
Let us assume centre = (x,y)
Radius r = 4-x = y = |4y - 3x|/5 = (3x-4y)/5 : since x>>y
3(4-y) - 4y = 5y
12 - 7y = 5y
y = 1
x = 3

Distance between centre C1 (1,2) & centre C2 (3,1) = \(\sqrt{(1-3)^2 + (2-1)^2}= \sqrt{2^2 + 1^2}=\sqrt{5}\)

IMO B

The diagonal divides the rectangle in two equal areas. Hence, the circles created with tangent to the sides and diagonal are congruent.
Let r be the radius of each circle.
The length of rectangle is 4 radius across and width is 3 radius across.
Length = 4 = 4r and Width = 3 = 3r
=> r = 1

The straight line distance between the circles forms the hypotenuse of a right angled triangle formed with centre of both circles.
The two sides of this right angled triangle are 2 (two radius across) and 1 (one radius across).
Hence, the distance between two circles = √(2)^2 + (1)^2 = √5

Answer: B

Let, the distance between two centers of the circles= d
& Let the radius of each circle= r

if they are in the same plane then distance should be 2, but they have some angle, so it should be more than 2 but less than 3

answer B

3r=3
So, r=1
Reqd distance=\(\sqrt{r^2+(4-2r)^2}\)
=\(\sqrt{1+4}\)
=\(\sqrt{5}\)
Answer is B.

We can divide the length of the rectangle into 3 parts, both the two external parts are r, instead the central part is 4-r-r = 3-2r
In the same way, the width of the rectangle can be divided into 3 parts which are: r, 3-2r, r

We draw the triangle that has as hypotenuse the line that connects the 2 circles and we know that the value of the line value is √(3-2r)^2 + (4-2r)^2
To find the line, we need to find the radius r

Given the problem's data, we know that the area of half the rectangle is 4x3/2=12
In each half rectangle, we can draw 3 different triangles, and the sums of their area must be 12 as well
The area of the first triangle has base 3 and height r => 3r/2
The area of the second triangle has base 4 and height r =>4r/2
The area of the third triangle has base 5 (Which is the diagonal of the rectangle), and height r => 5r/2
3r/2 + 4r/2 + 5r/2 = 12
3r+4r+5r=12
12r=12
r=1

Given the radius we can use the previous formula to find the distance between the two circles
√(3-2r)^2 + (4-2r)^2
√4+1 = √5

IMO Option B is the correct option.

The figure looks like it is drawn to scale.

The distance between the centers of these circles looks shorter than the side 3 of the rectangle.

The distance between the centers of these circles looks longer than the half of side 4 of the rectangle.

I suspect the distance between the centers of the circle is bigger than 2, but smaller than 3.

By visual estimation, it seems the diameter of each of the circle is very close to 2.

The line joining the center of the circles is slightly bigger than the diameter.

Therefore in my estimation, the line joining the center of the circles should be slightly bigger than 2.

Only Option B fits my estimations. So, I chose B.

[quote="Bunuel"]
Two circles are inscribed in a rectangle, as shown above. What is the distance between the centers of these circles?

A. 2
B. \(\sqrt{5}\)
C. 3
D. \(\sqrt{10}\)
E. \(2\sqrt{5}\)

Solution is attached.