Try this beautiful problem from Geometry based on ratio of Circles from AMC 10A, 2009, Problem 21.
Many Gothic cathedrals have windows with portions containing a ring of congruent circles that are circumscribed by a larger circle, In the figure shown, the number of smaller circles is four. What is the ratio of the sum of the areas of the four smaller circles to the area of the larger circle?

Geometry
Circle
Pythagoras
Answer: \(4(3-2 \sqrt{2})\)
AMC-10A (2009) Problem 21
Pre College Mathematics

We have to find out the ratio of the sum of the areas of the four smaller circles to the area of the larger circle. To find out the area any circle,we need radius.so at first we have to find out radius of two types circle.Can you find out the radius of two type circle i.e small circle and big circle..................
Can you now finish the problem ..........

Let the radius of the Four small circles be \(r\).Therfore from the above diagram we can say \(CD=DE=EF=CF=2r\). Now the quadrilateral \(CDEF\) in the center must be a square. Therefore from Pythagoras theorm we can say \(DF=\sqrt{(2r)^2 + (2r)^2}=2r\sqrt 2\). So \(AB=AD+DF+BF=r+2r\sqrt 2+r=2r+2r\sqrt 2\)
Therefore radius of the small circle is \(r\) and big circle is\(R=r+r \sqrt{2}=r(1+\sqrt{2})\)
Can you now finish the problem ..........

Therefore the area of the large circle is \(L=\pi R^{2}=\pi r^{2}(1+\sqrt{2})^{2}=\pi r^{2}(3+2 \sqrt{2}) \)and the The area of four small circles is \(S=4 \pi r^{2}\)
The ratio of the area will be \(\frac{S}{L}=\frac{4 \pi r^{2}}{\pi r^{2}(3+2 \sqrt{2})}\)
=\(\frac{4}{3+2 \sqrt{2}}\)
=\(\frac{4}{3+2 \sqrt{2}} \cdot \frac{3-2 \sqrt{2}}{3-2 \sqrt{2}}\)
=\(\frac{4(3-2 \sqrt{2})}{3^{2}-(2 \sqrt{2})^{2}}\)
=\(\frac{4(3-2 \sqrt{2})}{1}\)
=\(4(3-2 \sqrt{2})\)

In 2025, 8 students from Cheenta Academy cracked the prestigious Regional Math Olympiad. In this post, we will share some of their success stories and learning strategies. The Regional Mathematics Olympiad (RMO) and the Indian National Mathematics Olympiad (INMO) are two most important mathematics contests in India.These two contests are for the students who are […]

Cheenta Academy proudly celebrates the success of 27 current and former students who qualified for the Indian Olympiad Qualifier in Mathematics (IOQM) 2025, advancing to the next stage — RMO. This accomplishment highlights their perseverance and Cheenta’s ongoing mission to nurture mathematical excellence and research-oriented learning.

Cheenta students shine at the Purple Comet Math Meet 2025 organized by Titu Andreescu and Jonathan Kanewith top national and global ranks.

Celebrate the success of Cheenta students in the Stanford Math Tournament. The Unified Vectors team achieved Top 20 in the Team Round.