2025 AMC 8 Problems/Problem 18
Contents
[hide]Problem 18
The circle shown below on the left has a radius of 1 unit. The region between the circle and the inscribed square is shaded. In the circle shown on the right, one quarter of the region between the circle and the inscribed square is shaded. The shaded regions in the two circles have the same area. What is the radius , in units, of the circle on the right?
Solution
The area of the shaded region in the circle on the left is the area of the circle minus the area of the square, or . The shaded area in the circle on the right is
of the area of the circle minus the area of the square, or
, which can be factored as
. Since the shaded areas are equal to each other, we have
, which simplifies to
. Taking the square root, we have
~mrtnvlknv
Solution 2
We start with the first area. Since the square is inscribed, its diagonal is its side length is
its area is
, therefore the first area is
. The second area is
, found in a similar manner. Writing and solving the equation, we have
The answer is
~Tacos_are_yummy_1
Solution 3 (Using similarity)
Since the two figures are similar, and the 4 regions in the first circle are equal in area to one of the four regions in the second circle, the second figure's area is times the first figure's area. Therefore, the side lengths are multiplied by a factor of
, so the answer is
.
~alwaysgonnagiveyouup
Solution 4
(Really similar to Solution 1, but has a more larger explanation for the answer)
First find the area of the smaller circle with the known radius. Since the radius is , the area of the circle is
, and finding the side length of the square would equal to
. Squaring that gives us the area of the square which is
, and the shaded area of the first one is equal to
. The second area can be shortened to
, since we only need
th of the sections. We then make them equal, which makes this equation:
Simplifying this equation would be . We can factor out the
which makes
. Dividing out the
would give us
on the left side.
which leaves us to choice
.
~Imhappy62789
Solution 5
We are given two circles, each with an inscribed square. In the first circle, the radius is 1 unit, and the region between the circle and the inscribed square is shaded. In the second circle, one quarter of the region between the circle and the inscribed square is shaded. The shaded areas in both circles are the same. Our goal is to determine the radius
We know that the shaded region in the first circle (with radius 1) and the shaded region in the second circle (with radius
Let’s recall that the area between the circle and the inscribed square is proportional to the square of the radius of the circle because the square's side length scales with the radius of the circle.
So, in the first circle with radius 1:
and the shaded area is proportional to the entire area between the circle and the square, i.e., 1.
In the second circle with radius
and the shaded area is one-quarter of this total area, so it is proportional to
We are told that the shaded areas in both circles are the same, so we set up the following equation:
Multiply both sides of the equation by 4:
Taking the square root of both sides:
Thus, the radius of the second circle is
~aoum
Solution 6
Notice that if you shaded the rest of the region between the square and the circle, then the area would be times the area of the shaded region in the smaller circle. Since the radius changes by a factor of
,
, solving for gets you
. Hence, the answer is
~Avy11
Video Solution by Pi Academy (Clever)
https://youtu.be/Iv_a3Rz725w?si=E0SI_h1XT8msWgkK
Video Solution 1 by SpreadTheMathLove
https://www.youtube.com/watch?v=jTTcscvcQmI
Video Solution (A Clever Explanation You’ll Get Instantly)
https://youtu.be/VP7g-s8akMY?si=wti8JNhNkM77LtdZ&t=2146 ~hsnacademy
Video Solution by Thinking Feet
See Also
2025 AMC 8 (Problems • Answer Key • Resources) | ||
Preceded by Problem 17 |
Followed by Problem 19 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 | ||
All AJHSME/AMC 8 Problems and Solutions |
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