Mind Your Decisions, I am Presh Talwalkar. In square ABCD, E is the midpoint of AD. Construct a circle between the three points: E, B, and C.
Which shape has the larger perimeter, the square or the circle? I saw this problem on Math StackExchange. It was described as a surprisingly hard problem for high school students.
Pause if you'd like to give this problem a try, and when you're ready, keep watching to learn how to solve this problem. I'm going to present three different ways to solve this problem. Let's get started with the intersecting chords theorem.
To begin, we'll construct the perpendicular bisector of the top side of the square. Since point E, this top point is tangent to the circle; this perpendicular bisector is a diameter of the circle It also bisects the bottom side of the square. So this length is equal to Y, this length will also be equal to Y, and the entire side of the square will be equal to 2y.
If we write "r" for the radius of the circle, then 2r is the diameter of the circle, and the remaining distance on the diameter is equal to 2r - 2y. Now we can use the intersecting chords theorem. We must have: 2y(2r-2y) = y(y) Let's simplify this equation.
We'll first distribute, and then we'll group like terms. We'll then factor y, and we can then solve this equation. y is either equal to 0 or y is equal to 4/5 r.
Since we want y to be a positive value, so the square side has a positive length, We reject y=0. Therefore, Y = 4/5 r, which means r = 5/4 y. Let's now calculate the square's perimeter.
The side of the square is equal to 2y, which means the perimeter is equal to 4(2y), which is equal to 8y. Now, let's calculate the circle's perimeter or its circumference. That'll be equal to 2πr.
Now, r is equal to 5/4 multiplied by y, which means 2πr is equal to 5π/2 multiplied by y. And this is approximately equal to 7. 85y which is definitely less than 8y.
Therefore, the square has the larger perimeter. Now, let's proceed to the next method which uses a property of right triangles. We'll proceed like method 1 where we have the perpendicular bisector.
Notice this will also be perpendicular on this side of the square. We'll label the lengths as before. And now, if this is the center of the circle, this length will be equal to r, and this length will also be a radius of the circle.
Now, this distance is equal to x, what must x be equal to? Well, we want the entire side of the square to be equal to 2y, therefore we must have x + r = 2y, which means x = 2y - r. We now consider this right triangle.
By the Gogu theorem, we have (2y - r)^2 + y^2 = r^2 We can then simplify this equation, and just like before, we end up with a very similar equation. We have two possibilities, and we reject Y = 0. And this proceeds just like before where we calculate the perimeter of the square and the circle, and we're going to get that the square's perimeter is going to be larger than the circle's perimeter.
Now on to method 3 using similar triangles. Let's construct the following chords of the circle. We have this chord and this chord.
Now, this entire triangle will have an angle that's inscribed in a semicircle. Therefore, this is a right angle. Now, that means this is an altitude on the hypotenuse of this right triangle.
Therefore, it divides this right triangle into two similar triangles. This is the larger right triangle, and it has a ratio of its short leg to its long leg equal to 1/2 That's y/2y This will be similar to this right triangle, which must have the same ratio. In this triangle, the longer leg is equal to y, so the shorter leg must be equal to y/2 to get a ratio of 1/2.
We now have r is the radius which means that 2r = 2y + y/2 We then have r is equal to 5/4y, which means just as before we calculate the squares perimeter and the circle's perimeter and we're going to get that the square's perimeter is larger. I know there are many other ways to solve this problem It's a really fun one. Do check out the thread on Math StackExchange and you can share your solutions over there.
Thanks for making Mind Your Decisions one of the best channels on YouTube. As always, thanks for watching and thanks for your support!
