This Math Puzzle Went Viral! – Can YOU Solve it?

[00:00:00]Hello my lovelies. It's Suzanna, and today I want to show you how to solve this problem.

[00:00:06]A rectangle contains three circles. We can see the rectangle here, and inside we have these three circles.

[00:00:16]The circles are tangent to the sides of the rectangle and are also tangent to one another.

[00:00:25]So, these circles touch the sides of the rectangle here, everywhere, and they also touch each other here and here.

[00:00:38]What is the length of segment AB? We have to find this length here.

[00:00:46]Okay, we are given the diameters of these circles. I'm not a fan of using the diameters, so let's find the radius here of this circle, which would be half the diameter. So, half of three would be 1.5.

[00:01:07]Same here, if the diameter is four, half of it is two for my radius, and the same thing here, my radius, half of six is three.

[00:01:21]Usually, when you want to solve problems like this, these touching points here everywhere are really important to find a solution. So, let's start with this touching point here. I want to draw the radius here from the center to this touching point. This is my radius of length two here.

[00:01:46]The same thing here, from the center to this touching point, which is my radius of length three.

[00:01:54]And now, I have a line here of length 2 + 3, so of length 5.

[00:02:04]And the question is, why is this here actually a straight line? Because it consists of this part and of this part, but why does it transition here smoothly into one another?

[00:02:22]The reason is because these circles touch each other here.

[00:02:28]Because if we take a look at the tangent to this circle here at this point, if I draw the tangent, it would look like this.

[00:02:38]Then the radius is always perpendicular to this tangent that touches my circle at this point.

[00:02:49]And the same thing for this circle. It is the same tangent here at this touching point, and the radius is always automatically perpendicular to this tangent. And if I have a 90° angle here and here, this here has to be a straight line.

[00:03:12]Okay, so we have a straight line of length 5 here.

[00:03:17]But we were interested in the length of this horizontal line here, and our line is not horizontal, but we could draw a horizontal line here. So, if we go horizontally from here to here, we also get a right angle here. And now we've created a right triangle.

[00:03:42]We know the length of this side. We don't know the length of this side.

[00:03:48]Let's just call this part X. This would be actually a part of our segment. So, this would be X what we are trying to find right now.

[00:03:58]Uh what about the length of this side here?

[00:04:02]We can find it because we know the length of this part, which is three, and we know the length of this part, which is two. So, this is 3 - 2 then, which is equal to 1.

[00:04:19]And now, in our right triangle, we can use the Pythagorean theorem to find the length of X.

[00:04:26]For the Pythagorean theorem, we first need to know where our hypotenuse is.

[00:04:32]That is always the side that lies across the right angle. So, this is our hypotenuse.

[00:04:40]And the Pythagorean theorem then says, take one side, the one in our case, and square it, plus take the other side and square it, so X squared, and this is equal to the hypotenuse squared, so 5 squared.

[00:04:58]Let's calculate. 1 squared is equal to 1. 5 squared is equal to 25.

[00:05:06]And if we want to solve this equation for X now, let's bring the 1 to the other side by subtracting 1 on both sides of the equation.

[00:05:18]This is equal to zero. So, we only have X squared left here, and on the other side, 25 - 1 is equal to 24.

[00:05:30]To solve for X, we have to get rid of the square. So, let's take the square root on both sides of the equation, so that we get two solutions for our X. The first, X1, a positive solution, and a second x two negative solution.

[00:05:51]The positive solution is the square root of 24. We just leave it like this. We cannot really calculate it better right now. And we also get negative the square root of 24.

[00:06:08]But our x is the length of a side, so it has to be positive. So, we cannot use this negative solution here.

[00:06:18]We just take the square root of 24 for our x. So, we know the length of this part here now, which is the square root of 24 here and here.

[00:06:33]Okay, so we found this part already.

[00:06:35]Now, we only need to find the length of this part. So, let's go to new page and let's use this touching point now to find this length here.

[00:06:49]Because here we can do the same thing.

[00:06:51]From the center of the circle to this touching point, this is my radius of length 1.5.

[00:06:59]And here from the center to this touching point, this is my radius of length two.

[00:07:05]So, this line has a length of 1.5 + 2, which is 3.5 then.

[00:07:15]And I'm interested in this horizontal part, so let's draw this horizontal part here so that we get this right triangle this time.

[00:07:27]We don't know the length of this part. I call it y, so this will be my y.

[00:07:33]And what about the length of this part here?

[00:07:37]>> [sighs] >> Well, we know the length of this entire part, which is of length six. And we know the length of this part down here, which is two. And what about the length of this part here?

[00:07:52]This is my radius of my small circle here, which is of length 1.5.

[00:08:00]So, I have 1.5 here, which means that for this side, I have to take the six, subtract the two, which gives me four, and subtract the 1.5, which gives me 2.5 then for the length of this side.

[00:08:21]Now, we have a right triangle, and we can use the Pythagorean theorem again.

[00:08:27]The hypotenuse is across my right angle, so this is my hypotenuse, and for the Pythagorean theorem, I take one side then, it's the Y, and I square it, plus the other side, the 2.5, and I square it, and then I get the hypotenuse squared, so 3.5 squared.

[00:08:52]If you calculate these numbers, 2.5 squared is 6.25, and this squared is 12.25.

[00:09:08]And now we want to solve this equation for Y.

[00:09:12]So, let's subtract 6.25 on both sides of the equation, so that we have Y squared here. This is zero.

[00:09:26]And on the other side, 12.25 - 6.25 is equal to six.

[00:09:33]So, to solve for Y, we have to get rid of the square. So, let's take the square root on both sides and we get two solutions for our Y, a positive and a negative one.

[00:09:48]The square root of six is our positive solution and negative the square root of six here as well. Although, Y is the length of a side again, so we are not interested in the negative solution.

[00:10:06]We take the square root of six for our Y, so we know this length here now.

[00:10:15]And we're almost done. We are interested in this entire length here. So, now we only have to add these two square roots.

[00:10:26]So, we take the square root of six and add it to the square root of 24.

[00:10:34]Of course, you could use a calculator and get a decimal number as a result, but I want to find the exact value here first.

[00:10:46]Um so, let's add these two uh square roots. How's that possible?

[00:10:51]Not yet, actually, because we have different numbers here in our square root. We first have to see if there are any similarities between these numbers here.

[00:11:05]And the six is a part of the 24. I can write the 24 as 4 * 6, where I have the six in here.

[00:11:16]And I also have the six in the first square root.

[00:11:20]And now that I have a multiplication here, I am allowed to write this one big square root as two separate square roots that are multiplied by each other then.

[00:11:36]I have the four in my first square root and a six in my second.

[00:11:42]And the square root of four is something we can calculate because the four is a square number. So, the square root of four is equal to two. So, I'm going to erase this and write down a two.

[00:11:56]And now we have one square root of six here and I want to add two of the square root of six here. So, one of them plus two of them is three of the square root of six here. This is my exact value. And now if I want to use the calculator and calculate it, I get a result of 7.354, the length of the is segment. If you liked my video, please give it a thumbs up. It helps me a lot. I wish you wonderful day and I hope to see you in one of my next videos. Take care.