Question number 24. A square fits snugly between the horizontal line and two touching circles, both of radius 1. What is the length of the side of the square here that's fitting in between the two circles and the line? Let me label that side here x, so we're going to be looking for x, and I will draw in a rectangle here that has vertices at the points where the circles touch the line and at the centers of the circles. So this rectangle over here has dimensions 1, that's the radius, and then that's twice the radius, so that is 2. And let me draw in then a radius of the circle, so that would be another radius, and from this point on, let me draw here a line like so, so I will have this right triangle over here. We know also that the hypotenuse of that right triangle is 1, and then I can copy some dimensions. This dimension here is x, I will call this one y, and here I will have z. So looking at that right triangle, we know that 1 is going to be z squared plus y squared, and looking at the radius of the circle here on the side, we also know that 1 is equal to y plus x. Now looking at the dimension over here, we know that 2 is going to be equal to 2z plus x. So if we run along that line, we have distance z, then x, and then here z again, so 2 is 2z plus x. And with these equations, we are able to solve for x, so let's do that now. Now we can solve for x. So we have that 1 is equal to z squared plus y squared, and then we have a relationship between z and x, and we have a relationship between y and x. So first, replacing for y, y is equal to 1 minus x, quantity squared, and then replacing for z, z is equal to 2 minus x divided by 2, quantity squared, and then we have 1 minus x here, quantity squared, all of this is still equal to 1. So here, that's the equation we have to solve for x, so let's do that. Let's expand, we have a 1 over 4 here, then 4 minus 4x plus x squared, plus 1 minus 2x plus x squared, and so that gives us 1 is equal to 1 minus x plus 1 fourth x squared, plus 1 minus 2x plus x squared, let's multiply everywhere by 4, so we have 4 is equal to 4 minus 4x plus x squared, plus 4 minus 8x plus 4x squared, and then we can subtract 4 from both sides, we obtain 5x squared, we have minus 12x, and then we have plus 4, and this factors into 5x minus 2, that's one factor, and the second factor is x minus 2. So there are two solutions, x is equal to 2, which is not possible, that is larger than the radius of the circle, and so the other solution is that x has to be here equal to 2 fifths. So x is equal to 2 fifths, we have solved for x, that is the side of the square here, and that would be answer A.

geometry

square

circle

algebra

solution