Question number 29. An oval inscribed in an 8 by 4 rectangle forms two pairs of identical arcs. Two smaller arcs here on the left and on the right and two larger ones on the top and on the bottom. Intersecting arcs are tangent to the same line at the point of intersection. So we have four of these points and they're labeled here with little notches in blue. We also know that the oval has vertical and horizontal symmetry and the question is what is the radius of the larger arc? So in other words what is the radius of the circle here that is used to make this arc? And I can draw in that circle to illustrate exactly which circle we're talking about. So roughly something like this. Here is a circle and it matches here the arc and I can draw in a smaller circle here too to illustrate how the smaller arc is made and at the point of tangency where they intersect, so over here for example, the tangent line I can approximate will be tangent to both. So for example a line like that. Okay and so we have the radius of the larger circle I can draw that in and I'll draw it through the center of the smaller circle and then I can label it. So we have, let me just call it radius here R like that so that would be radius R and then we can connect the centers of these smaller arcs like that with the line and make a right triangle here by connecting the center of the large circle to this line and that makes a right triangle here and by labeling the sides of this right triangle we can use the Pythagorean formula on it to solve for the radius capital R. So we know that the leg here is equal to the quantity, well we have R and then we have to subtract a little bit and what we're subtracting is exactly half the width of this rectangle so 2 because the distance here I will draw a line in blue from the center here along the same line to the arc that distance that I just labeled here that is distance 2 and so overall R plus 2 minus 2 is the radius here. So we have one leg the hypotenuse is we can read that off the whole radius is R but we're going the radius 1 out to the blue arc so the hypotenuse here we can label as R minus 1 that's over here R minus 1 is the hypotenuse and finally what we're missing is the distance here between the centers of the small arcs and the whole width of the rectangle is 8 this would be 8 divided by 2 and then we have to subtract 1 so that's 8 divided by 2 minus 1 so that distance is 3 and now we can use the Pythagorean formula so we have a right triangle with known lengths for the legs and the Pythagorean formula tells us that the square of the hypotenuse which is R minus 1 quantity squared is equal to the sum of the squares of the sides so that's 3 squared plus quantity R minus 2 squared and here R is the radius of the large arc and then we solve so let's multiply out on both sides here we have on the left hand side R squared minus 2R plus 1 on the right hand side 9 plus R squared minus 4R plus 4 and then we simplify the square terms here on both sides cancel and then we have on the left hand side 1 minus 2R on the right hand side we have 13 minus 4R and then collecting terms we would have a 2R on the left hand side a 12 on the right hand side dividing by 2 finally gives us the answer so R is equal to 6 and that we know is the radius of the large arcs and so we take that as our answer a 6

oval

rectangle

Pythagorean theorem

radius

geometry