# Senior Problems

Take a shot at these senior-ly problems. ### Solution to Problem V-1-S.1:

Solution to Problem V-1-S.1: Let x,y,12 be the sides of the triangle. Then x^2+y^2 = 12, and we have to maximize p = x+y + 12 = x+ sqrt(144-x^2) + 12. dp/dx = 0 for x = 6 and d^2p/dx^2 < 0 for x = 6. Hence the greatest perimeter is 24 which we obtain for x = 6.

Solution to Problem V-1-S.2: Let E and F be the midpoints of BC and AD respectively. Clearly r_2 = |AE| = |EC| = |EP| = 10. |EF| = |AB| = 8. Hence |FP| = 6 since EP is the hypotenuse of the right angled triangle EFP. Let us assume that P lies on ED. Then |AP| = 16, |DP| = 4, |BP| = 8sqrt(5), |CP| = 4sqrt(5), r_1 =|BP|/2 = 4sqrt(5), r_3 = |CP| /2 = 2sqrt(5). ### Ayan, I'm afraid there are

Ayan, I'm afraid there are some errors in your working. These are mainly to do with assumptions that you are making but there are also some errors. Let me point some of these out to you and do try the solutions again.
V-1 S.1 '12' need not be the hyptenuse. Error in working, please check the derivative and also, do remember that the sides are integer sides.
V-1 S.2 Please redraw the diagram, your assumptions for r_2 are wrong. Also, how did you assume that P is on ED, you are given that P is on AD.
Do keep trying! we are happy to help you if you get stuck ### You are correct Sneha. I don

You are correct Sneha. I don't know why I assumed that the hypotenuse is 12. Probably because of trying to think while typing the power signs :)

If x is the length of the hypotenuse and y is the length of the third side, then
x^-y^2 = 144 and we have to maximize x+y. Given x,y are integer, x-y has to be at least 2 (since 1 makes them fractional) and hence x = 37, y = 35 and the perimeter is 84.

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