2008 iTest Problems/Problem 32

Revision as of 20:53, 4 August 2018 by Rockmanex3 (talk | contribs) (Solution)

Problem

A right triangle has perimeter $2008$, and the area of a circle inscribed in the triangle is $100\pi^3$. Let A be the area of the triangle. Compute $\lfloor A\rfloor$.

Solution

We know the perimeter and area of incircle, so we can find the semiperimeter of the triangle and radius of the incircle and plug the values into the area formula $A = sr.$


The semiperimeter of the triangle is $\tfrac{2008}{2} = 1004.$ The radius of the incircle is $\sqrt{100 \pi} = 10\pi$. That means the area of the triangle is $1004 \cdot 10\pi = 10040\pi.$


By noting that $\pi \approx 3.14159265$ (or by using a calculator), the value $10040\pi$ is around $31541.59,$ so $\lfloor A \rfloor = \boxed{31541}$.

See Also

2008 iTest (Problems)
Preceded by:
Problem 31
Followed by:
Problem 33
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 76 77 78 79 80 81 82 83 84 85 86 87 88 89 90 91 92 93 94 95 96 97 98 99 100
Invalid username
Login to AoPS