2007 iTest Problems/Problem 16

Revision as of 20:01, 17 June 2018 by Rockmanex3 (talk | contribs) (Solution to Problem 16)

Problem

How many lattice points lie within or on the border of the circle in the $xy$-plane defined by the equation \[x^2+y^2=100\]

$\text{(A) }1\qquad \text{(B) }2\qquad \text{(C) }4\qquad \text{(D) }5\qquad \text{(E) }41\qquad \text{(F) }42\qquad \text{(G) }69\qquad \text{(H) }76\qquad \text{(I) }130\qquad \\ \text{(J) }133\qquad \text{(K) }233\qquad \text{(L) }311\qquad \text{(M) }317\qquad \text{(N) }420\qquad \text{(O) }520\qquad \text{(P) }2007$

Solution

Use casework to divide the problem into two cases -- points on the coordinate axes and points not on the coordinate axes.

  • For points on the axes, there are $10$ points on each ray plus the origin, making a total of $41$ points.
  • For points not on the axes, symmetry can be used by focusing on one quadrant then multiplying by four because the equation is a circle where the center is the origin. Because the points are within the circle, $x^2 + y^2 \le 100$. If $x = 9$, then$y \le 4$. If $x = 8$, then $y \le 6$. If $x = 7$, then $y \le 7$. If $x = 6$ or $x = 5$, then $y \le 8$. Finally, if $x \le 4$, then $y \le 9$. Altogether, there are a total of $9(4) + 8(2) + 7 + 6 + 4 = 69$ points in the first quadrant, so there are a total of $69 \cdot 4 = 276$ points not on the coordinate axes.

In total, there are $276 + 41 = \boxed{\textbf{(M) } 317}$ points within the circle.

See Also

2007 iTest (Problems, Answer Key)
Preceded by:
Problem 15
Followed by:
Problem 17
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 TB1 TB2 TB3 TB4