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Difference between revisions of "2007 iTest Problems/Problem 16"
(Solution to Problem 16)
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* 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, <math>x^2 + y^2 \le 100</math>.  If <math>x = 9</math>, then<math>y \le 4</math>.  If <math>x = 8</math>, then <math>y \le 6</math>.  If <math>x = 7</math>, then <math>y \le 7</math>.  If <math>x = 6</math> or <math>x = 5</math>, then <math>y \le 8</math>.  Finally, if <math>x \le 4</math>, then <math>y \le 9</math>.  Altogether, there are a total of <math>9(4) + 8(2) + 7 + 6 + 4 = 69</math> points in the first quadrant, so there are a total of <math>69 \cdot 4 = 276</math> points not on the coordinate axes.
 
* 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, <math>x^2 + y^2 \le 100</math>.  If <math>x = 9</math>, then<math>y \le 4</math>.  If <math>x = 8</math>, then <math>y \le 6</math>.  If <math>x = 7</math>, then <math>y \le 7</math>.  If <math>x = 6</math> or <math>x = 5</math>, then <math>y \le 8</math>.  Finally, if <math>x \le 4</math>, then <math>y \le 9</math>.  Altogether, there are a total of <math>9(4) + 8(2) + 7 + 6 + 4 = 69</math> points in the first quadrant, so there are a total of <math>69 \cdot 4 = 276</math> points not on the coordinate axes.
  
−
In total, there are <math>276 + 41 = \boxed{317}</math> points within the circle.
+
In total, there are <math>276 + 41 = \boxed{\textbf{(M) } 317}</math> points within the circle.
  
 
==See Also==
 
==See Also==

Revision as of 19:01, 17 June 2018

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
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