Difference between revisions of "2007 iTest Problems/Problem 16"

Latest revision as of 19:28, 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.

$[asy] import graph; size(9.22 cm); real lsf=0.5; pen dps=linewidth(0.7)+fontsize(10); defaultpen(dps); pen ds=black; real xmin=-1.2,xmax=10.2,ymin=-1.2,ymax=10.2; pen cqcqcq=rgb(0.75,0.75,0.75), evevff=rgb(0.9,0.9,1), zzttqq=rgb(0.6,0.2,0); /*grid*/ pen gs=linewidth(0.7)+cqcqcq+linetype("2 2"); real gx=1,gy=1; for(real i=ceil(xmin/gx)*gx;i<=floor(xmax/gx)*gx;i+=gx) draw((i,ymin)--(i,ymax),gs); for(real i=ceil(ymin/gy)*gy;i<=floor(ymax/gy)*gy;i+=gy) draw((xmin,i)--(xmax,i),gs); Label laxis; laxis.p=fontsize(10); xaxis(xmin,xmax,defaultpen+black,Ticks(laxis,Step=1.0,Size=2,NoZero),Arrows(6),above=true); yaxis(ymin,ymax,defaultpen+black,Ticks(laxis,Step=1.0,Size=2,NoZero),Arrows(6),above=true); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); draw(Arc((0,0),10,-5,95)); for (int i=1; i < 10; ++i) { for (int j=1; j < 10; ++j) { if (i*i + j*j <= 100) { dot((i,j)); } } } [/asy]$

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

2007 iTest (Problems, Answer Key)
Preceded by: Problem 15	Followed by: Problem 17
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Difference between revisions of "2007 iTest Problems/Problem 16"

Latest revision as of 19:28, 17 June 2018

Problem

Solution

See Also

@@ Line 24: / Line 24: @@
 * For points on the axes, there are <math>10</math> points on each ray plus the origin, making a total of <math>41</math> 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, <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.
+<asy>
+import graph; size(9.22 cm); real lsf=0.5; pen dps=linewidth(0.7)+fontsize(10); defaultpen(dps); pen ds=black;
+real xmin=-1.2,xmax=10.2,ymin=-1.2,ymax=10.2;
+pen cqcqcq=rgb(0.75,0.75,0.75), evevff=rgb(0.9,0.9,1), zzttqq=rgb(0.6,0.2,0);
+/*grid*/ pen gs=linewidth(0.7)+cqcqcq+linetype("2 2"); real gx=1,gy=1;
+for(real i=ceil(xmin/gx)*gx;i<=floor(xmax/gx)*gx;i+=gx) draw((i,ymin)--(i,ymax),gs); for(real i=ceil(ymin/gy)*gy;i<=floor(ymax/gy)*gy;i+=gy) draw((xmin,i)--(xmax,i),gs);
+Label laxis; laxis.p=fontsize(10);
+xaxis(xmin,xmax,defaultpen+black,Ticks(laxis,Step=1.0,Size=2,NoZero),Arrows(6),above=true); yaxis(ymin,ymax,defaultpen+black,Ticks(laxis,Step=1.0,Size=2,NoZero),Arrows(6),above=true);
+clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle);
+draw(Arc((0,0),10,-5,95));
+for (int i=1; i < 10; ++i)
+{
+  for (int j=1; j < 10; ++j)
+  {
+    if (i*i + j*j <= 100)
+    {
+      dot((i,j));
+    }
+  }
+}
+</asy>
+In total, there are <math>276 + 41 = \boxed{\textbf{(M) } 317}</math> points within the circle.
 ==See Also==