Difference between revisions of "1957 AHSME Problems/Problem 43"

Latest revision as of 10:10, 27 July 2024

Problem

We define a lattice point as a point whose coordinates are integers, zero admitted. Then the number of lattice points on the boundary and inside the region bounded by the $x$ -axis, the line $x = 4$ , and the parabola $y = x^2$ is:

$\textbf{(A)}\ 24 \qquad \textbf{(B)}\ 35\qquad \textbf{(C)}\ 34\qquad \textbf{(D)}\ 30\qquad \textbf{(E)}\ \infty$

Solution 1

$[asy] path p = (0,0){right}..(1,1)..(2,4)..(3,9)..(4,16); // Shaded Region fill(p--(4,0)--cycle,lightred); // x-Axis draw((-4,0)--(16,0), arrow=Arrows); label("$x$",(18,0)); // y-Axis draw((0,-4)--(0,16), arrow=Arrows); label("$y$",(0,18)); // y=x^2 draw(p); // x=4 draw((4,-5)--(4,20), arrow=Arrows(TeXHead)); [/asy]$

We want to find the number of lattice points in and on the boundary of the shaded region in the diagram. To do this, we will look at the integer values of $x$ from $0$ to $4$ . At a given value of $x$ , the amount of lattice points in the region is $x^2+1$ , because all of the integers from $0$ up to and including $x^2$ are in the region. Thus, evaluating this expression at at $x=0,1,2,3,$ and $4$ and adding the results together, we see that the number of lattice points is $1+2+5+10+17=35$ , so our answer is $\boxed{\textbf{(B) }35}$ .

Solution 2 (not as fast, but cool theorems)

Let us construct a simple polygon with vertices at $(0,0), (1,1), (2,4), (3,9), (4,16),$ and $(4,0)$ . Notice that this polygon overestimates the area of the shaded region in the diagram (because the parabola has positive concavity) yet does not have any more lattice points, because it matches the parabola at integral values of $x$ , and lattice points can only exist when $x$ is an integer. After using the Shoelace Theorem, we see that the area of the polygon is $\tfrac{114-70}2=22$ . Let $B$ be the number of lattice points on the boundary of the polygon, and let $I$ be the number of lattice points inside the polygon. We know that $B=5+17+5-3=24$ by counting the lattice points on the parabola, on the right edge, and on the bottom edge while taking care to subtract out the three points we counted twice (namely, $(0,0), (4,16),$ and $(4,0)$ ). Now, recalling that the polygon has area $22$ , we can use Pick's Theorem to solve for $I$ as follows: \begin{align*} I + \frac B 2 -1 &= 22 \\ I + \frac{24}{2} &= 23 \\ I + 12 &= 23 \\ I &= 11 \end{align*} Because the total amount of lattice points inside the polygon (and, consequently, inside the shaded region) is $I+B$ , our answer is $11+24=\boxed{\textbf{(B) }35}$ .

1957 AHSC (Problems • Answer Key • Resources)
Preceded by Problem 42	Followed by Problem 44
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Difference between revisions of "1957 AHSME Problems/Problem 43"

Latest revision as of 10:10, 27 July 2024

Contents

Problem

Solution 1

Solution 2 (not as fast, but cool theorems)

See Also

@@ Line 10: / Line 10: @@
 \textbf{(E)}\ \infty</math>
-== Solution ==
+== Solution 1 ==
-<math>\boxed{\textbf{(B) }35}</math>.
+<asy>
+path p = (0,0){right}..(1,1)..(2,4)..(3,9)..(4,16);
+// Shaded Region
+fill(p--(4,0)--cycle,lightred);
+// x-Axis
+draw((-4,0)--(16,0), arrow=Arrows);
+label("$x$",(18,0));
+// y-Axis
+draw((0,-4)--(0,16), arrow=Arrows);
+label("$y$",(0,18));
+// y=x^2
+draw(p);
+// x=4
+draw((4,-5)--(4,20), arrow=Arrows(TeXHead));
+</asy>
+We want to find the number of lattice points in and on the boundary of the shaded region in the diagram. To do this, we will look at the integer values of <math>x</math> from <math>0</math> to <math>4</math>. At a given value of <math>x</math>, the amount of lattice points in the region is <math>x^2+1</math>, because all of the integers from <math>0</math> up to and including <math>x^2</math> are in the region. Thus, evaluating this expression at at <math>x=0,1,2,3,</math> and <math>4</math> and adding the results together, we see that the number of lattice points is <math>1+2+5+10+17=35</math>, so our answer is <math>\boxed{\textbf{(B) }35}</math>.
+== Solution 2 (not as fast, but cool theorems) ==
+Let us construct a simple polygon with vertices at <math>(0,0), (1,1), (2,4), (3,9), (4,16),</math> and <math>(4,0)</math>. Notice that this polygon overestimates the area of the shaded region in the diagram (because the parabola has [[convex function|positive concavity]]) yet does not have any more lattice points, because it matches the parabola at integral values of <math>x</math>, and lattice points can only exist when <math>x</math> is an integer.
+After using the [[Shoelace Theorem]], we see that the area of the polygon is <math>\tfrac{114-70}2=22</math>. Let <math>B</math> be the number of lattice points on the boundary of the polygon, and let <math>I</math> be the number of lattice points inside the polygon. We know that <math>B=5+17+5-3=24</math> by counting the lattice points on the parabola, on the right edge, and on the bottom edge while taking care to subtract out the three points we counted twice (namely, <math>(0,0), (4,16),</math> and <math>(4,0)</math>). Now, recalling that the polygon has area <math>22</math>, we can use [[Pick's Theorem]] to solve for <math>I</math> as follows:
+\begin{align*}
+I + \frac B 2 -1 &= 22 \\
+I + \frac{24}{2} &= 23 \\
+I + 12 &= 23 \\
+I &= 11
+\end{align*}
+Because the total amount of lattice points inside the polygon (and, consequently, inside the shaded region) is <math>I+B</math>, our answer is <math>11+24=\boxed{\textbf{(B) }35}</math>.
 == See Also ==
-{{AHSME 50p box|year=1957|num-b=40|num-a=42}}
+{{AHSME 50p box|year=1957|num-b=42|num-a=44}}
 {{MAA Notice}}
 [[Category:AHSME]][[Category:AHSME Problems]]