1998 AHSME Problems/Problem 29: Difference between revisions

Latest revision as of 23:54, 29 January 2019

Problem

A point $(x,y)$ in the plane is called a lattice point if both $x$ and $y$ are integers. The area of the largest square that contains exactly three lattice points in its interior is closest to

$\mathrm{(A) \ } 4.0 \qquad \mathrm{(B) \ } 4.2 \qquad \mathrm{(C) \ } 4.5 \qquad \mathrm{(D) \ } 5.0 \qquad \mathrm{(E) \ } 5.6$

Solution

$[asy] real e = 0.1; dot((0,-1)); dot((1,-1)); dot((-1,0)); dot((0,0)); dot((1,0)); dot((2,0)); dot((-1,1)); dot((0,1)); dot((1,1)); dot((0,2)); dot((-1,-1)); dot((2,2)); dot((1,2)); dot((2,1)); dot((2,-1)); dot((-1,2)); draw((0.8, -1.4+e)--(1.8-e, 0.6)--(-0.2, 1.6-e)--(-1.2+e, -0.4)--cycle); [/asy]$ The best square's side length is slightly less than $\sqrt 5$ , yielding an answer of $\boxed{\textbf{(D) }5.0}$

1998 AHSME (Problems • Answer Key • Resources)
Preceded by Problem 28	Followed by Problem 30
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1998 AHSME Problems/Problem 29: Difference between revisions

Latest revision as of 23:54, 29 January 2019

Problem

Solution

See also

@@ Line 5: / Line 5: @@
-== Solution 1 ==
+== Solution ==
-Sadly, I don't actually have a solution. However, after doing some work on Geogebra, I have convinced myself that the answer is almost certainly A
+<asy>
+real e = 0.1;
+dot((0,-1));
-== Solution 2 ==
+dot((1,-1));
-Apply Pick's Theorem. 4 lattice points on the border edges, 3 points in the interior. <math>A = I + \frac{B}{2} -1</math>, implying that <math>max(A) = 4</math>, <math>\boxed{A}</math>
+dot((-1,0));
+dot((0,0));
+dot((1,0));
+dot((2,0));
+dot((-1,1));
+dot((0,1));
+dot((1,1));
+dot((0,2));
+dot((-1,-1));
+dot((2,2));
+dot((1,2));
+dot((2,1));
+dot((2,-1));
+dot((-1,2));
+draw((0.8, -1.4+e)--(1.8-e, 0.6)--(-0.2, 1.6-e)--(-1.2+e, -0.4)--cycle);
+</asy>
+The best square's side length is slightly less than <math>\sqrt 5</math>, yielding an answer of <math>\boxed{\textbf{(D) }5.0}</math>
 == See also ==
 {{AHSME box|year=1998|num-b=28|num-a=30}}
 {{MAA Notice}}