Difference between revisions of "1998 AHSME Problems/Problem 29"

(Solution 1)
(Solution 1)
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draw((0.8, -1.4+e)--(1.8-e, 0.6)--(-0.2, 1.6-e)--(-1.2+e, -0.4)--cycle);
 
draw((0.8, -1.4+e)--(1.8-e, 0.6)--(-0.2, 1.6-e)--(-1.2+e, -0.4)--cycle);
 
</asy>
 
</asy>
The best square's side length is slightly less than <math>\sqrt 5</math>, yielding an answer of <math>\textbf{(D) }5.0</math>
+
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>
  
 
== Solution 2 ==
 
== Solution 2 ==

Revision as of 00:51, 7 February 2017

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 1

[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)); 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 $\boxed{\textbf{(D) }5.0}$

Solution 2

Apply Pick's Theorem. 4 lattice points on the border edges, 3 points in the interior. $A = I + \frac{B}{2} -1$, implying that $max(A) = 4$, $\boxed{A}$ (This is incorrect, because the vertices are not necessarily lattice points. The key idea is in fact to consider the lines on which the sides lie, and in fact not the vertices.

See also

1998 AHSME (ProblemsAnswer KeyResources)
Preceded by
Problem 28
Followed by
Problem 30
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