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

(Solution 1)
(Solution 2)
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== Solution 2 ==
 
== Solution 2 ==
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>
+
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> (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 ==
 
== See also ==
 
{{AHSME box|year=1998|num-b=28|num-a=30}}
 
{{AHSME box|year=1998|num-b=28|num-a=30}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Revision as of 21:35, 7 December 2014

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

The answer is actually (D)

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
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30
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