1994 AIME Problems/Problem 12

Problem

A fenced, rectangular field measures $24$ meters by $52$ meters. An agricultural researcher has 1994 meters of fence that can be used for internal fencing to partition the field into congruent, square test plots. The entire field must be partitioned, and the sides of the squares must be parallel to the edges of the field. What is the largest number of square test plots into which the field can be partitioned using all or some of the 1994 meters of fence?

Solution 1

Suppose there are $n$ squares in every column of the grid, so there are $\frac{52}{24}n = \frac {13}6n$ squares in every row. Then $6|n$ , and our goal is to maximize the value of $n$ .

Each vertical fence has length $24$ , and there are $\frac{13}{6}n - 1$ vertical fences; each horizontal fence has length $52$ , and there are $n-1$ such fences. Then the total length of the internal fencing is $24\left(\frac{13n}{6}-1\right) + 52(n-1) = 104n - 76 \le 1994 \Longrightarrow n \le \frac{1035}{52} \approx 19.9$ , so $n \le 19$ . The largest multiple of $6$ that is $\le 19$ is $n = 18$ , which we can easily verify works, and the answer is $\frac{13}{6}n^2 = \boxed{702}$ .

Solution 2

Assume each partitioned square has a side length of $1$ (just so we can get a clear image of what the formula will look like). The amount of internal fencing that is required to partition the field is clearly $52*(24+1) + 24(52+1)$ . (If you are confused, just draw the square out). This is clearly greater than $1994$ , so the actual side length that we are looking for is greater than $1$ .

Now we can convert this into an equation. The equation is simply $(\frac{52}{x})(\frac{24}{x}+1) + (\frac{24}{x})(\frac{52}{x}+1)$ (The intutition comes from considering partioning the field into side lengths of $1$ and then partitioning those squares). This is equivalent to $\frac{2496}{x^2} + \frac{76}{x}$ , which should be less than or equal to $1994$ .

Now we can just find possible lengths of the square that are greater than $1$ and test them out. A viable side length would mean that $\frac{24}{x}, \frac{52}{x} \in$ N. Since $\gcd(24,52) = 4$ , then the smallest value greater than $1$ that we satisfies the conditions has $4$ in the numerator, and hence $3$ in the denominator. Test out $x=\frac{4}{3}$ . This will equate to something less than $1994$ , and hence the smallest square length that is plausible is $\frac{4}{3}$ .

Now the rest is elementary, we do $\frac{52}{\frac{4}{3}} * \frac{24}{\frac{4}{3}} \Rightarrow 39*18 = \boxed{702}$

1994 AIME (Problems • Answer Key • Resources)
Preceded by Problem 11	Followed by Problem 13
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15
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1994 AIME Problems/Problem 12

Contents

Problem

Solution 1

Solution 2

See also