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Problem 447

Geometry Difficulty 2.7 Multiple choice CEMC Gauss (Grade 8) · Canada · 2015

In square $PQRS$ , $M$ is the midpoint of $PS$ and $N$ is the midpoint of $SR$ .

If the area of $\triangle SMN$ is 18, then the area of $\triangle QMN$ is

Pick one

Official solution

We begin by choosing the largest number in each row, $5,10,15,20,25$ , and calling this list $L$ . The sum of the five numbers in $L$ is $5+10+15+20+25=75$ and this sum satisfies the condition that no two numbers come from the same row.

However, the numbers in $L$ are taken from columns 1 and 5 only, and the numbers must be chosen so that no two come from the same column.

Thus, the largest of the five answers given, 75, is not possible.
Note: In assuring that we take one number from each row, this choice of numbers, $L$ , is the only way to obtain a sum of 75 (since we chose the largest number in each row).

Of the five answers given, the next largest answer is 73.

Since $L$ uses the largest number in each row and has a sum of 75, we can obtain a sum of 73 either by replacing one of the numbers in $L$ with a number that is two less, or by replacing two of the numbers in $L$ with numbers that are each one less.

For example, the list $3,10,15,20,25$ (one change to $L$ ) has sum 73 as does the list $4,9,15,20,25$ (two changes to $L$ ).

That is, to obtain a sum of 73 while choosing exactly one number from each row, we must choose at least three of the numbers from $L$ .

However, since two numbers in $L$ lie in column 1 and three numbers from $L$ lie in column 5, it is not possible to choose at least three numbers from $L$ so that no two of the numbers are from the same column.

Any other replacement would give a sum less than 73, which would require the replacement of a number with a larger number in another row to compensate. This is impossible since each row is represented in $L$ by the largest number in the row.
Therefore, it is not possible to obtain a sum of 73.

Of the five answers given, the next largest answer is 71.

By choosing the numbers, $3,9,14,20,25$ we obtain the sum $3+9+14+20+25=71$ while satisfying the condition that no two numbers come from the same row and no two numbers come from the same column.

Thus, 71 is the largest possible sum that satisfies the given conditions.
Note: There are other choices of five numbers which also give a sum of 71 and satisfy the given conditions.

Source: CEMC, University of Waterloo, licensed CC-BY-NC-4.0. Statement reproduced verbatim; metadata (topic, difficulty, ordering) added by this project.