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Algebra Difficulty 4.8 AIME Find the answer Canada

Two joggers each run at their own constant speed and in opposite directions from one another around an oval track. They meet every 36 seconds. The first jogger completes one lap of the track in a time that, when measured in seconds, is a number (not necessarily an integer) between 80 and 100. The second jogger completes one lap of the track in a time, $t$ seconds, where $t$ is a positive integer. The product of the smallest and largest possible integer values of $t$ is

Pick one

Solution

We can determine which triangle has the greatest area by using a fixed side length of 4 for each of the identical squares and using this to calculate the unknown areas.
We begin by constructing $\triangle PVU$ and noticing that it is contained within square $QABP$ , as shown.

[[IMAGE0]]

The area of $\triangle PVU$ is determined by subtracting the areas of triangles $PQV$ , $VAU$ and $PBU$ from the area of square $QABP$ .

Since $QA=8$ and $AB=8$ , then the area of square $QABP$ is $8\times8=64$ .

Since $PQ=8$ and $QV=2$ , then the area of $\triangle PQV$ is $\frac12\times8\times2=8$ .

Since $VA=6$ and $AU=6$ , then the area of $\triangle VAU$ is $\frac12\times6\times6=18$ .

Since $PB=8$ and $UB=2$ , then the area of $\triangle PBU$ is $\frac12\times8\times2=8$ .
Therefore, the area of $\triangle PVU$ is $64-8-18-8=30$ .

Next, we construct $\triangle PXZ$ and then construct rectangle $CDSP$ by drawing $CD$ parallel to $PS$ through $X$ . Further, $X$ is the midpoint of the side of a square and so $C$ and $D$ are also midpoints of the sides of their respective squares.

[[IMAGE1]]

The area of $\triangle PXZ$ is determined by subtracting the areas of triangles $PCX$ , $XDZ$ and $PSZ$ from the area of rectangle $CDSP$ .

Since $CD=12$ and $DS=6$ , then the area of rectangle $CDSP$ is $12\times6=72$ .

Since $PC=6$ and $CX=8$ , then the area of $\triangle PCX$ is $\frac12\times6\times8=24$ .

Since $XD=4$ and $DZ=4$ , then the area of $\triangle XDZ$ is $\frac12\times4\times4=8$ .

Since $PS=12$ and $ZS=2$ , then the area of $\triangle PSZ$ is $\frac12\times12\times2=12$ .
Therefore, the area of $\triangle PXZ$ is $72-24-8-12=28$ .

Construct $\triangle PVX$ and notice that it is contained within square $QABP$ , as shown.

[[IMAGE2]]

The area of $\triangle PVX$ is determined by subtracting the areas of triangles $PQV$ , $VAX$ and $PBX$ from the area of square $QABP$ .

As we previously determined, the area of square $QABP$ is $64$ and the area of $\triangle PQV$ is $8$ .

Since $VA=6$ and $AX=2$ , then the area of $\triangle VAX$ is $\frac12\times6\times2=6$ .

Since $PB=8$ and $XB=6$ , then the area of $\triangle PBX$ is $\frac12\times8\times6=24$ .
Therefore, the area of $\triangle PVX$ is $64-8-6-24=26$ .

Construct $\triangle PYS$ and the perpendicular from $Y$ to $E$ on $PS$ , as shown.

[[IMAGE3]]

Since $PS=12$ and $YE=4$ ( $YE$ is parallel to $RS$ and thus equal in length to the side of the square), then the area of $\triangle PYS$ is $\frac12\times12\times4=24$ .

Construct $\triangle PQW$ , as shown.

[[IMAGE4]]

Since $PQ=8$ and $QW=6$ , then the area of $\triangle PQW$ is $\frac12\times8\times6=24$ .

The areas of the 5 triangles are $30,28,26,24,$ and 24. The triangle with greatest area, 30, is $\triangle PVU$ .

Want a route through all this instead of an archive? The track puts 2,444 problems in a working order, from Junior Challenge level to the IMO shortlist.

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