Simplify
This chapter may feel a bit less concrete than some of the others. The main idea is simple, though: when you work the problems in this chapter (and the drills afterward), look for a way to simplify what you’re given into something basic and easy to compute.
A key AMC skill is turning a problem that looks long, abstract, or information-heavy into a small set of usable facts.
- Sometimes a wordy problem hides a very short solution.
- Sometimes the math isn’t stated directly, but the needed calculation is still precise.
- Sometimes there’s a viewpoint or method that makes the whole problem straightforward.
In each case, the job is the same: translate the situation into a simpler form you can work with.
Here are two AMC problems that become much more accessible once you reduce them to a clean equation. After you try these, keep the same mindset for the quiz questions: if your current approach feels too slow or messy, there’s probably a simpler interpretation.
Example: The question below is from 2016 AMC 8
In an All-Area track meet, sprinters enter a -meter dash competition. The track has lanes, so only sprinters can compete at a time. At the end of each race, the five non-winners are eliminated, and the winner will compete again in a later race. How many races are needed to determine the champion sprinter?
A.
B.
C.
D.
E.
Answer: C.
To end with runner, we must eliminate runners.
Each race eliminates runners (everyone except the winner). So the number of races is the number of groups of eliminations needed to reach :
This problem can be simplified into the equation: .
Example: The question below is from 2020 AMC 10A
A point is chosen at random within the square in the coordinate plane whose vertices are and . The probability that the point is within units of a lattice point is . (A point is a lattice point if and are both integers.) What is to the nearest tenth?
A.
B.
C.
D.
E.
Answer: B.
This problem can be distilled down to the basic ratio where is the solution.
Here’s the key observation behind that equation. The plane is tiled by squares whose corners are lattice points. In each such square, the region within distance of a lattice point consists of four quarter-circles (one at each corner). Those four quarter-circles add up to the area of one full circle of radius .
So, within a typical square:
- area within distance of a lattice point
- total area of the square
The problem says this probability is , so we set
.
This simplifies to just .