2022 AMC 10 A Problem 25
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To help you master 2022 AMC 10 A Problem 25, this page includes:
- The original AMC problem to try, plus the correct answer
- Clear, step-by-step solutions using different methods
- Key ideas, useful formulas, and takeaways you can apply to similar problems
- Related problems from past AMC exams, plus new practice problems that build the same skills
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2022 AMC10 A Problem 25
25.
Let , , and be squares that have vertices at lattice points (i.e. points whose coordinates are both integers) in the coordinate plane, together with their interiors. The bottom edge of each square is on the -axis. The left edge of and the right edge of are on the -axis, and contains as many lattice points as does . The top two vertices of are in , and contains of the lattice points contained in . See the figure (not drawn to scale). The fraction of lattice points in that are in is times the fraction of lattice points in that are in . What is the minimum possible value of the edge length of plus the edge length of plus the edge length of ?
Solution
Lattice-Point Counting and Integer-Equation Method
Count lattice points by rows and columns, including the boundary. The ratio for and gives a common integer parameter, while the intersection ratio fixes the horizontal placement of .
An axis-aligned square of integer edge length contains lattice points. When two sets overlap, their union contains the sum of their counts minus the count of their intersection.
Step 1: Parameterize the sizes of the two large squares
Let the edge lengths of and be positive integers and . Each has one more lattice row and one more lattice column than its edge length. Because the point-count ratio is the square of a positive rational ratio, the corresponding numbers of rows are in the ratio .
Step 2: Count the lattice points in the union
The squares share the part of the -axis from height zero through the top of . Since is shorter, this common boundary contains one point for every integer height from zero through . Inclusion–exclusion then gives the union count.
Step 3: Show that the top of T cannot lie above S
Let be the edge length of . If its top were above the top of , both top vertices would have to lie in , because at that height the union has no points to the left of the -axis. Then the overlap with could only occur along the -axis, while the overlap with would contain at least as many points. This contradicts the condition derived below that the overlap with is twelve times as large. Therefore the top of is no higher than the top of , so every point of lies in the continuous region .
Step 4: Translate the intersection-fraction condition
Let and be the numbers of lattice points in the intersections with and , respectively. Substitute the known total point counts. The stated factor of then simplifies to a direct ratio between the two intersection counts.
Step 5: Count the shared axis inside T
Both intersection counts are positive: otherwise their ratio condition and the placement of the top vertices would be impossible. Thus reaches both sides of the -axis and contains that axis. Its common portion with both and consists of one vertical lattice column containing one point at each integer height. Since lies in the union, inclusion–exclusion counts all points of .
Step 6: Determine the horizontal placement of T
Write the horizontal interval of as running from the integer coordinate to the integer coordinate , where . In , there are lattice columns; in , there are lattice columns. Every such column has the same number of lattice points because has full height in both larger squares.
Step 7: Introduce the placement parameter
Let the number of lattice columns of in be . Dividing the formula for by the positive number of points in each column relates to the height. The left-side column ratio then determines both endpoints and the edge length of .
Step 8: Use the one-fourth condition
The point count of is one fourth of the union count. Substituting the two integer parameters produces the equation that all legal configurations must satisfy.
Step 9: Find the first possible value of the placement parameter
Treat the equation as a quadratic in . Its discriminant must be a perfect square. For each smaller positive value of , the resulting number lies strictly between the consecutive squares displayed, so none can produce an integer . The first successful value is .
Step 10: Prove that this solution gives the global minimum
The preceding check eliminates every smaller positive value of . Any value at least eight already forces a sum larger than the candidate, even if only the weak lower bound on is used. Thus no later solution of the integer equation can improve the result.
Step 11: Construct the attaining configuration
Use the successful parameters to obtain all three edge lengths. Place to the right of the axis and to the left as required. Place with the calculated horizontal endpoints; its top is below the top of , so both top vertices are in the union. This gives a legal configuration attaining the bound.
For the construction, and , giving the required ratio . Their union has points, while has , exactly one fourth as many. Also, has points and has points, so . The minimum is therefore , corresponding to choice .
Similar Problems in Real Exams
The following five AMC 10 A problems are closely related to AMC 10 A 2022 Question 25. These real AMC 10 A problems test the same key concept—Number Theory. Click any problem to learn and practice, then continue to the adapted practice section later in this article for more problems to try.
2025 AMC 10 A Problem 5
2025 AMC 10 A Problem 17
2024 AMC 10 A Problem 3
2024 AMC 10 A Problem 4
2024 AMC 10 A Problem 5
Adapted Problems to Practice
What Formulas Do You Need for This Problem?
- Lattice points in an integer-sided square:
is a nonnegative integer; the sides are parallel to the coordinate axes. There are possible integer -coordinates and independently possible integer -coordinates.
- Inclusion–exclusion for two sets:
and are finite sets; points in their intersection have been counted once in each individual count. The lattice points on the shared -axis would otherwise be counted twice.
- Column counting in a rectangular intersection:
The region must contain the same complete set of lattice heights in every counted column. Each part of lying in or extends from height through height , so every included column has points.
Key Takeaway: Remember This Strategy for Next Time
A figure contains integer-sided, axis-aligned squares and asks about lattice points.
Try: Replace an edge length by the row-and-column count , so the square contains points.
Why: Counting coordinate choices converts geometric information into integer equations.
Two regions meet along a full boundary segment.
Try: Use inclusion–exclusion and count the shared boundary points explicitly.
Why: Boundary lattice points belong to both closed regions and would otherwise be counted twice.
An integer equation is quadratic in one variable and the problem asks for the least solution.
Try: Complete the square or examine the discriminant, then test the smallest parameter values and prove all later values are too large.
Why: An integer solution forces a perfect-square condition, which efficiently eliminates impossible candidates while supporting a complete minimum proof.
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