Competition · AMC preparation · step 4 of 4

AMC 10 · 2019A · #6

Grade 4 geometry-2d

Archetype Small-Domain Constrained Search └ Sub-archetype Factor-Pair Enumeration

perpendicular-bisectorline-symmetrycaseworksimilar-figures caseworkidentify-subproblems ↑ Prerequisites: perpendicular-bisector

📏 Short solution 💡 2 insights

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Problem

For each of five quadrilateral types, decide whether some point in the plane is equally far from all four of its corners. Count how many types have such a point.

Pick an answer.

(A)

(B)

(C)

(D)

(E)

AMC 10 2019 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.

How to solve

Strategy Draw a Diagram

Draw each of the five shapes (Tool #1) and check whether all four corners can sit on one circle. Tool #7 turns the question into five small yes/no checks, one per shape. Tool #3 sweeps the list and tallies the yes-count, which directly gives the multiple-choice answer.

1STEP 1

Reframe as points on a circle

Reframe: one point equidistant from the 4 corners means the 4 corners lie on a circle. The real question — which shapes are cyclic?

PA = PB = PC = PD ⇔ A, B, C, D lie on a circle centered at P

2STEP 2

Check the square

Square: its two diagonals cross at the center, which sits the same distance from all four corners. Yes.

Square center → all 4 vertices equidistant

3STEP 3

Check the rectangle

Rectangle: its diagonals are equal and bisect each other, so their crossing point is equidistant from all four corners. Yes.

Rectangle diagonals are equal and bisect each other → midpoint is equidistant from all 4 corners

4STEP 4

Check the rhombus

Rhombus (not a square): its diagonals are perpendicular but unequal, so the center is nearer two corners than the other two. No.

Rhombus diagonals unequal → center is closer to two corners than the other two

5STEP 5

Check the parallelogram

Tilted parallelogram: its diagonals are unequal, so no crossing point reaches all four corners equally. No.

Tilted parallelogram diagonals unequal → no equidistant point

6STEP 6

Check the isosceles trapezoid

Isosceles trapezoid: left-right symmetry lets a point on the midline slide until the top and bottom corner distances match. Yes.

Symmetric trapezoid → a point on the axis of symmetry is equidistant from all 4 corners

7STEP 7

Count the shapes that work

Yes for square, rectangle, and isosceles trapezoid — that is 3 of the 5 types.

Yes count = 3 → (C)

Answer

Cross-check with a quick test using the diagonals: for a rectangle the two diagonals are equal in length AND bisect each other, so their crossing is the center of the circumscribed circle. The same is not true for a non-rectangle rhombus or a tilted parallelogram. For the isosceles trapezoid, the axis of symmetry guarantees a circle through all four corners (this is a well-known property). So exactly 3 types work — matches (C).

💡Key takeaway

This AMC 10 problem only needs Grade 4 shape-spotting you already know: a quadrilateral has a single point equally far from all four corners exactly when its corners sit on one circle. Squares, rectangles, and isosceles trapezoids pass this test; rhombi and tilted parallelograms fail. Answer: 3.

Reframe as points on a circle
Check the square
Check the rectangle
Check the rhombus
Check the parallelogram
Check the isosceles trapezoid
Count the shapes that work

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