AMC 10 · 2019 · #6

Grade 4 geometry-2d

Archetype Small-Domain Constrained Search

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

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Problem

For how many of the following types of quadrilaterals does there exist a point in the plane of the quadrilateral that is equidistant from all four vertices of the quadrilateral?

a square
a rectangle that is not a square
a rhombus that is not a square
a parallelogram that is not a rectangle or a rhombus
an isosceles trapezoid that is not a parallelogram

$\textbf{(A) } 0 \qquad\textbf{(B) } 2 \qquad\textbf{(C) } 3 \qquad\textbf{(D) } 4 \qquad\textbf{(E) } 5$

Pick an answer.

(A)

(B)

(C)

(D)

(E)

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Toolkit + CCSS Solution

Understand

Restated: 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.

Givens: Five shape types: (1) a square, (2) a rectangle that is not a square, (3) a rhombus that is not a square, (4) a parallelogram that is not a rectangle or a rhombus, (5) an isosceles trapezoid that is not a parallelogram; We need a single point equidistant from all $4$ vertices; Choices: (A) $0$, (B) $2$, (C) $3$, (D) $4$, (E) $5$

Unknowns: How many of the five types admit such an equidistant point

Understand

Restated: 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.

Plan

Primary tool: #1 Draw a Diagram

Secondary: #3 Eliminate Possibilities, #7 Identify Subproblems

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.

Execute — Answer: C

#7 Identify Subproblems 4.G.A.2 Step 1

Reframe: a point equidistant from $4$ vertices means those $4$ vertices lie on a circle centered at that point.
So the question becomes: which shapes have all $4$ corners on one circle?

$$PA = PB = PC = PD \;\Leftrightarrow\; A, B, C, D \text{ lie on a circle centered at } P$$

💡 Equal distances from one point means all corners share one circle.

#1 Draw a Diagram 3.G.A.1 Step 2

Square: draw a square and its two diagonals.
The diagonals cross at the center, and all four corners are the same distance from that center.
Yes.

$$\text{Square center} \Rightarrow \text{all 4 vertices equidistant}$$

💡 A square's center sits the same distance from each corner.

#1 Draw a Diagram 3.G.A.1 Step 3

Rectangle (not a square): the two diagonals of any rectangle are equal in length and cut each other in half, so the crossing point is the same distance from all four corners.
Yes.

$$\text{Rectangle diagonals are equal and bisect each other} \Rightarrow \text{midpoint is equidistant from all 4 corners}$$

💡 Equal diagonals that bisect each other share one center point.

#1 Draw a Diagram 3.G.A.1 Step 4

Rhombus (not a square): draw a long thin rhombus.
Its diagonals are perpendicular but unequal lengths, so the crossing point is closer to two corners than the other two.
No.

$$\text{Rhombus diagonals unequal} \Rightarrow \text{center is closer to two corners than the other two}$$

💡 A slanted rhombus has a fat axis and a thin axis — the center is not equally far from all corners.

#1 Draw a Diagram 3.G.A.1 Step 5

Parallelogram (not a rectangle or rhombus): a tilted parallelogram has unequal diagonals, so its center fails too.
No.

$$\text{Tilted parallelogram diagonals unequal} \Rightarrow \text{no equidistant point}$$

💡 Tilt destroys the balance — no single point sits at equal distance.

#1 Draw a Diagram 4.G.A.3 Step 6

Isosceles trapezoid (not a parallelogram): draw one with parallel top and bottom and equal slanted sides.
By symmetry across the vertical midline, a point on that midline is equally far from the left-and-right pairs; sliding it up or down can balance the top pair with the bottom pair.
Yes.

$$\text{Symmetric trapezoid} \Rightarrow \text{a point on the axis of symmetry is equidistant from all 4 corners}$$

💡 Left-right symmetry lets one point on the axis balance the top and bottom corners.

#3 Eliminate Possibilities K.MD.B.3 Step 7

Tally the yes-shapes: square, rectangle, isosceles trapezoid.
That is $3$ types out of $5$.

$$\text{Yes count} = 3 \Rightarrow \textbf{(C)}$$

💡 Count the shapes marked yes.

[1] #7 4.G.A.2 Reframe: a point equidistant from $4$ vertices means those $4$ vertices lie on a

[2] #1 3.G.A.1 Square: draw a square and its two diagonals. The diagonals cross at the center,

[3] #1 3.G.A.1 Rectangle (not a square): the two diagonals of any rectangle are equal in length

[4] #1 3.G.A.1 Rhombus (not a square): draw a long thin rhombus. Its diagonals are perpendicula

[5] #1 3.G.A.1 Parallelogram (not a rectangle or rhombus): a tilted parallelogram has unequal d

[6] #1 4.G.A.3 Isosceles trapezoid (not a parallelogram): draw one with parallel top and bottom

[7] #3 K.MD.B.3 Tally the yes-shapes: square, rectangle, isosceles trapezoid. That is $3$ types

Review

Reasonableness: 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).

Alternative: Tool #3 (Eliminate Possibilities) plus the rule that a quadrilateral has a single point equidistant from all corners exactly when opposite angles sum to $180^\circ$. Square ($90 + 90$): yes. Rectangle: yes. Rhombus: opposite angles are equal but not $90^\circ$, so they sum to $2A \ne 180$ unless $A = 90$ (square): no. Parallelogram: same reasoning, no. Isosceles trapezoid: base angles equal, opposite angles sum to $180^\circ$: yes. Count $= 3$.

CCSS standards used (min grade 4)

4.G.A.2 Classify two-dimensional figures based on presence of parallel or perpendicular lines (Recognizing the five quadrilateral types (square, rectangle, rhombus, parallelogram, isosceles trapezoid) by their parallel-side and right-angle structure.)
3.G.A.1 Understand that shapes in different categories share attributes (Using each shape's defining attributes (diagonals equal? perpendicular? bisecting each other?) to decide whether the corners share one circle.)
4.G.A.3 Recognize a line of symmetry for a two-dimensional figure (Using the isosceles trapezoid's axis of symmetry to locate a point equidistant from all four corners.)
K.MD.B.3 Classify objects into given categories and count the numbers in each (Counting the number of shape types that admit an equidistant point (yes pile vs. no pile).)

⭐ 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: $3$.

Problem

Toolkit + CCSS Solution

Understand

Plan

Execute — Answer: C

Review

CCSS standards used (min grade 4)

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