AMC 10 · 2024 · #18

Grade 10 geometry-2d
rotation-isometrythirty-sixty-ninety-triangleisosceles-trianglemodular-arithmeticspatial-visualization spatial-visualizationpattern-recognitioncasework ↑ Prerequisites: pythagorean-theoremangle-sum-triangleisosceles-triangle
📏 Long solution 💡 4 insights 📊 Diagram
Problem
A card is a rectangle with sides 1 and 2+√(3). A second identical card is laid on the first so that a diagonal of the new card sits exactly on top of the diagonal AC of the first. A third card is laid on the second the same way, then a fourth on the third, and so on, with each placement turning the card clockwise. Find how many cards in total have been used the first time a corner of a newly placed card sits exactly on the point B of the first card.

Pick an answer.

(A)
6
(B)
8
(C)
10
(D)
12
(E)
$\text{No new vertex will land on }B.$
How to solve
Strategy Visualize Spatial Relationships

Tool #17 (Visualize Spatial Relationships) is the primary move because the whole problem is one mental rotation repeated. Lining up diagonals forces every card to share a single center, and every corner of every card is half a diagonal away from that center, so the entire growing stack lives on one circle. Once that is seen, a card is no longer a rectangle to be re-drawn — it is just four marks on a circle, and a placement just turns those four marks. Tool #1 (Draw a Diagram) supplies the coordinates that pin down the size of that turn, which comes from an angle hidden inside the rectangle. Tool #5 (Look for a Pattern) then takes over: the turn is the same every time, so the four marks step around the circle by a fixed angle and the picture must repeat. Tool #3 (Eliminate Possibilities) finishes it — there are only four corners that could land on B, so test all four and keep the earliest.

1STEP 1

Every card shares one center

Sharing a diagonal means sharing a centre.

diagonal=√(1²+(2+√(3))²)=√(8+4√(3))=√(6)+√(2), OA=OB=OC=OD=(√(6)+√(2))/2
2STEP 2

Find the hidden 30-60-90 triangle

The root three summons a 60 degree angle.

CP=√(1²+(√(3))²)=2, ∠ DCP=60°, ∠ DPC=30°
3STEP 3

The diagonal cuts off 15 degrees

An isosceles triangle delivers 15 degrees.

PA=(2+√(3))-√(3)=2=PC, ∠ APC=150°, ∠ PCA=∠ PAC=(180°-150°)/2=15°, ∠ ACD=75°
4STEP 4

Read the four corner directions

The corners sit at 75, 105, 255 and 285 degrees.

O=(1/2,(2+√(3))/2); A:75°, B:105°, C:255°, D:285°
5STEP 5

One placement turns things 30 degrees

Reflecting rotates the whole set by 30 degrees.

θ↦ 150°-θ: {75,105,255,285}↦{75,45,255,225}={75,105,255,285}-30°
6STEP 6

Track the marks around the circle

Compute which placement lands each corner on B.

θ₀-30(n-1)≡ 105 (mod 360): 105→ n=13, 75→ n=12, 255→ n=6, 285→ n=7
7STEP 7

Take the earliest hit

The earliest is the 6th.

255°-5 × 30°=105° → n=6 → (A)
Answer
6
A direct simulation confirms it. With O as origin, the corner directions of cards 1 through 6 are {75,105,255,285}, {45,75,225,255}, {15,45,195,225}, {345,15,165,195}, {315,345,135,165}, {285,315,105,135}. The value 105° first reappears at card 6, and every corner is (√(6)+√(2))/2≈ 1.932 from O, so the coordinates agree to the last digit. Answer (E) is impossible because 30° divides 360°, so the marks are forced to cycle and a hit must occur. The wrong numbers are the shapes of one specific slip: if you insist that the corner named B on the new card must be the one that lands on B, you get n=13, and if you only follow the corner named A you get n=12, which is choice (D) and the most tempting trap. The problem says a vertex, so all four corners are eligible, and the corner that started life at C gets there first. Choices (B) and (C) match no corner at all, since 8 and 10 do not solve any of the four congruences.
💡Key takeaway

Lining up diagonals pins every card to the same center, so each new card is just a 30° turn — and any of the four corners, not only the one named B, is allowed to be the one that lands on B.

  • Every card shares one center
  • Find the hidden 30-60-90 triangle
  • The diagonal cuts off 15 degrees
  • Read the four corner directions
  • One placement turns things 30 degrees
  • Track the marks around the circle
  • Take the earliest hit