AMC 10 · 2005 · #19
Grade 8 geometry-2dThree one-inch squares are placed with their bases on a line. The center square is lifted out and rotated 45∘, as shown. Then it is centered and lowered into its original location until it touches both of the adjoining squares. How many inches is the point B from the line on which the bases of the original squares were placed?
Pick an answer.
AMC 10 2005 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.
Try it yourself first — the explanation is most useful after you’ve attempted it.
Toolkit + CCSS Solution
Understand
Restated: Two unit squares sit on a line with a one-inch gap between them where a third square used to be. That third square is turned $45^{\circ}$ into a diamond and lowered into the gap until it rests on the two squares beside it. Find how high the top corner $B$ of the diamond ends up above the base line.
Givens: Each square has side length $1$ inch; The two outer squares keep their bases on the line, tops at height $1$; The middle square is rotated exactly $45^{\circ}$ and lowered straight down until it touches both outer squares; $B$ is the top corner of the rotated (diamond) square; Answer choices: (A) $1$, (B) $\sqrt{2}$, (C) $\frac{3}{2}$, (D) $\sqrt{2}+\frac{1}{2}$, (E) $2$
Unknowns: The height of point $B$ above the base line
Understand
Restated: Two unit squares sit on a line with a one-inch gap between them where a third square used to be. That third square is turned $45^{\circ}$ into a diamond and lowered into the gap until it rests on the two squares beside it. Find how high the top corner $B$ of the diamond ends up above the base line.
Givens: Each square has side length $1$ inch; The two outer squares keep their bases on the line, tops at height $1$; The middle square is rotated exactly $45^{\circ}$ and lowered straight down until it touches both outer squares; $B$ is the top corner of the rotated (diamond) square; Answer choices: (A) $1$, (B) $\sqrt{2}$, (C) $\frac{3}{2}$, (D) $\sqrt{2}+\frac{1}{2}$, (E) $2$
Plan
Primary tool: #1 Draw a Diagram
Secondary: #17 Visualize Spatial Relationships, #7 Identify Subproblems
This is a picture problem, so Tool #1 (Draw a Diagram) carries it: mark the two resting corners and the diamond's own corners to see exactly where things touch. Tool #17 (Visualize Spatial Relationships) is needed to see that the tilted square is too wide to drop through and instead hangs on the two top corners of its neighbours. Tool #7 (Identify Subproblems) then splits the height of $B$ into two easy pieces — how high the diamond's bottom corner sits, plus the diamond's vertical diagonal.
Execute — Answer: D
4.G.A.1 Step 1 Mark the gap and its corners
- When the middle square is lifted out, the two remaining squares leave a gap exactly $1$ inch wide (the width of the square that was removed).
- The inner top corners of those two squares are the ledges the diamond will land on.
- They sit $1$ inch apart, and each is at height $1$ above the base line.
💡 The removed square leaves a hole its own width, so the two corners that box in that hole are one inch apart.
8.G.A.1 Step 2 The diamond is too wide to fall through
- Turning the unit square $45^{\circ}$ makes a diamond whose horizontal width is its diagonal, $\sqrt{2}\approx 1.41$ inch.
- That is wider than the $1$-inch gap, so the diamond cannot drop through.
- Its two lower slanted edges instead catch on the two inner top corners, and its bottom corner hangs down between them.
💡 A tilted square is fatter across than a straight one, so it wedges on the corners instead of sliding down.
7.G.B.5 Step 3 Spot the 45-45-90 triangle
- Join the diamond's bottom corner to the two corners it rests on.
- The two lower edges of the diamond meet at the bottom corner in a $90^{\circ}$ angle (it is still a corner of the square), and each edge tilts $45^{\circ}$ from horizontal because the square was turned $45^{\circ}$.
- So this triangle is a right isosceles ($45$-$45$-$90$) triangle whose hypotenuse is the $1$-inch segment joining the two resting corners.
💡 A square corner tilted 45 degrees gives two edges that each slope at 45 degrees — a perfect right isosceles triangle.
8.G.B.7 Step 4 Find the height of the bottom corner
- In a $45$-$45$-$90$ triangle the height from the right angle down to the hypotenuse is exactly half the hypotenuse.
- Here the hypotenuse is $1$, so the bottom corner sits $\tfrac12$ inch below the line joining the two resting corners.
- That line is at height $1$, so the diamond's bottom corner is at height $1-\tfrac12=\tfrac12$.
💡 In an isosceles right triangle the peak reaches down to the middle of the base, exactly half the base away.
8.G.B.7 Step 5 Add the vertical diagonal to reach B
- Point $B$ is the top corner of the diamond, straight above the bottom corner.
- The segment from the bottom corner to $B$ is a diagonal of the unit square, and since the square was turned exactly $45^{\circ}$ this diagonal is vertical.
- Its length is $\sqrt{1^2+1^2}=\sqrt{2}$.
- So the height of $B$ is $\tfrac12+\sqrt{2}$, which is choice (D).
💡 Stack the bottom corner's height and the straight-up diagonal to get all the way to the top corner.
4.G.A.1 When the middle square is lifted out, the two remaining squares leave a gap exac 8.G.A.1 Turning the unit square $45^{\circ}$ makes a diamond whose horizontal width is i 7.G.B.5 Join the diamond's bottom corner to the two corners it rests on. The two lower e 8.G.B.7 In a $45$-$45$-$90$ triangle the height from the right angle down to the hypoten 8.G.B.7 Point $B$ is the top corner of the diamond, straight above the bottom corner. Th Review
Reasonableness: Check with coordinates. Put the gap between $x=1$ and $x=2$, both resting corners at height $1$: points $(1,1)$ and $(2,1)$. The diamond is centred at $x=1.5$; its lower-right edge has slope $-1$ through $(2,1)$ and its lower-left edge has slope $+1$ through $(1,1)$. These meet at the bottom corner $(1.5,\,0.5)$, matching the $\tfrac12$ height. Adding the vertical diagonal $\sqrt{2}$ puts $B$ at $(1.5,\,0.5+\sqrt{2})\approx(1.5,\,1.91)$. That is a bit under $2$, so it beats choices (A) $1$, (B) $\sqrt2\approx1.41$, (C) $1.5$ and stays under (E) $2$ — only (D) $\sqrt2+\tfrac12\approx1.91$ fits.
Alternative: Skip the triangle fact and use the diagonals directly. The diamond's horizontal diagonal is $\sqrt2$ long and its ends stick out past the gap; the two edges cross the corners $(1,1)$ and $(2,1)$. Since the horizontal diagonal sits at the diamond's centre height $h_c$, an edge of slope $-1$ from the bottom corner $(1.5,\,h_c-\tfrac{\sqrt2}{2})$ passing through $(1,1)$ gives $h_c=\tfrac12+\tfrac{\sqrt2}{2}$. Then $B=h_c+\tfrac{\sqrt2}{2}=\tfrac12+\sqrt2$, the same answer.
CCSS standards used (min grade 8)
4.G.A.1Draw points, lines, line segments, rays, angles, and identify in figures (Reading the figure to see the gap is $1$ inch wide and the resting corners sit at height $1$.)8.G.A.1Verify experimentally the properties of rotations, reflections, and translations (Seeing that the $45^{\circ}$ rotation makes the square $\sqrt2$ inches wide, too wide to drop through the $1$-inch gap.)7.G.B.5Use facts about supplementary, complementary, vertical, and adjacent angles (Recognising the right angle at the diamond's bottom corner and the $45^{\circ}$ tilt of its lower edges, giving a $45$-$45$-$90$ triangle.)8.G.B.7Apply the Pythagorean theorem to determine unknown side lengths in right triangles (Getting the $\tfrac12$-inch drop in the isosceles right triangle and the $\sqrt2$ length of the square's diagonal.)
⭐ A square turned on its point is wider than it looks, so it hangs on the corners — then just stack the little drop and the straight-up diagonal to find the top.
⭐ A square turned on its point is wider than it looks, so it hangs on the corners — then just stack the little drop and the straight-up diagonal to find the top.
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