Question

type the correct answer in the box. use numerals instead of words. squares abcd and efgh share a common center on a coordinate plane, as shown in the figure. $overline{eh}$ is parallel to diagonal $overline{ac}$. the number of lines of reflection about which the combined figure can reflect onto itself is

Explanation:

Step1: Analyze the squares' symmetry

Both squares ABCD and EFGH are squares with a common center. A square has 4 lines of reflection symmetry (2 diagonals, 2 lines through midpoints of opposite sides). But since EH is parallel to AC, we check the combined figure. The lines of reflection must map both squares onto themselves.

Step2: Identify valid reflection lines

The diagonals of ABCD (AC and BD) and the lines through midpoints of ABCD's sides (parallel to its sides) are considered. For the combined figure, the lines that are symmetry axes for both squares: the two diagonals of ABCD (since EFGH is rotated 45 degrees? Wait, no, EH is parallel to AC, so EFGH is oriented such that its sides are parallel to ABCD's diagonals. So the symmetry lines of the combined figure: the two diagonals of ABCD (AC, BD) and the two lines perpendicular to those diagonals (through the center, parallel to AB and AD? Wait, no. Wait, a square has 4 lines of symmetry. But when we have two squares with EH parallel to AC, the combined figure's symmetry lines: let's think. The large square ABCD has 4 lines (2 diagonals, 2 midlines). The small square EFGH, since EH is parallel to AC, its sides are parallel to ABCD's diagonals. So EFGH's symmetry lines are the same as ABCD's diagonals? No, EFGH is a square, so it has 4 lines too. But the combined figure: the lines that are symmetry for both. Let's see: the two diagonals of ABCD (AC, BD) – reflecting over AC: EH is parallel to AC, so EH maps to itself? Wait, EH is parallel to AC, so reflecting over AC would map EH to a line parallel to AC, and since EFGH is centered, it should map EFGH onto itself. Similarly for BD. Then the other two lines: the midlines of ABCD (horizontal and vertical, if ABCD is axis-aligned). Reflecting over the horizontal midline: would EFGH map onto itself? EFGH is rotated 45 degrees, so its sides are at 45 degrees. Reflecting over a horizontal line (midline of ABCD) would rotate EFGH? No, reflection over horizontal line: a square rotated 45 degrees would have its symmetry lines at 45 degrees. Wait, maybe I made a mistake. Wait, the problem says EH is parallel to AC. So AC is a diagonal of ABCD, EH is a side of EFGH, parallel to AC. So EFGH is a square with sides parallel to ABCD's diagonals. So the combined figure: the symmetry lines are the two diagonals of ABCD (AC, BD) and the two lines perpendicular to those diagonals (i.e., the lines parallel to AB and AD, through the center). Wait, no, a square has 4 lines. Wait, maybe the combined figure has 4 lines? No, wait, let's count. Wait, the large square has 4, the small square has 4, but the combined figure: the lines that are symmetry for both. Let's take an example: reflect over AC (diagonal of ABCD). ABCD maps to itself, EFGH: since EH is parallel to AC, EH is on a line parallel to AC, so reflecting over AC would map EH to a line parallel to AC, and since EFGH is centered, EFGH maps to itself. Similarly for BD. Then reflect over the midline of AB and CD (horizontal line through center). ABCD maps to itself, EFGH: its sides are at 45 degrees, so reflecting over horizontal line would flip EFGH vertically, but since it's centered, does it map to itself? Wait, EFGH is a square with sides parallel to diagonals of ABCD, so its vertices are at (center + (a,a), (a,-a), (-a,-a), (-a,a)) if ABCD is from (-b,-b) to (b,b). Then reflecting over horizontal line (y=0) would map (a,a) to (a,-a), (a,-a) to (a,a), (-a,-a) to (-a,a), (-a,a) to (-a,-a) – which is a rotation? No, reflection over y=0 swaps (x,y) and (x,-y). So (a,a) becomes (a,-a), which is a vertex of EFGH. So EFGH maps to itself. Similarly for reflection over vertical line (x=0). So the combined figure has 4 lines of symmetry? Wait, no, wait the problem: "the number of lines of reflection about which the combined figure can reflect onto itself". Wait, maybe I was wrong earlier. Wait, a square has 4 lines. But let's check the diagram: ABCD is a square, EFGH is a square inside, with EH parallel to AC. So the combined figure: the symmetry lines are the 4 lines of the square ABCD, because EFGH is symmetric with respect to those lines. Wait, but EFGH is rotated 45 degrees? No, EH is parallel to AC, so EFGH is oriented so that its sides are parallel to ABCD's diagonals. So EFGH's sides are at 45 degrees to ABCD's sides. So the symmetry lines of the combined figure: the two diagonals of ABCD (AC, BD) and the two lines perpendicular to those diagonals (i.e., the lines parallel to AB and AD, through the center). Wait, that's 4 lines. Wait, but let's think again. A square has 4 lines of symmetry. So the combined figure, being two squares with EH parallel to AC, should have 4 lines? No, wait, maybe the answer is 4? Wait, no, wait the problem: maybe I made a mistake. Wait, let's see: the large square has 4 lines, the small square has 4 lines, but the combined figure's symmetry lines are the intersection of their symmetry lines. The large square's symmetry lines: 2 diagonals, 2 midlines. The small square's symmetry lines: since its sides are parallel to large square's diagonals, its symmetry lines are the same as large square's diagonals and the midlines (but rotated 45 degrees? No, a square's symmetry lines are always 4: 2 diagonals, 2 midlines. So if the small square is rotated 45 degrees relative to the large square, then their symmetry lines are the 4 lines that are symmetry for both. Wait, no, if the small square is rotated 45 degrees, then the only symmetry lines common to both are the 2 diagonals of the large square (which are the midlines of the small square) and the 2 midlines of the large square (which are the diagonals of the small square). So total 4 lines. Wait, but let's count: reflecting over AC (diagonal of large square): large square maps to itself, small square: since EH is parallel to AC, EH is on a line parallel to AC, so reflecting over AC would map EH to a line parallel to AC, and since small square is centered, it maps to itself. Similarly for BD. Reflecting over the midline of AB and CD (horizontal line): large square maps to itself, small square: its sides are at 45 degrees, so reflecting over horizontal line would flip it vertically, but since it's a square, it maps to itself (because it's symmetric over that line? Wait, no, a square rotated 45 degrees: its horizontal midline (of the large square) is a diagonal of the small square. So reflecting over the large square's horizontal midline is reflecting over the small square's diagonal. So the small square, being a square, is symmetric over its diagonals. So yes, reflecting over the large square's horizontal midline (small square's diagonal) maps the small square to itself. Similarly for the vertical midline (other diagonal of small square). So total 4 lines of symmetry. Wait, but maybe the answer is 4? Wait, no, wait the problem: "the number of lines of reflection about which the combined figure can reflect onto itself". Let me check again. The large square has 4, the small square has 4, but the combined figure: the lines that are symmetry for both. Since EH is parallel to AC, the small square is oriented such that its sides are parallel to large square's diagonals. So the symmetry lines of the combined figure are the 4 lines: 2 diagonals of large square (which are midlines of small square) and 2 midlines of large square (which are diagonals of small square). So total 4? Wait, no, maybe I'm overcomplicating. Wait, a square has 4 lines of symmetry. So the combined figure, being two squares with the small one rotated 45 degrees, still has 4 lines of symmetry? Wait, no, let's take a concrete example. Let ABCD be a square with vertices at (0,0), (1,0), (1,1), (0,1). Then AC is from (0,0) to (1,1), BD from (0,1) to (1,0). EFGH is a square with EH parallel to AC, so EH is a side from (0.4,0.6) to (0.6,0.4) (for example), so EFGH has vertices at (0.5,0.5 + a), (0.5 + a,0.5), (0.5,0.5 - a), (0.5 - a,0.5). Then reflecting over AC (y=x): (0.5,0.5 + a) maps to (0.5 + a,0.5), which is a vertex of EFGH. Similarly for other points. Reflecting over BD (y=1 - x): (0.5,0.5 + a) maps to (0.5 - a,0.5), which is a vertex. Reflecting over y=0.5 (horizontal midline): (0.5,0.5 + a) maps to (0.5,0.5 - a), vertex. Reflecting over x=0.5 (vertical midline): (0.5 + a,0.5) maps to (0.5 - a,0.5), vertex. So all 4 lines work. So the number of lines is 4? Wait, but maybe the answer is 4? Wait, no, wait the problem says "the number of lines of reflection about which the combined figure can reflect onto itself". Wait, maybe I made a mistake. Wait, the large square has 4, the small square has 4, but the combined figure: are all 4 lines valid? Let's see: yes, because reflecting over any of the 4 lines of the large square maps the large square to itself, and since the small square is symmetric with respect to those lines (because its sides are parallel to large square's diagonals, so its diagonals are the large square's midlines), so reflecting over large square's midlines (which are small square's diagonals) maps the small square to itself. So total 4 lines. Wait, but maybe the answer is 4? Wait, no, wait the diagram: EFGH is inside ABCD, with EH parallel to AC. So the combined figure's symmetry lines: 4. Wait, but let's check the problem again. The user says "Type the correct answer in the box. Use numerals instead of words." So maybe the answer is 4? Wait, no, wait a square has 4 lines, but maybe the combined figure has 4? Wait, no, maybe I'm wrong. Wait, let's think of the two squares: ABCD is a square, EFGH is a square rotated 45 degrees (since EH is parallel to AC, which is a diagonal). So the combined figure is a square with a square rotated 45 degrees inside, sharing the center. The symmetry lines of such a figure: the 4 lines of the square (2 diagonals, 2 midlines). So the number is 4? Wait, no, wait the problem might be that the small square is oriented such that only 2 lines? Wait, no, a square has 4 lines. Wait, maybe the answer is 4. But let's confirm. Yes, a square has 4 lines of reflection symmetry, and the combined figure here, being two squares with the small one rotated 45 degrees but centered, still has 4 lines of symmetry. So the number is 4.

Step1: Recall square symmetry

A square has 4 lines of reflection symmetry (2 diagonals, 2 midlines).

Step2: Analyze combined figure

Both squares (ABCD, EFGH) share the center, with EH ∥ AC (so EFGH is rotated 45° relative to ABCD). The symmetry lines of the combined figure are the 4 lines of symmetry of ABCD (since EFGH, as a square, is symmetric about these lines: its diagonals align with ABCD’s midlines, and its midlines align with ABCD’s diagonals).

Answer:

4