Constructions and Tilings

For the upper pair, use any radius greater than half XY; the intersection is equidistant from X and Y. Repeat below with another such radius. Both centres lie on the perpendicular bisector, so their joining line is that bisector.

Draw two valid arc pairs using different radii on the same side. Their two intersection points lie on the same perpendicular bisector.

A point lies on the perpendicular bisector exactly when its distances to X and Y are equal. Equal-radius arcs enforce those equal distances.

Set compass radius greater than half XY. Draw intersecting arcs from X and Y above and below; join the intersections and use them as centres for the two eye arcs.

The rope ends stay fixed at X and Y and the rope is folded equally, so each pulled midpoint has equal distance to the endpoints. Two such points determine the perpendicular bisector.

Because OX=OY and AX=AY, A lies on the perpendicular bisector of XY. That bisector passes through O at a right angle to the line.

Mark OX=OY. The equal-radius arcs locate a point A also equidistant from X and Y. OA is the perpendicular bisector and contains O.

Triangles OAC and OBC have OA=OB, AC=BC and common side OC. They are congruent by SSS, so the two angles at O are equal.

A full turn is 360°. Divide equally among 8 gaps: 360°÷8=45°.

OA=OB, AC=BC and OC is common regardless of which arc intersection is chosen. Therefore corresponding angles at O are equal.

Each bisection halves an available angle. The resulting angles are multiples of 90°/2ⁿ; 65.5° is not obtained by that simple sequence.

Draw an arc centred at A to mark B on the ray. With the same compass width, draw an arc from B to meet the first at C. AB=BC=CA, so triangle ABC is equilateral and ∠CAB=60°.

The angle-bisector construction halves an angle while preserving its vertex and arms.

Around the centre, 6×60°=360°. At each outer vertex, two 60° triangle angles meet on a straight line, giving an interior angle 180°−60°=120°.

Add known angles: 40+60+50+30+40+90=310°. Subtract from 360° to get 50°.

A regular hexagon divides the circle into six 60° central angles. Each chord then equals the radius, so all six sides are 5 cm.

Check one tip triangle: its base angles and vertex angle are each 60° in the constructed regular pattern. Rotational symmetry gives the same result for all six.

PA=PB because A and B lie on the circle centred at P. Thus P lies on the perpendicular bisector of AB. Since A and B are on l, that bisector meets l at 90°.

The equal supports and their midpoint symmetry make the two sides mirror images. The arc radius controls the arch’s point and width.

Each sector triangle has all sides equal to the radius. Adjacent sectors share a side, leaving six equal outer sides and six 120° interior angles.

Each domino covers 2 squares. Since both dimensions are even, pair adjacent squares in each row.

Pair row 1 with row 2 and row 3 with row 4 in every column: 14 dominoes cover all 28 squares.

A sum of groups of 2 is even. An odd area cannot be fully covered by 2-square tiles.

An even row count permits vertical pairing in each column; an even column count permits horizontal pairing in each row. If both are odd, mn is odd and cannot be split into pairs.

Dominoes cover pairs, so area must be even. But geometry and adjacency also matter; use a checkerboard colouring or a forced-placement argument to prove impossibility.

Colour the grid alternately. Adjacent squares have opposite colours. If the region has unequal colour counts, no arrangement of dominoes can cover it.

A domino always covers one square of each colour. Removing a square can leave unequal colour totals; then no tiling exists. The removed square’s position matters.

A valid tiling must cover every unit exactly once, with no overlap or gap. The printed diagrams are required to decide these two specific cases.

A regular polygon tessellates when an integer number of its interior angles fit exactly around each vertex.

Start with the long edges and pointed corners. Fit the large triangles first, then place the other five pieces in the spaces left. Turn pieces or flip the parallelogram if needed.