Geometry & Angles

After four peaks the zig-zag is at the end of the diameter, so after two peaks it must be at the centre \(O\) of the circle. The triangle \(OPR\) is isosceles since \(OP\) and \(OR\) are both radii, hence angle \(PRO = \alpha\) and angle \(POR = 180° - 2\alpha\). Angle \(OTR = 180° - \alpha\) (angles on a straight line), and hence angle \(TRO = 3\alpha - 180°\) (angles in triangle \(TRO\)).

Triangle \(OTS\) is isosceles because its base angles are equal, and hence \(ST = SO\). Therefore triangles \(OPR\) and \(OTS\) are congruent because they both have two sides which equal the radius of the circle, with angle \(180° - 2\alpha\) between (SAS). Hence \(OT = PR = TR\) (since \(PR = RT\) in isosceles triangle \(PRT\)). Hence triangle \(OTR\) is isosceles, and its base angles are equal, that is \(180° - 2\alpha = 3\alpha - 180°\). Solving this gives \(\alpha = 72°\).

The diagram shows exactly one third of the shaded area in the original diagram. It is made up of the quadrilateral \(UOPX\), together with a sector of the outer circle, \(POQ\), where \(O\) is the centre of the original circle.
Since \(P\) and \(Q\) are two of the six points equally spaced around the circle, \(\angle POQ = \tfrac{1}{6} \times 360° = 60°\). The outer circle has radius 2 cm, so the area, in cm², of sector \(POQ\) is \(\tfrac{1}{6} \times \pi \times 2^2 = \tfrac{2\pi}{3}\).

First we prove that the four white triangles are congruent. Note that they each have a right angle. Also angles \(HGA\) and \(FGE\) are equal (vertically opposite). The quadrilateral \(ACEG\) is a parallelogram since each of its sides come from the rectangles. Hence angles \(FGE\) and \(GAC\) are equal (corresponding), and angles \(GAC\) and \(ECD\) are equal (corresponding). Therefore each triangle has a right-angle, and an angle equal to \(HGA\), and therefore they each have the same angles. Moreover, they each have a corresponding side of length 3 cm. By labelling triangle \(ABC\) with lengths 3, \(x\) and \(y\), Pythagoras’ Theorem gives \(x^2 + 3^2 = y^2\) [1]. The triangles \(ABC\) and \(CDE\) are congruent, so \(CD = CB = x\). Since \(AD = 9\), we can see \(x + y = 9\) [2].

Equation [2] rearranges to \(x = 9 - y\), so \(x^2 = (9 - y)^2 = 81 - 18y + y^2\). Substituting this into [1] gives \(81 - 18y + y^2 + 3^2 = y^2\), so \(18y = 90\) and \(y = 5\). Thus \(x = 9 - 5 = 4\). The area of each white triangle is \(\tfrac{1}{2} \times 3x = \tfrac{1}{2} \times 3 \times 4 = 6\) cm². Thus the overlap is \(3 \times 9 - 2 \times 6 = 15\) cm².

The diagram shows the top left-hand corner of the original diagram. The centre of the semicircle shown is \(Q\). Also, \(U\) and \(S\) are the points where the edges of the bigger square touch the semicircle shown. Therefore both \(QU\) and \(QS\) are radii of the semicircle and \(\angle TSQ = \angle TUQ = 90°\). Also \(\angle UTS\) is a right angle as it is the corner of a square. Therefore \(UQST\) is a square. Hence \(QS = ST\).

The routes of the hare and the tortoise form a right-angled triangle as shown on the diagram. The hare travels 5 times as fast, but arrives at the same time as the tortoise, so has travelled five times further, giving
\[x + y = 25. \ [1]\]
Pythagoras’ Theorem gives \(x^2 + 5^2 = y^2\), so \(y^2 - x^2 = 25\), and this factorises to give \((y + x)(y - x) = 25\). However, \(x + y = 25\) by equation [1]. So \(25(y - x) = 25\) and then \(y - x = 1\), and \(y = x + 1\). Substituting this into equation [1] gives \(x + x + 1 = 25\), and hence \(x = 12\) and \(y = 13\).

The diagrams show one of the twenty-five congruent right-angled triangles and the two such triangles at vertex R.
Let the two acute angles in each of the triangles be \(x°\) and \(y°\).

On the diagram is a pentagon outlined, and four of its angles are known. The sum of the angles in a pentagon is 540°. The missing angle is then 540° −270° − 70° − 60° − 90° = 50°.

To turn the figure given in the question into a hexagon by adding the smallest number of triangles, two triangles should be added to create each vertex of the hexagon (shaded dark grey) and one triangle added to create each edge (shaded light grey) as shown on the right for one edge. Since a hexagon has six vertices and six edges, the smallest number of triangles required is 6 × (2 + 1) = 18.

In the diagram, \(P\), \(Q\) and \(R\) are the vertices of the triangle; \(S\) and \(T\) are both centres of a semicircle; \(U\) is a point where two semicircles touch and \(V\) and \(W\) are points where a semicircle touches the triangle. So angles \(PVS\) and \(TWS\) are right angles, since a tangent to a circle is perpendicular to the radius of that circle at the point of contact. Also, the line joining the centres of two touching semicircles passes through their point of contact, so \(SUT\) is a straight line and has length 2. Note that \(\angle VPS = 60°\), and so \(\angle PSV = 30°\).
Note also that \(ST = 2\), \(TW = 1\) and \(\angle TWS\) is a right angle.
Therefore \(\sin \angle WST = \dfrac{1}{2}\) and so \(\angle WST = 30°\). However \(\angle TQS = 60°\) and so \(\angle STQ\) is a right angle.
From triangle \(SPV\), we see that \(\dfrac{SV}{PS} = \dfrac{1}{PS} = \sin 60° = \dfrac{\sqrt{3}}{2}\) and so
\(PS = \dfrac{2}{\sqrt{3}}\). Similarly, from triangle \(SQT\), we see that \(SQ = \dfrac{4}{\sqrt{3}}\).
Therefore \(PQ = \dfrac{6}{\sqrt{3}} = \dfrac{6\sqrt{3}}{3} = 2\sqrt{3}\).

Let \(PQ\) have length \(4x\). Then, as \(\tan \angle QPR = \dfrac{3}{2}\), \(QR\) has length \(6x\). Therefore, as \(S\) is the midpoint of side \(QR\), \(QS\) has length \(3x\). By Pythagoras’ Theorem in triangle \(PQS\), \(PS^2 = PQ^2 + QS^2 = (4x)^2 + (3x)^2 = 16x^2 + 9x^2 = 25x^2\). Hence \(PS = \sqrt{25x^2} = 5x\).
So \(\sin \angle QPS = \dfrac{QS}{PS} = \dfrac{3x}{5x} = \dfrac{3}{5}\).

Since the total shaded area is half the area of the larger equilateral triangle, the area of that triangle is six times the area of each of the shaded equilateral triangles . Now when two similar figures have side-lengths in the ratio \(1 : k\), the ratio of their areas is \(1 : k^2\). In this case, the ratio of the areas is \(1 : 6\) so the ratio of the side-lengths is \(1 : \sqrt{6}\).
Hence the the length of the sides of the larger triangle is \(\sqrt{6}\) cm.

Let \(O\) be the centre of the circle, and \(P\), \(Q\), \(P'\) and \(Q'\) the vertices of the square. The triangles \(OPQ\) and \(OP'Q'\) are congruent since they are right-angled and have two equal sides (\(PQ = P'Q'\) since they are edges of a square, and \(OP = OP'\) because each is a radius). Hence \(OQ = OQ'\), and \(O\) is thus the midpoint of the edge of the square.

Let the side of the square be \(2x\). And \(OP = 1\) since it is a radius. And by Pythagoras’ Theorem on triangle \(OPQ\) we have \((2x)^2 + x^2 = 1^2\), so \(4x^2 + x^2 = 1\), and \(5x^2 = 1\). This gives \(x^2 = \tfrac{1}{5}\). The area of the square is \((2x)^2 = 4x^2 = 4 \times \tfrac{1}{5} = \tfrac{4}{5}\).

The four triangles on the left will fold to form one square-based pyramid (without the base). The four triangles on the right will fold to make another pyramid, with the two pyramids hinged at the dashed edge. When these two pyramids are folded at this edge, the bottom end of \(x\) will coincide with the right-hand end of E; so \(x\) will coincide with E.

In the diagram shown, \(T\) is the foot of the perpendicular from \(R\) to \(PQ\) produced. Consider triangle \(QTR\). Angle \(QTR = 90°\) and, because the sum of angles on a straight line = 180°, angle \(RQT = (180 - 2 \times 60)° = 60°\). So \(\sin 60° = \dfrac{RT}{QR}\) and hence
\(RT = QR \times \sin 60° = 2 \times \dfrac{\sqrt{3}}{2} = \sqrt{3}\). Also \(\cos 60° = \dfrac{QT}{QR}\), so
\(QT = QR \times \cos 60° = 2 \times \tfrac{1}{2} = 1\). Therefore \(PT = 7\).

By drawing radii and chords as shown, we can see that the triangles are equilateral and therefore each of the angles is 60°. Hence the left and right circles have each lost 120° (one third) of their circumferences, and the central circle has one-third \(\left(\tfrac{1}{6} + \tfrac{1}{6}\right)\) of its circumference remaining. Since \(1 - \tfrac{1}{3} + 1 - \tfrac{1}{3} + \tfrac{1}{3} = \tfrac{5}{3}\), the perimeter of the shape is \(\tfrac{5}{3} \times 2\pi R = \tfrac{10\pi R}{3}\).

When Cathie cut the paper as described in the question, her cuts divided the original paper as shown in the diagram. It can then be seen that the five pieces shaded are squares.

The diagram shows the original diagram shaded as on a chess board. Note that as the counter starts on an unshaded cell then, in whichever line it is reflected, one reflection will move it into a shaded cell. Similarly, a second reflection will move the counter from a shaded cell into an unshaded cell. Finally, a third reflection will move the counter from an unshaded cell into a shaded cell. It is left to the reader to show that it is possible to reach each of the shaded cells in precisely three reflections . Hence the number of cells the counter could occupy after three moves is equal to the number of shaded cells, that is 8.

The diagram of the square and the triangle is shown. Consider the triangles \(PQT\) and \(RQT\). They share side \(QT\); also \(PQ = RQ\) because they are sides of a square; and \(PT = RT\) because they are sides of the equilateral triangle, so by SSS the triangles \(PQT\) and \(RQT\) are congruent. Therefore \(\angle PQT = \angle RQT\). Also \(\angle PQT + \angle RQT + 90° = 360°\) so \(\angle RQT = \tfrac{1}{2}(360° - 90°) = 135°\).

The triangle \(PQR\) is congruent to the triangle \(TQS\) since they are right-angled triangles with sides of length 3 and 2. Hence the angle \(PQR\) is also \(\alpha°\) and then \(\alpha + \beta + \alpha = 90\). One can check that the other statements are false.

At each of the points indicated by a dot, four quadrilaterals meet, so that at least four colours are needed.
The diagram shows that four colours is possible, so the minimum number of colours Khelen needs is four.

\(PS\) is a diameter, so \(\angle PRS = 90°\) (angle in a semicircle). Let \(U\) be the point on \(QT\) for which \(SU\) is perpendicular to \(QT\). Hence \(RSUT\) is a rectangle, and \(SU = TR = 3\).
In triangle \(TPQ\), we have \(\angle PTQ = 90°\) and \(\angle QPT = 60°\), so \(\angle TQP = 30°\). Also, \(\angle PQS = 90°\) (angle in a semicircle) so \(\angle UQS = 90° - \angle TQP = 90° - 30° = 60°\).
Also \(\sin \angle UQS = \dfrac{US}{QS} = \dfrac{3}{QS}\). So
\[QS = \frac{3}{\sin \angle UQS} = \frac{3}{\frac{1}{2}\sqrt{3}} = 3 \times \frac{2}{\sqrt{3}} = 2\sqrt{3}.\]

Since triangle \(ABC\) is isosceles with \(AB = BC\) and we are given that \(LB = AK\), the other parts of the equal sides must themselves be equal. Hence \(LC = BK = AC\). Draw in line \(KC\) as shown to form triangles \(ACK\) and \(LCK\). Since \(AK = KL\), \(AC = LC\) and \(KC\) is common to both, triangles \(ACK\) and \(LCK\) are congruent and hence \(\angle KAC = \angle CLK\).
Let the size of \(\angle LBK\) be \(x°\). Since \(KL = LB\), triangle \(KLB\) is isosceles and hence \(\angle BKL = x°\). Since an exterior angle of a triangle is equal to the sum of the interior opposite angles, \(\angle KLC = 2x°\) and hence \(\angle KAC = 2x°\). Since the base angles of an isosceles triangle are equal, \(\angle ACL = 2x°\). Therefore, since angles in a triangle add to 180°, when we consider triangle \(ABC\), we have \(x° + 2x° + 2x° = 180°\) and hence \(x = 36\). Therefore the size of \(\angle ABC\) is 36°.

In the diagram shown, \(F\) is the foot of the perpendicular from \(D\) to \(AB\), \(G\) is the foot of the perpendicular from \(D\) to \(AC\) and \(H\) is the foot of the perpendicular from \(E\) to \(AB\). So \(DF = DG = EH = \sqrt{3}\). Consider triangles \(AFD\) and \(AGD\). Edge \(AD\) is common to both triangles, \(DF = DG\) and angles \(AFD\) and \(AGD\) are both right angles. So the triangles are congruent (RHS). Therefore \(\angle DAF = \angle DAG = 60° \div 2 = 30°\).
So in triangle \(AFD\), \(\tan 30° = \dfrac{DF}{AF}\).
Therefore \(AF = \sqrt{3} \div \left(1/\sqrt{3}\right) = \sqrt{3} \times \sqrt{3} = 3\). The same argument shows that \(BH = 3\). It is clear that \(FH = DE\), so \(AB - DE = AF + BF = 3 + 3 = 6\).
Therefore the difference between the lengths of the edges of the two triangles equals 6.

The diagram shows that the hexagon is an equilateral triangle of side 11 which has an equilateral triangle of side 1 removed from the bottom left corner, an equilateral triangle of side 2 removed from the bottom right corner and an equilateral triangle of side 6 removed from the top.

The diameter of the non-overlapping circles is the difference between the radius of the large circle and the radius of the small circle, namely \(9 - 1 = 8\). Thus the radius of each circle is 4 cm, and the centre of each circle is 1 + 4 = 5 cm from the centre of the inner circle. The closest the circles could be is when they are touching. The diagram shows two of these circles and the inner circle of radius 1 cm. We need to establish the size of angle \(CAB\). Since \(5^2 + 5^2 < 8^2\), we deduce that \(\angle CAB > 90°\), and hence it is not possible to fit 4 circles in the annulus.
We can also show that \(\angle CAB < 120°\). For if \(\angle CAB = 120°\), then \(\angle CAD = 60°\) and \(AD = 2\tfrac{1}{2}\).
But \(AD = \sqrt{5^2 - 4^2} = 3\), so \(\angle CAD < 60°\) and \(\angle CAB < 120°\). Hence three circles can fit in the annulus, and that is the maximum.

Therefore obtuse angle \(CDA = (60 + 108)° = 168°\) and reflex angle \(CDA = (360 - 168)° = 192°\). The sum of the interior angles of a quadrilateral is 360°. Therefore, in quadrilateral \(ABCD\), \(\angle BAD = (360 - 192 - 72 - 90)° = 6°\). Angle \(BAE\) is a right angle. So \((6 + 60 + x)° = 90°\). Therefore \(x = 90 - 66 = 24\).

The diagrams show that all four options could be the shape of the region where two triangles overlap.

Add to the diagram three lines parallel to two of the sides of the rectangle, creating angles \(a\), \(b\), \(c\), \(d\), \(e\) and \(f\) as shown.
Since alternate angles formed by parallel lines are equal, we have \(a = 26°\) and \(f = 10°\). Since \(a + b = 33°\) and \(e + f = 14°\), we have \(b = 7°\) and \(e = 4°\). Similarly, since alternate angles are equal, we have \(c = b = 7°\) and \(d = e = 4°\). Therefore \(\theta = c + d = 7° + 4° = 11°\).

From the diagram, it can be seen that the distance between the centres of the two circles is 11 cm – 2 × the radius of a circle or equivalently 11 cm – the diameter of each circle. Since this diameter is 7 cm, the distance between the centres is 11 cm – 7 cm = 4 cm.

As \(PQ = PR\), \(\angle PRQ = \angle PQR = x°\). Also, as \(PR = PS\), \(\angle PRS = \angle PSR = x°\). Therefore, \(\angle QRS = (2x)°\). The sum of the interior angles of quadrilateral \(PQRS\) is 360°. Therefore \(x + x + 2x + x = 360\), so \(5x = 360\). Hence \(x = 72\).

We start by drawing the line segment \(IG\). Let \(P\) be the point on \(IG\) such that \(PN\) is parallel to \(FH\). The angle \(PNM\) is alternate to \(NMH\) so \(\angle PNM = \alpha\). Also, the triangle \(PNI\) is similar to the triangle \(GHI\) (the angles of each triangle are clearly the same); moreover since \(N\) is the midpoint of \(HI\), \(PN = \tfrac{1}{2}GH\). Also \(IP = \tfrac{1}{2}IG\), so \(PG = \tfrac{1}{2}IG\). Since \(MG = \tfrac{1}{2}FG\), the triangle \(PMG\) is similar to \(IFG\), and in particular, \(PM = \tfrac{1}{2}IF\). However, we know \(IF\) is equal in length to \(GH\) so we have \(PN = \tfrac{1}{2}GH = \tfrac{1}{2}IF = PM\), so triangle \(MNP\) is isosceles and \(\angle PMN = \angle PNM = \alpha\).
Since triangles \(PMG\) and \(IFG\) are similar, we have \(\angle IFG = \angle PMG = \alpha + \alpha = 2\alpha\).

The diagram shows part of the diagram in the question. Point \(O\) is the centre of the circle and points \(A\), \(B\), \(C\), \(D\), \(E\), \(F\), are as shown. Consider triangle \(BOF\): \(BF\) is equal in length to the side of the square so \(BF = 2\). Also \(OB = OF = 2\) as they are both radii of the circle. So triangle \(BOF\) is equilateral.
Therefore \(\angle BOF = 60\)°, so the area of sector \(OBF = \frac{60}{360} \times \pi \times 2^2 = \frac{2\pi}{3}\).
By Pythagoras' Theorem: \(OA = \sqrt{2^2 - 1^2} = \sqrt{3}\).
So the area of triangle \(OBF = \frac{1}{2} \times 2 \times \sqrt{3} = \sqrt{3}\). Therefore the area of the segment bounded by arc \(BF\) and line segment \(BF = \frac{2\pi}{3} - \sqrt{3}\). The area of rectangle \(BFEC = BF \times AD = 2 \times (\sqrt{3} - x)\).
The shaded region has area 2, so the area of the above segment + area of rectangle \(BFEC\) = area of the given square minus 2 = 4 − 2 = 2.
Hence \(\frac{2\pi}{3} - \sqrt{3} + 2\sqrt{3} - 2x = 2\). So \(x = \frac{\pi}{3} + \frac{\sqrt{3}}{2} - 1\).

Since \(VWXYZ\) is a pentagon, the sum of its interior angles is 540°. Now \(\angle ZVW\) is an interior angle of a regular pentagon and so is 108°. Both \(\angle VWX\) and \(\angle XYZ\) are 90°; and the reflex angle \(\angle YZV\) = 90° + 108°.
Therefore 540° = 108° + 90° + 90° + 90° + 108° + \(x\)°.
Hence \(x = 540 - 486 = 54\).

As shown in the diagram, the perpendicular from \(A\) to \(BC\) meets \(BC\) at \(D\). Because \(AD\) is parallel to two sides of the lower rectangle, \(\angle BAD = 43\)° and \(\angle DAC = 29\)° (corresponding angles in both cases). The angles at point \(A\) sum to 360°, so \(43 + 29 + 90 + x + 90 = 360\) and hence \(x = 108\).

The diagram shows part of the figure, to which have been added \(A\) and \(C\), centres of two of the quarter-circle arcs, \(B\) and \(D\), points of intersection of two arcs, and \(E\), the centre of the small circle. In cm, the radii of each arc and the small circle are \(R\) and \(r\) respectively. Firstly, note that \(\angle BCD\) is a right angle as arc \(BD\) is a quarter of a circle. Therefore, by Pythagoras' Theorem \(R^2 + R^2 = 2^2\) so \(R = \sqrt{2}\). Consider triangle \(ACE\): from the symmetry of the figure we deduce that \(\angle AEC = \frac{1}{4} \times 360\)° = 90°. So, by Pythagoras' Theorem \((R + r)^2 + (R + r)^2 = (2R)^2 = 4R^2\). Therefore \((R + r)^2 = 2R^2 = 2 \times 2 = 4\).
Hence \(R + r = 2\), so \(r = 2 - R = 2 - \sqrt{2}\).

In the diagram, \(O\) is the centre of the lower semicircle, \(A\) and \(C\) are the points of intersection of the two semicircles and \(B\) is the point at the centre of the rectangle and also of the overlap. Now \(OA\) is a radius of the semicircle so \(OA\) has length 5 cm. Also \(OB\) is half the height of the rectangle so has length 4 cm. Angle \(ABO\) is a right angle. So triangle \(ABO\) is a (3, 4, 5) triangle and hence \(AC = 2 \times 3\) cm = 6 cm.

The lengths of \(WX\), \(XZ\) and \(ZW\) are all equal (each being the diagonal of a square face), hence triangle \(WXZ\) is equilateral and angle \(ZWX\) is 60°.
The other angles are all 90°, so the total of all four angles is \(90° + 90° + 90° + 60° = 330°\).

Let the length of the side of the square be \(2x\) cm. Then, using Pythagoras' Theorem in the triangle shown, \((2x)^2 + x^2 = 1^2\). So \(4x^2 + x^2 = 1\). Therefore \(x^2 = \frac{1}{5}\) and the area of the square is \(4x^2\) cm² = \(\frac{4}{5}\) cm².

When the net is folded up to form the rhombicuboctahedron, the left-hand edge of the square marked \(X\) is joined to the right-hand edge of the square marked \(E\) so that the eight squares at the centre of the net form a band around the solid. In this band, the square opposite square \(P\) is the square which is four squares away from \(P\), that is square \(D\). So if the square marked \(P\) is placed face down on a table, then the square marked \(D\) will be facing up.

Each square has side-length 1 unit, so by Pythagoras' Theorem the diagonals have length \(\sqrt{1^2 + 1^2} = \sqrt{2}\). The distance between the two circles consists of a whole diagonal and two part diagonals (from the corner of a square to the circle). This is the same as the length of two whole diagonals, minus a diameter, that is \(2\sqrt{2} - 1\).

The sum of the interior angles of a quadrilateral is 360°, so \(x + 5x + 2x + 4x = 360\), that is \(12x = 360\). Therefore \(x = 30\) and the angles of the quadrilateral, taken in order, are 30°, 150°, 60° and 120°. The diagram shows the shape of the quadrilateral.
Since 30 + 150 = 180, we see that \(AB\) and \(DC\) are parallel. Since it has no equal angles, it is not a rhombus or a parallelogram so it is a trapezium.

Let the lengths of the sides of the equilateral triangles that are cut off be \(x\) cm, \(y\) cm and \(z\) cm, as shown in the diagram.
The length of a side of the large equilateral triangle is \(\tfrac{1}{3} \times 60\) cm = 20 cm. The perimeter of the irregular hexagon is 40 cm. Therefore we have

\(40 = x + (20 - x - y) + y + (20 - y - z) + z + (20 - z - x)\). Hence \(40 = 60 - (x + y + z)\) and therefore \(x + y + z = 20\). Therefore the sum of the perimeters of the triangles cut off is \((3x + 3y + 3z)\) cm = 60 cm.

The angles at a point sum to 360° so
\(75 + z = 360\) and \(y + x = 360\).
Therefore \(75 + z + y + x = 720\).
The sum of the interior angles of a hexagon is 4 × 180° = 720°.
Therefore

By plotting the points, it is easy to check that \(BCDE\) is a square. Since any three vertices of a square determine the fourth vertex, it is impossible to make a square using three of these points and the point \(A\).

In the diagram \(V\) is the point where the perpendicular from \(P\) meets \(TS\). Let the side of the square \(RSTU\) be \(x\). So the radius of the arc from \(R\) to \(T\) is \(x\). Therefore \(SP\) has length \(x\), \(PV\) has length \(x - 8\) and \(VS\) has length \(x - 1\). Applying Pythagoras' Theorem to triangle \(PVS\):
\((x - 8)^2 + (x - 1)^2 = x^2\). So \(x^2 - 16x + 64 + x^2 - 2x + 1 = x^2\).
Therefore \(x^2 - 18x + 65 = 0\), so \((x - 5)(x - 13) = 0\).
Hence \(x = 5\) or \(x = 13\), but \(x > 8\) so the length of the side of the square \(RSTU\) is 13.

Label the vertices of the right-angled triangle as \(A\), \(B\), \(C\), with angle \(ABC = 90°\). Let \(M\) be the point where the angle bisector of \(CAB\) meets the side \(BC\). Let \(N\) be the point where the perpendicular from \(M\) meets the side \(AC\). Let \(x\) be the length of the bisector \(AM\).
Then triangles \(ABM\) and \(ANM\) are congruent (they have all three angles the same, and one corresponding side \(AM\) in common). Thus \(MN = 1\).
If we reflect triangle \(MNC\) in the line \(NC\), then we have an equilateral triangle with all sides equal to length 2. Hence \(\angle NCM = 30°\) (half of 60°). Thus \(\angle CAB = 180° - 90° - 30° = 60°\), and \(\alpha = 30°\). Then triangles \(MNC\) and \(MNA\) are congruent (all angles the same, and common length \(MN\)), so \(AM = MC = 2\).
An alternative solution can be obtained using the Angle Bisector Theorem. This gives us \(AB : AC = BM : CM = 1 : 2\). Suppose, then, that \(AB = x\) and \(AC = 2x\). By Pythagoras' Theorem applied to triangle \(ABC\), \((2x)^2 = x^2 + 3^2\). Therefore \(x^2 = 3\). Therefore, by Pythagoras' Theorem applied to triangle \(MBA\), \(AM^2 = x^2 + 1^2 = 3 + 1 = 4\). Therefore \(AM = \sqrt{4} = 2\).

The sum of the exterior angles of a convex polygon equals 360°. The angle marked \(p\)° is the exterior angle of a regular pentagon. So \(p = 360 \div 5 = 72\). The angle sum of a triangle equals 180°, so \(q = 180 - 90 - 72 = 18\). The angle marked \(r\)° is the interior angle of a regular pentagon, so \(r = 180 - 72 = 108\). The angles marked \(q\)°, \(r\)° and \(x\)° lie along a straight line, so \(x = 180 - (q + r)\) \(= 180 - (18 + 108) = 54\).

Let each side of the octagon have length \(x\). The octagon may be divided into eight triangles by joining the centre of the circle to the vertices of the octagon. One such triangle is shown. Each of these triangles has one side of length 1 (the radius of the smaller circle), one side of length 2 (the radius of the larger circle) and one side of length \(x\). So all eight triangles are congruent. Therefore \(\angle AOB = 360^\circ \div 8 = 45^\circ\).
Let \(D\) be the foot of the perpendicular from \(B\) to \(AO\). Then triangle \(BDO\) is an isosceles right-angled triangle.
Let \(OD = DB = y\). Applying Pythagoras' Theorem to triangle \(BDO\):
\(y^2 + y^2 = 1\). So \(y = \tfrac{1}{\sqrt{2}}\).
Applying Pythagoras' Theorem to triangle \(ADB\):

So the length of the perimeter is \(8x = 8\sqrt{5 - 2\sqrt{2}}\).
(Note that the length of AB may also be found by applying the Cosine Rule to triangle OAB.)

\(A\), \(B\), \(C\), \(D\), \(E\) are the centres of the five semicircles. Note that \(AC\) joins the centres of two touching semicircles and therefore passes through the point of contact of the semicircles. So \(AC\) has length 2 + 2 = 4.
This also applies to all of the other sides of triangles \(ACD\) and \(BED\). Hence both triangles are equilateral. So each of the nine arcs which make up the perimeter of the shaded shape subtends an angle of 60° at the centre of a semicircle.
So the length of the perimeter of the shaded figure is 9 × \(\tfrac{1}{6}\) × 2 × π × 2 = 6π.

Clearly \(AD\) lies along one of the lines of symmetry of the figure. So \(\angle FDA = \angle EDA = x^\circ\). Triangle \(DEF\) is equilateral so \(\angle EDF = 60^\circ\).
The angles which meet at a point sum to 360°, so \(x + x + 60 = 360\).
Therefore \(x = 150\).

Let the base \(AB\) of the triangle be the side of length 8 cm and let \(AC\) be the side of length 6 cm. So \(C\) must lie on the circle with centre \(A\) and radius 6 cm as shown. The area of the triangle is to be 7 cm², so the perpendicular from \(C\) to \(AB\) (or to \(BA\) produced) must be of length \(\tfrac{7}{4}\) cm.
The diagram shows the four possible positions of \(C\). However, since\(\angle BAC_1 = \angle BAC_3\) and \(\angle BAC_2 = \angle BAC_4\), these correspond to exactly two possibilities for the length of the third side \(AC\). The diagrams below show the two possibilities.

The radius of each disc in the figure is equal to half the side-length of the square, i.e. \(\tfrac{1}{\pi}\). Because the corners of a square are right-angled, the square hides exactly one quarter of each disc.
So three-quarters of the perimeter of each disc lies on the perimeter of the figure. Therefore the length of the perimeter is \(4 \times \tfrac{3}{4} \times 2\pi \times \tfrac{1}{\pi} = 6\).

The sum of the interior angles of a pentagon is 540° so \(\angle ABC = 540^\circ \div 5 = 108^\circ\). Each interior angle of an equilateral triangle is 60°, so \(\angle FBC = 60^\circ\).
Therefore \(\angle ABF = 108^\circ - 60^\circ = 48^\circ\). As \(ABCDE\) is a regular pentagon, \(BC = AB\). However, \(BC = FB\) since triangle \(BFC\) is equilateral.
So triangle \(ABF\) is isosceles with \(FB = AB\).
Therefore \(\angle FAB = \angle AFB = (180^\circ - 48^\circ) \div 2 = 66^\circ\).

In the diagram, \(T\) is the foot of the perpendicular from \(Q\) to \(SR\) produced. Angles \(PQR\) and \(QRT\) are alternate angles between parallel lines so \(\angle QRT = 60^\circ\). Triangle \(QRT\) has interior angles of 90°, 60°, 30° so it may be thought of as being half of an equilateral triangle of side 1 unit, since the length of \(QR\) is 1 unit. So the lengths of \(RT\) and \(QT\) are \(\tfrac{1}{2}\) and \(\tfrac{\sqrt{3}}{2}\) units respectively.
Applying Pythagoras' Theorem to \(\Delta QST\), \(SQ^2 = ST^2 + QT^2 = \left(\tfrac{5}{2}\right)^2 + \left(\tfrac{\sqrt{3}}{2}\right)^2 = \tfrac{25}{4} + \tfrac{3}{4} = 7\).
So the length of \(SQ\) is \(\sqrt{7}\) units.

As triangle \(ABC\) is equilateral, \(\angle BAC = 60^\circ\). Since the grey triangles are equilateral, \(\angle ADE = 60^\circ\), so the triangle \(ADE\) is equilateral. The length of the side of this triangle is equal to the length of \(DE\) = (5 + 2 + 5) cm = 12 cm. So \(AF = AD - FD\) = (12 − 5) cm = 7 cm. By a similar argument, we deduce that \(BD\) = 7 cm, so the length of the side of triangle \(ABC\) = (7 + 5 + 7) cm = 19 cm.

Opposite angles of a parallelogram are equal, so \(\angle QPS = 50^\circ\). Therefore \(\angle QPT = 112^\circ\) and, as triangle \(QPT\) is isosceles, \(\angle PQT = (180^\circ - 112^\circ) \div 2 = 34^\circ\).
As \(PQRS\) is a parallelogram, \(\angle PQR = 180^\circ - 50^\circ\)
\(= 130^\circ\).
So \(\angle TQR = 130^\circ - 34^\circ = 96^\circ\).

The diagram shows that the interior angles of the polygon may be divided up to form the interior angles of six triangles. So their sum is 6 × 180°.

Points \(A\), \(B\), \(C\), \(D\) and \(E\) are, respectively: the point where the large semicircle and the circle touch, the centre of the circle, the centre of the left-hand semicircle, the centre of the large semicircle and the centre of the right-hand semicircle. The radius of the circle is \(r\) m. In triangle \(BCD\), \(CD\) has length 1m, \(BD\) has length \((2 - r)\) m, since \(AD\) is a radius of the semicircle of diameter 4m, and BC has length \((1 + r)\) m, since it is the sum of the radii of the left-hand semicircle and the circle. Therefore, by Pythagoras' Theorem: \((1 + r)^2 = 1^2 + (2 - r)^2\), that is \(1 + 2r + r^2 = 1 + 4 - 4r + r^2\). So \(6r = 4\) and the radius of the circle is \(\tfrac{2}{3}\)m.
(Note that triangle BCD is a 3,4,5 triangle with sides \(\tfrac{3}{3}\)m, \(\tfrac{4}{3}\)m and \(\tfrac{5}{3}\)m.)

As can be seen from the diagram, Manchester is 40km west of Leeds and 25 km south of it.
Therefore, by Pythagoras' Theorem, the distance in km from Leeds to Manchester as the crow flies is
\(\sqrt{25^2 + 40^2} = 5\sqrt{5^2 + 8^2} = 5\sqrt{89}\).
Now \(\sqrt{81} = 9\) and \(\sqrt{100} = 10\), so \(9 < \sqrt{89} < 10\).
This means that the required distance is between 45km and 50km and, of the options given, only 47km (corresponding to the approximation \(\sqrt{89} \approx 9.4\)) lies in this interval.
(Please note that the distances given in this problem are all approximate.)

In a triangle, an exterior angle is equal to the sum of the two interior, opposite angles. Repeatedly applying this theorem:
\(p = 150 - 80 = 70\);
\(q = p - 40 = 30\); \(r = q - 20 = 10\).
Therefore \(x = 180 - r = 170\).

Alternate angles \(BDF\) and \(DFG\) are equal, so lines \(BD\) and \(FG\) are parallel. Therefore \(\angle BCA = \angle FGC = 80^\circ\) (corresponding angles).
Consider triangle \(ABC\): \(x + 70 + 80 = 180\), so \(x = 30\).

Let the length of the side of the regular octagon be \(x\) units and let \(A\), \(B\), \(C\), \(D\), \(E\), \(F\) be the points shown. So \(AC = CE = x\).
Now \(\angle ACE\) = 135° (interior angle of regular octagon), so \(\angle ACB\) = 45° and hence triangle \(ABC\) is an isosceles right-angled triangle with \(AB = BC\).
Also, by Pythagoras' Theorem: \(AB^2 + BC^2 = AC^2 = x^2\) so \(AB = BC = \tfrac{\sqrt{2}}{2}x\).
Similarly, \(EF = \tfrac{\sqrt{2}}{2}x\).
Therefore \(BF = \left(\tfrac{\sqrt{2}}{2}x + x + \tfrac{\sqrt{2}}{2}x\right)\) units \(= x(1 + \sqrt{2})\) units.
But we are given that \(BF = (1 + \sqrt{2})\) units so \(x = 1\).
Now the area of the octagon formed by the overlap of the squares is equal to the area of one of these squares minus the sum of the area of four triangles, each of which is congruent to triangle \(CDE\).
Thus, in square units, the required area is

If the first triangle selected to be shaded is a corner triangle, then the final figure will have at least one axis of symmetry provided that the second triangle selected is one of five triangles. For example, if A is chosen first then there will be at least one axis of symmetry in the final figure if the second triangle selected is B, D, E, G or H. The same applies if an inner triangle is selected first: for example, if B is chosen first then there will be at least one axis of symmetry in the final figure if the second triangle selected is A, C, F, G or H.
So, irrespective of which triangle is selected first, the probability that the final figure has at least one axis of symmetry is \(\dfrac{5}{7}\).

Let \(Y\) and \(Z\) be the points shown. The interior angle of a regular hexagon is 120°, so \(\angle XZY\) = 120° − 90° = 30°. The side of the square has the same length as the side of the regular hexagon, so \(YZ = XZ\). Hence triangle \(XYZ\) is isosceles and \(\angle ZXY = \angle ZYX = \tfrac{1}{2}\)(180° − 30°) = 75°.

As can be seen from the diagram, the square whose vertices are the centres of the original four circles has side of length 2 units and this distance is equal to the diameter of the circle through \(X\), \(Y\), \(Z\) and \(T\).

In the diagram on the right, triangle \(ABC\) represents the garden, \(CD\) represents the fence and \(E\) is the foot of the perpendicular from \(D\) to \(AC\).
The two sections of the garden have the same perimeter so \(AD\) is 10 m longer than \(DB\). Hence \(AD\) = 30 m and \(DB\) = 20 m. As \(\angle AED\) and \(\angle ACB\) are both right angles, triangles \(AED\) and \(ACB\) are similar.
So \(\tfrac{AE}{AC} = \tfrac{AD}{AB} = \tfrac{30}{50}\). Hence \(AE = \tfrac{3}{5} \times 30\) m = 18 m. So \(EC\) = (30 − 18) m = 12 m.
Also, \(\tfrac{ED}{CB} = \tfrac{AD}{AB} = \tfrac{30}{50}\). Hence \(ED = \tfrac{3}{5} \times 40\) m = 24 m.
Finally, by Pythagoras' Theorem: \(CD^2 = EC^2 + ED^2 = (12^2 + 24^2)\) m² = 5 × 12² m².
So the length of the fence is \(12\sqrt{5}\) m.

The sum of the interior angles of a triangle is 180°, so \(\angle PRQ\) = (180 − \(\alpha\) − \(\beta\))°; hence \(\angle QRM\) = \(\left(90 - \tfrac{\alpha}{2} - \tfrac{\beta}{2}\right)\)°. As \(RN\) is perpendicular to \(PQ\), \(\angle NRQ\) = (90 − \(\beta\))°. So \(\angle MRN\) = \(\left(90 - \tfrac{\alpha}{2} - \tfrac{\beta}{2}\right)\)° − (90 − \(\beta\))° = \(\left(\tfrac{\beta}{2} - \tfrac{\alpha}{2}\right)\)°.

Let the vertices of the quadrilateral be \(A\), \(B\), \(C\), \(D\) as shown. Then, by Pythagoras' Theorem:
\(AC^2 = AD^2 + DC^2 = (7^2 + 9^2)\) cm² = 130 cm².
Similarly, \(AB^2 + BC^2 = AC^2\), so
\(AB^2 = (130 - 9)\) cm² = 121 cm². Therefore \(AB\) has length 11 cm and the area, in cm², of quadrilateral \(ABCD\) is \(\tfrac{1}{2} \times 9 \times 7 + \tfrac{1}{2} \times 3 \times 11 = 48\).

Notice that a single copy of the logo consists of four dots which lie in a straight line plus two other dots which lie on the perpendicular bisector of this line. These two dots are not evenly spaced above and below the line of four dots. Of the options given, only A has two lines of four dots with two more dots in the correct positions relative to each line.

Let the radius of the semicircle be \(r\) and let the perpendicular height of the isosceles triangle be \(h\). Then \(\tan x\)° = \(\tfrac{h}{r}\).
Now the area of the semicircle = \(\tfrac{1}{2}\pi r^2\), whilst the area of the triangle = \(\tfrac{1}{2} \times 2r \times h = rh\). So \(rh = \tfrac{1}{2}\pi r^2\), giving \(\tfrac{h}{r} = \tfrac{\pi}{2}\).

From the symmetry of the figure, the two circles must be concentric. Let their centre be \(O\). Let the radius of the semicircles be \(r\). Then the radius of the outer circle is \(2r\) and, by Pythagoras' Theorem, the radius of the inner shaded circle is \(\sqrt{r^2 + r^2}\), that is \(\sqrt{2}r\).
So the radii of the two circles are in the ratio \(\sqrt{2} : 2\), that is \(1 : \sqrt{2}\), and hence the ratio of their areas is \(1 : 2\).

Any line which passes though the centre of the square divides the square into two congruent shapes. An example is shown on the right.

Let \(r\) be the radius of the circle. Then, in the smaller square:

and in the larger square:

So \(\tfrac{5x}{4} = \tfrac{y}{2}\) and we deduce that \(x : y = 2 : 5\).

Let \(T\) be the centre of the semicircle with diameter \(QR\) and let \(OT\) produced meet the circumference of the larger semicircle at \(U\).
By symmetry, we note that \(OT\) is perpendicular to \(QR\).
As \(TR = TO = TQ\) (radii of semicircle), triangles \(ORT\) and \(OQT\) are both isosceles, right-angled triangles. So \(QOR\) is a right angle. By Pythagoras' Theorem: \(QR^2 = OQ^2 + OR^2 = 2^2 + 2^2 = 8\). So \(QR = \sqrt{8} = 2\sqrt{2}\) and the radius of semicircle \(QOR\) is \(\sqrt{2}\).
The area of the shaded region is equal to the area of semicircle \(QOR\) plus the area of the quadrant bounded by \(OQ\), \(OR\) and arc \(QUR\) less the area of triangle \(OQR\). So the required area is \(\tfrac{1}{2}\pi(\sqrt{2})^2 + \tfrac{1}{4}\pi 2^2 - \tfrac{1}{2} \times 2 \times 2 = \pi + \pi - 2 = 2\pi - 2\).

The diagram shows that the two pieces will fit together to form a right-angled triangle which has base 8 and height 6. The length of the hypotenuse = \(\sqrt{6^2 + 8^2}\), that is 10, so the perimeter of the triangle is 24.

In fact, the second shadow corresponds to a cuboid which is half a cube, as shown alongside.

Each exterior angle of a regular hexagon = 360° ÷ 6 = 60°, so when sides \(HB\) and \(IC\) are produced to meet at \(A\), an equilateral triangle, \(ABC\), is created. Let the sides of this triangle be of length \(x\). As \(BC\), \(DE\) and \(FG\) are all parallel, triangles \(ABC\), \(ADE\) and \(AFG\) are all equilateral. So \(DE = DA = p + x\); \(FG = FA = q + p + x\).

The perimeter of trapezium \(BCED = x + p + x + 2p\) \(= 2x + 3p\); the perimeter of trapezium \(DEGF =\) \((p + x) + (q + p + x) + 2q\) \(= 2x + 2p + 3q\); the perimeter of hexagon \(FGIKJH = 2((q + p + x) + 2r)\) \(= 2x + 2p + 2q + 4r\).

So \(2x + 3p = 2x + 2p + 3q\); hence \(p = 3q\). Also \(2x + 2p + 3q\) \(= 2x + 2p + 2q + 4r\); hence \(q = 4r\). So \(p : q : r\) \(= 12r : 4r : r\) \(= 12 : 4 : 1\).

The diagram shows points \(A\) and \(B\), which are the centres of the two circles, and \(C\), the point on \(BQ\) such that \(AC\) is parallel to \(PQ\). As radii \(PA\) and \(QB\) are both perpendicular to tangent \(PQ\), \(APQC\) is a rectangle. So \(\angle ACB\) is a right angle. The length of \(AB = 1 + 4 = 5\); the length of \(BC = 4 - 1 = 3\). So, by Pythagoras' Theorem, \(AC = \sqrt{5^2 - 3^2} = 4\), which, therefore, is also the length of \(PQ\).

Let \(ABCD\) be the cross-section of one of the stones, as shown. As \(AD = BC\), \(ABCD\) is an isosceles trapezium with \(\angle ADC = \angle BCD\) (the proof of which is left to the reader). If \(\angle ADC = \angle BCD = \theta\), then \(2\theta\) is the interior angle of a regular 20-sided polygon, namely (180 − 360/20)°, which equals 162°. So \(\theta\) is 81°.

Considering the angles at \(B\) and \(E\):
\(\angle CBE\) = (180 − 75 − 60)° = 45°;
\(\angle DEB\) = (180 − 65 − 60)° = 55°. Therefore
\(\angle GHB\) = (45 + 55)° = 100° (exterior angle theorem) and, using the same theorem, \(\angle HGC\) = (100 − 60)° = 40°.

Let the sizes of the interior angles of the quadrilateral at the centre of the figure be \(a\)°, \(b\)°, \(c\)° and \(d\)°. Then \(a + b + c + d = 360\). Each of these angles is vertically opposite to one of the unmarked angles in the four triangles, as shown. So the sum of the marked angles plus the four angles of the quadrilateral is equal to the sum of the interior angles of four triangles, that is 720°. Hence the sum of the marked angles is 720° − 360° = 360°.

Triangles \(ABG\), \(GFE\) and \(EFA\) are similar, so \(AF : FE = EF : FG = GB : BA = 1 : 2\). Thus if \(AF = a\), then \(FE = 2a\), \(FG = 4a\), and the shaded area is \(8a^2\). By Pythagoras' Theorem, \(AE = \sqrt{5}a\), so \(AD = 2\sqrt{5}a\) and the area of the square is \(20a^2\). Thus the required fraction is 8/20, which is 2/5.

[Alternatively, as we have shown that \(FG = 4AF\), we can divide parallelogram \(AGCE\) into 10 congruent triangles, 8 of which make up rectangle \(EFGH\). So the area of the rectangle is 4/5 of the area of the parallelogram, which in turn is half the area of square \(ABCD\).]