I'm Up and Down, and Round and Round: questions with answers

Folding a circle so its boundary overlaps leaves a crease through the centre, and every diameter works. A circle has infinitely many diameters, so it has infinitely many lines of symmetry. That is what separates it from a regular 10-gon, whose 10 axes can be counted.

The longest chord of a circle is the diameter, which is $2 \times 14 = 28$ cm. The radius 14 is only half of it -- the radius runs from the centre to the circle, while the diameter runs all the way across.

A chord cannot be longer than the diameter, so the radius must be at least 12 ÷ 2 = 6 cm. Every radius of at least 6 cm is possible, with a centre on the perpendicular bisector of $AB$. Only 6 cm among the listed radii meets this condition, and it differs from 8 cm.

A circle through $A$ and $B$ has its centre on the perpendicular bisector of $AB$, and one through $B$ and $C$ has its centre on the perpendicular bisector of $BC$. For collinear points those two bisectors are perpendicular to the SAME line, so they are parallel and never meet. With no point equidistant from all three, there is no circle -- which is the same as saying a line cannot cut a circle three times.

130 degrees is more than a right angle, so this triangle is obtuse-angled, and the circumcentre of an obtuse-angled triangle falls outside it. Only an acute-angled triangle keeps its circumcentre inside; a right-angled one puts it on the boundary, at the midpoint of the hypotenuse.

The circumcentre is the midpoint of the hypotenuse. The radius is therefore half the hypotenuse: $14 \div 2 = 7$ cm. The side of length 6 cm joins two vertices; it is not a radius. The right-angle vertex cannot be the centre because it lies on the circumcircle.

A chord that subtends 180 degrees at the centre is the diameter, 10 cm. The chapter's comment after Theorem 8 says the diameter is the greatest chord. So every chord that is not a diameter is shorter than 10 cm. The chord subtending 120 degrees is not a diameter, so it is less than 10 cm. It is not 5 cm either. By Theorem 2, equal chords subtend equal angles. So another chord of 5 cm would subtend 60 degrees, not 120. Doubling the angle from 60 degrees to 120 degrees therefore does not double the chord.

A chord of 16 cm in a circle of radius 8 cm is twice the radius, so it is the diameter and passes through the centre. The two radii to its ends point in opposite directions, so the angle is the straight angle, 180 degrees. Doubling the chord did not double the angle: 60 doubled would be 120, and the true answer is 180.

$CM$ is perpendicular to $AB$, so triangles $CMA$ and $CMB$ are congruent by RHS. They have the equal radii $CA$ and $CB$, the common side $CM$, and right angles at $M$. So $AM = MB$, and $CM$ bisects $AB$. The segment from $C$ to the midpoint of $AB$ is the perpendicular one, so any bisecting segment from $C$ must meet $AB$ at a right angle. Here $\angle CNA = 70^\circ$ and $\angle CPA = 120^\circ$, and neither is $90^\circ$. So $CN$ and $CP$ meet $AB$ away from its midpoint. Exactly one of the three segments bisects $AB$.

They are equal, each 11 cm. Triangles $CMA$ and $CMB$ have equal radii $CA = CB = 15$ cm, share $CM$, and both have a right angle at $M$, so they are congruent and $AM = MB$. Nothing about which side of the centre the chord lies on enters the argument.

A perpendicular from the centre to a chord bisects that chord. A radius, the perpendicular distance and half the chord form a right triangle. By Pythagoras, $AB = 2\sqrt{ 30^2 - 4^2 } = 2\sqrt{ 884 }$ cm and $CD = 2\sqrt{ 30^2 - 8^2 } = 2\sqrt{ 836 }$ cm. Half of $AB$ is $\sqrt{ 884 }$ cm. The squared lengths are $(CD)^2 = 3344$ and $(AB/2)^2 = 884$, in square centimetres. Since $3344 > 884$ and both lengths are positive, $CD > AB/2$. Also $836 < 884$, so $CD < AB$. Therefore $AB/2 < CD < AB$: CD is shorter, but its decrease from AB is less than half of AB. Doubling the distance does not halve the chord. CD cannot be twice as long as AB or equal to AB, because it is farther from the centre and its computed length is smaller.

Half of AB is 6 cm and AB is 2 cm from the centre, so $r^2 = 6^2 + 2^2 = 40$. For CD, half its length squared is $40 - 1^2 = 39$. That is more than 36 and less than 144. So half of CD lies between 6 cm and 12 cm, and CD lies between 12 cm and 24 cm. Halving the distance does not double the chord, because the relation runs through squares.

The arc from $A$ to $B$ not containing $D$ subtends the straight angle, 180 degrees, at the centre. Halving it gives 90 degrees at $D$, so the angle in a semicircle is a right angle. Answering 180 keeps the central angle unhalved, and a triangle cannot hold a 180-degree angle; answering 45 halves 90 rather than 180.

Angle $APB$ is subtended by the major arc $AB$. The same arc subtends the reflex angle $AOB$, which equals $2 \times 134^\circ = 268^\circ$. The smaller angle is $360^\circ - 268^\circ = 92^\circ$.

$B$ and $D$ are the other pair of opposite angles, so angle $D = 180 - 100 = 80$ degrees. Taking opposite angles as equal gives 100 itself; making the adjacent pair supplementary works from $A$ instead and returns 70, which is angle $C$ over again.