Elementary Mathematics 2018 (II)

Step-by-step solution

1. 1
   For a cube of edge $a$, a face diagonal is $a\sqrt{2}$, and the space diagonal is the hypotenuse of a right triangle with legs $a\sqrt{2}$ and $a$: $\sqrt{2a^2+a^2}=a\sqrt{3}$.

   Why: Pythagoras is applied twice - once in a face, once in the vertical plane through that face diagonal.

   Pythagoras' theoremSpace diagonal of a cube and a cuboid
2. 2
   $a\sqrt{3}=l\Rightarrow a=\frac{l}{\sqrt{3}}\Rightarrow a^2=\frac{l^2}{3}$

   Why: Only $a^2$ is needed, so the surd never has to be carried further.

3. 3
   Total surface $=6a^2=6\times \frac{l^2}{3}=2l^2$

   Why: A cube has six congruent square faces.

   Volume and wall area of a cuboid

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Step-by-step solution

1. 1
   Triangle: side $\frac{P}{3}$, so $T=\frac{\sqrt{3}}{4}{\left(\frac{P}{3}\right)}^2=\frac{\sqrt{3}}{36}{P}^2\approx 0.0481P^2$

   Why: Writing every area over the same $P^2$ is what makes the three comparable at all.

   Area and altitude of an equilateral triangle
2. 2
   Square: side $\frac{P}{4}$, so $S=\frac{P^2}{16}=0.0625P^2$

   Why: The square's area in terms of its perimeter is $\frac{P^2}{16}$ directly.

   Perimeter and area of a square
3. 3
   Circle: $2\pi r=P$ gives $r=\frac{P}{2\pi }$, so $C=\pi r^2=\frac{P^2}{4\pi }\approx 0.0796P^2$

   Why: The $\pi$ in the circumference and the $\pi$ in the area partly cancel, leaving $\frac{1}{4\pi }$.

   Area and circumference of a circle
4. 4
   $0.0481<0.0625<0.0796$, so $T<S<C$.

   Why: Comparing three decimals settles the order without any of the $P^2$ ever being evaluated.

   Comparing two fractions

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Step-by-step solution

1. 1
   For similar triangles the ratio of areas is the square of the ratio of corresponding sides, so that ratio is $\sqrt{\frac{7-4\sqrt{3}}{7+4\sqrt{3}}}$.

   Why: The question gives areas and asks for sides, so a square root has to be taken somewhere.

   Similar triangles have proportional sides
2. 2
   $7-4\sqrt{3}=4-2(2)(\sqrt{3})+3=(2-\sqrt{3}{)}^2$ and likewise $7+4\sqrt{3}=(2+\sqrt{3}{)}^2$

   Why: Looking for $p+q=7$ with $pq=12$ gives $p=4$, $q=3$, so both surds are perfect squares in disguise.

   Denest $\sqrt{a\pm 2\sqrt{b}}$
3. 3
   Ratio of sides $=\frac{2-\sqrt{3}}{2+\sqrt{3}}$

   Why: Both $2-\sqrt{3}$ and $2+\sqrt{3}$ are positive, so no sign question arises.

4. 4
   Rationalise: $\frac{2-\sqrt{3}}{2+\sqrt{3}}\times \frac{2-\sqrt{3}}{2-\sqrt{3}}=\frac{(2-\sqrt{3}{)}^2}{4-3}=7-4\sqrt{3}$

   Why: The denominator $(2+\sqrt{3})(2-\sqrt{3})=1$, so the answer is simply the numerator squared.

   Rationalise a denominator using the conjugate

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Step-by-step solution

1. 1
   Let the radius be $r$ and the chord be $AB=\sqrt{3}r$. The perpendicular from the centre $O$ bisects $AB$, so in the right triangle it makes, $\sin \frac{\angle AOB}{2}=\frac{AB/2}{r}=\frac{\sqrt{3}}{2}$.

   Why: The perpendicular from the centre to a chord bisects it, which is what turns the chord length into an angle.

   The perpendicular from the centre bisects a chordSine, cosine and tangent in a right triangle
2. 2
   $\frac{\angle AOB}{2}=60^{\circ }$, so the central angle is $\angle AOB=120^{\circ }$.

   Why: $\frac{\sqrt{3}}{2}$ is the sine of $60^{\circ }$.

   Exact trigonometric ratios of the standard angles
3. 3
   A point on the major arc sees $AB$ at half the central angle: $\frac{120^{\circ }}{2}=60^{\circ }$.

   Why: The angle at the centre is twice the angle at the circumference on the remaining arc.

   The angle at the centre is twice the angle at the circumference
4. 4
   A point on the minor arc, together with $A$, $B$ and the major-arc point, forms a cyclic quadrilateral, so its angle is $180^{\circ }-60^{\circ }=120^{\circ }$.

   Why: Angles in opposite segments are supplementary, which is the quickest way from one arc to the other.

5. 5
   $k=\frac{120^{\circ }}{60^{\circ }}=2$

   Why: The question asks for the minor-arc angle as a multiple of the major-arc one, in that order.

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Step-by-step solution

1. 1
   The exterior angle at $B$ is $180^{\circ }-\angle B$, so its bisector makes $\angle DBC=\frac{180^{\circ }-\angle B}{2}=90^{\circ }-\frac{\angle B}{2}$ with $BC$.

   Why: $AB$ is produced, so the exterior angle at $B$ sits on the straight line through $A$ and $B$ and is $180^{\circ }-\angle B$.

   Angles on a straight line and vertically opposite anglesWhere two external bisectors meet: $90^{\circ }-\frac{1}{2}\angle A$
2. 2
   In the same way $\angle DCB=90^{\circ }-\frac{\angle C}{2}$.

   Why: $AC$ is produced too, so the configuration at $C$ is the mirror image of the one at $B$.

3. 3
   $\angle BDC=180^{\circ }-\left(90^{\circ }-\frac{\angle B}{2}\right)-\left(90^{\circ }-\frac{\angle C}{2}\right)=\frac{\angle B+\angle C}{2}$

   Why: Closing the angle sum in $△DBC$ makes the two $90^{\circ }$ cancel against the $180^{\circ }$.

   Angle sum of a triangle
4. 4
   $\angle B+\angle C=180^{\circ }-\angle BAC=130^{\circ }$, so $\angle BDC=65^{\circ }$ - equivalently $90^{\circ }-\frac{\angle BAC}{2}$.

   Why: Only the sum of the base angles is needed, which is why $\angle BAC$ alone determines the answer.

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Step-by-step solution

1. 1
   Let the heights be $h$ and $3h$, and the radii $3r$ and $r$, in that order.

   Why: A ratio fixes the quantities only up to a common factor, and those factors will cancel.

   Introduce a common factor $k$ for a ratio
2. 2
   $\frac{V_1}{V_2}=\frac{\frac{1}{3}\pi (3r{)}^{2}h}{\frac{1}{3}\pi r^2(3h)}=\frac{9r^{2}h}{3r^{2}h}=3$

   Why: The radius enters squared and the height only to the first power, so the two ratios do not simply cancel.

   Volume of a right circular cone
3. 3
   The ratio of volumes is $3:1$.

   Why: The cone with the wider base wins, because squaring the radius ratio beats the height ratio.

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1. 1
   $C$, $O$, $D$ lie on one straight line, so $\angle AOC+\angle AOD=180^{\circ }$.

   Why: $CD$ is a line through $O$, and $OA$ stands on it, making the two angles a linear pair.

   Angles on a straight line and vertically opposite angles
2. 2
   $5\angle AOD+\angle AOD=180^{\circ }\Rightarrow \angle AOD=30^{\circ }$ and $\angle AOC=150^{\circ }$

   Why: Substituting the given relation turns the linear pair into one equation in one unknown.

3. 3
   $\angle BOD=\angle AOC=150^{\circ }$ and $\angle BOC=\angle AOD=30^{\circ }$, so the four angles are $30^{\circ },30^{\circ },150^{\circ },150^{\circ }$.

   Why: Where two lines cross, each angle is repeated by the one vertically opposite it.

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Step-by-step solution

1. 1
   Take $A=(-a,0)$ and $B=(a,0)$, so the midpoint of $AB$ is the origin, and let $P=(x,y)$.

   Why: Choosing the axes along $AB$ and through its midpoint costs nothing and makes the two distances symmetric in $x$.

2. 2
   $P{A}^2+P{B}^2=\left[(x+a{)}^2+y^2\right]+\left[(x-a{)}^2+y^2\right]=2x^2+2y^2+2a^2$

   Why: The two cross terms $\pm 2ax$ cancel, which is exactly why the midpoint was chosen as origin.

   Square of a sum and of a difference
3. 3
   Setting this equal to a constant $c$ gives $x^2+y^2=\frac{c-2a^2}{2}$, itself a constant.

   Why: $a$ is fixed because $A$ and $B$ are fixed, so the whole right-hand side is a number.

4. 4
   So $P$ keeps a fixed distance from the midpoint of $AB$: the locus is a circle centred at that midpoint.

   Why: A constant distance from a fixed point is the definition of a circle - provided $c>2a^2$, so the constant is positive.

   A circle is the locus of points at a fixed distance from a fixed point

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Step-by-step solution

1. 1
   $AC$ is the hypotenuse, so $A{C}^2=A{B}^2+B{C}^2$.

   Why: This is the only relation the right angle gives, so it has to be the starting point.

   Pythagoras' theorem
2. 2
   Each leg is strictly shorter than the hypotenuse: $A{C}^2=A{B}^2+B{C}^2>A{B}^2$ gives $AC>AB$, and likewise $AC>BC$.

   Why: Both legs are positive, so dropping either one strictly decreases the sum.

3. 3
   $A{C}^3=AC\cdot A{C}^2=AC\cdot A{B}^2+AC\cdot B{C}^2>AB\cdot A{B}^2+BC\cdot B{C}^2=A{B}^3+B{C}^3$

   Why: Multiplying the identity by $AC$ and then replacing each $AC$ factor by the smaller leg makes the right-hand side strictly smaller.

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Step-by-step solution

1. 1
   Square of side $2a$: area $=(2a{)}^2=4a^2$

   Why: The side is given as $2a$, not $a$, which is what makes the answer come out without fractions.

   Perimeter and area of a square
2. 2
   The circle touches all four sides, so its diameter is the side $2a$ and its radius is $a$: area $=\pi a^2$.

   Why: A circle touching all four sides of a square is its inscribed circle, whose diameter equals the side.

   The largest circle that fits inside a square
3. 3
   Region between them $=4a^2-\pi a^2=(4-\pi )a^2$

   Why: 'Bounded externally by the square and internally by the circle' is the square with the circle removed.

   Cut a compound plane figure into standard pieces

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Step-by-step solution

1. 1
   $\angle A=90^{\circ }$ gives $B{C}^2=p^2+q^2$; and an angle in a semicircle is a right angle, so the semicircle on $BC$ passes exactly through $A$.

   Why: That is what makes the two shaded crescents well defined: their inner boundary is the arc $BAC$ of the big semicircle.

   Pythagoras' theoremThe angle at the centre is twice the angle at the circumference
2. 2
   Semicircle on $AB$: $\frac{1}{2}\pi {\left(\frac{p}{2}\right)}^2=\frac{\pi p^2}{8}$; on $AC$: $\frac{\pi q^2}{8}$; on $BC$: $\frac{\pi (p^2+q^2)}{8}$.

   Why: Each semicircle has the side as its diameter, so the radius is half the side in every case.

   Area and circumference of a circle
3. 3
   Shaded $=\left(\text{semicircle on }AB\right)+\left(\text{semicircle on }AC\right)+\left(\text{area of }△ABC\right)-\left(\text{semicircle on }BC\right)$

   Why: The big semicircle covers the triangle together with the two regions that are NOT shaded, so adding the triangle back and removing the big semicircle leaves exactly the two crescents.

   Cut a compound plane figure into standard pieces
4. 4
   $=\frac{\pi p^2}{8}+\frac{\pi q^2}{8}+\frac{pq}{2}-\frac{\pi (p^2+q^2)}{8}=\frac{pq}{2}$

   Why: The three $\pi$ terms cancel precisely because of Pythagoras, leaving the area of the triangle alone.

   Area of a triangle from base and height

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Step-by-step solution

1. 1
   $A$ lies on the circle and on the segment $OT$, so $OA$ is a radius and $OT=OA+AT=6+4=10$ cm.

   Why: The figure shows $A$ between $O$ and $T$, so the two pieces simply add.

2. 2
   $PT$ is a tangent at $P$, so it is perpendicular to the radius $OP$: $\angle OPT=90^{\circ }$.

   Why: This is the one fact that turns the figure into a right triangle, and without it nothing can be computed.

   The two tangents from an external point
3. 3
   $PT=\sqrt{O{T}^2-O{P}^2}=\sqrt{100-36}=\sqrt{64}=8$ cm

   Why: In right triangle $OPT$ the hypotenuse is $OT$, so the tangent length is the remaining leg.

   Pythagoras' theorem

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1. 1
   $AD$ is a diameter of length $18$ cm and $AB=BC=CD$, so each part is $6$ cm and $BD=12$ cm.

   Why: The three equal parts fix every radius in the figure from the single given radius of $9$ cm.

2. 2
   Upper half of the big circle $=\frac{1}{2}\pi (9{)}^2=\frac{81\pi }{2}$

   Why: The diameter $AD$ cuts the big circle into two equal halves, and the shading above the line sits inside the upper one.

   Area and circumference of a circle
3. 3
   Semicircle on $BD$ (radius $6$) $=\frac{1}{2}\pi (6{)}^2=18\pi$; semicircle on $AB$ (radius $3$) $=\frac{1}{2}\pi (3{)}^2=\frac{9\pi }{2}$

   Why: The semicircle on $BD$ is the hole in the upper crescent; the one on $AB$ is the extra piece shaded below the line.

4. 4
   Shaded $=\frac{81\pi }{2}-18\pi +\frac{9\pi }{2}=45\pi -18\pi =27\pi$

   Why: The two halves combine to $\frac{90\pi }{2}=45\pi$ before the $BD$ semicircle is taken out.

   Cut a compound plane figure into standard pieces

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1. 1
   The figure marks $\angle BAC=30^{\circ }$ (at $A$, between $AB$ and $AC$) and $\angle DBC=70^{\circ }$ (at $B$, between $BD$ and $BC$).

   Why: Both marked angles stand on chords of the circle, which is the only way the data can be used.

2. 2
   $\angle BAC$ and $\angle BDC$ both stand on the chord $BC$, with $A$ and $D$ on the same arc, so $\angle BDC=\angle BAC=30^{\circ }$.

   Why: Angles in the same segment are equal - the corollary that carries a marked angle to a vertex where it is useful.

   The angle at the centre is twice the angle at the circumference
3. 3
   In $△BDC$: $\angle BCD=180^{\circ }-\angle DBC-\angle BDC=180^{\circ }-70^{\circ }-30^{\circ }=80^{\circ }$

   Why: Both other angles of that triangle are now known, so the third closes immediately.

   Angle sum of a triangle

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Step-by-step solution

1. 1
   In the figure the chord $FD$ meets the diameter $AB$ at $E$; the other segments drawn are the radii $OF$ and $OC$ and the chords $CE$ and $CD$. So $C$, $E$, $D$ form a genuine triangle with $\angle ECD=\angle EDC=32^{\circ }$.

   Why: Reading the collinear triple correctly is the whole of the setup: it is $F$, $E$, $D$ that are in a line, not $C$, $E$, $D$.

2. 2
   $\angle CEF$ and $\angle CED$ are angles on the straight line $FD$ at $E$, so $\angle CEF$ is the exterior angle of $△CED$ at $E$: $\angle CEF=\angle ECD+\angle EDC=32^{\circ }+32^{\circ }=64^{\circ }$.

   Why: An exterior angle equals the sum of the two remote interior angles, which avoids computing $\angle CED$ at all.

   Angle sum of a triangle
3. 3
   $\angle EDC$ is the same angle as $\angle FDC$, because $E$ lies on $DF$; it is an angle at the circumference standing on the chord $CF$.

   Why: $D$ is on the circle, so once $\angle FDC$ is identified the chord $CF$ is the one it subtends.

4. 4
   The angle at the centre on that same chord is twice it: $\angle COF=2\times 32^{\circ }=64^{\circ }$. So the two angles are $64^{\circ }$ and $64^{\circ }$.

   Why: $O$ is the centre and $C$, $F$ are on the circle, so the doubling applies directly.

   The angle at the centre is twice the angle at the circumference

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Step-by-step solution

1. 1
   $△ABR\sim △PQR$ pairs $A$ with $P$, $B$ with $Q$ and $R$ with $R$, so $\frac{AB}{PQ}=\frac{BR}{QR}=\frac{AR}{PR}$.

   Why: The order of the letters in a similarity statement is what fixes which side matches which; nothing else does.

   Similar triangles have proportional sides
2. 2
   $\frac{AB}{PQ}=\frac{6}{3}=2$, so $△ABR$ is twice $△PQR$.

   Why: One matched pair of sides is enough to pin the scale factor.

3. 3
   $QR$ corresponds to $BR$: $QR=\frac{BR}{2}=\frac{8.2}{2}=4.1$ cm

   Why: $QR$ belongs to the smaller triangle, so the known length is divided, not multiplied.

4. 4
   $AR$ corresponds to $PR$: $AR=2\times PR=2\times 5.2=10.4$ cm

   Why: $AR$ belongs to the larger triangle, so here the known length is multiplied.

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Step-by-step solution

1. 1
   $\angle ABC=90^{\circ }$ and $BD\perp AC$, so in $△ADB$ and $△BDC$ the angle at $D$ is right and $\angle DAB=\angle DBC$ (both complete $\angle ABD$ to $90^{\circ }$). By AA, $△ADB\sim △BDC$.

   Why: The altitude from the right angle always creates two triangles similar to the original and to each other; that is the fact this question is built on.

   Similar triangles have proportional sides
2. 2
   Corresponding sides give $\frac{AD}{BD}=\frac{BD}{DC}$, so $B{D}^2=AD\cdot DC$.

   Why: $AD$ and $BD$ are the legs about $D$ in the first triangle, $BD$ and $DC$ the matching legs in the second.

3. 3
   $B{D}^2=9\times 4=36\Rightarrow BD=6$ cm

   Why: Only the positive root is a length.

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Step-by-step solution

1. 1
   The big semicircle has radius $54$. Each small semicircle has diameter $54$, so radius $27$, with centres $27$ cm either side of the mid-point $O$ of the big diameter.

   Why: The two marked $54$ cm pieces fix every radius on the base line.

2. 2
   Let the top circle have radius $\rho$, with its centre on the vertical through $O$ at height $k$ (by symmetry). Touching the big arc from inside gives $k=54-\rho$; touching a small semicircle from outside gives $\sqrt{27^2+k^2}=27+\rho$.

   Why: Touching internally puts the centres $54-\rho$ apart; touching externally puts them $27+\rho$ apart, and that second distance is a hypotenuse.

   Distance between the centres of touching circlesPythagoras' theorem
3. 3
   $729+(54-\rho {)}^2=(27+\rho {)}^2\Rightarrow 729+2916-108\rho =729+54\rho \Rightarrow 2916=162\rho \Rightarrow \rho =18$

   Why: The ${\rho }^2$ terms cancel on both sides, leaving a linear equation.

   Square of a sum and of a difference
4. 4
   Shaded $=\frac{1}{2}\pi (54{)}^2-2\times \frac{1}{2}\pi (27{)}^2-\pi (18{)}^2=1458\pi -729\pi -324\pi =405\pi$ cm${}^2$

   Why: The hatching is the part of the big semicircle left after the two small semicircles and the full circle are removed.

   Area and circumference of a circleCut a compound plane figure into standard pieces

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Step-by-step solution

1. 1
   $XP\parallel AC$, so $\angle XPB=\angle ACB=60^{\circ }$, and $\angle B=60^{\circ }$ too: $△XBP$ is equilateral, giving $BX=XP=BP$. Call this common length $a$.

   Why: A line parallel to one side of an equilateral triangle cuts off another equilateral triangle, because all the angles are $60^{\circ }$.

   Angles made by a transversal of two parallel linesBase angles of an isosceles triangle
2. 2
   In the same way $YQ\parallel AB$ makes $△YQC$ equilateral: $YC=YQ=QC$. Call this $c$.

   Why: The same argument at the other end of $BC$, with $\angle C=60^{\circ }$.

3. 3
   $XY\parallel BC$ makes $△AXY$ equilateral, so $AX=AY=XY$. Since $AX=30-a$ and $AY=30-c$, this forces $a=c$ and $XY=30-a$.

   Why: $X$ and $Y$ must be at equal distances from $A$ for $XY$ to be parallel to $BC$, which is what ties the two ends together.

   Similar triangles have proportional sides
4. 4
   $XY+XP+YQ=(30-a)+a+a=30+a=40\Rightarrow a=10$

   Why: Two of the three lengths are $a$ and the third is $30-a$, so the sum is linear in $a$ and the $30$ falls out.

5. 5
   $PQ=BC-BP-QC=30-a-c=30-10-10=10$ cm

   Why: $P$ and $Q$ both lie on $BC$ with $B$, $P$, $Q$, $C$ in that order, so the three pieces add to $30$.