Elementary Mathematics 2018 (II)

Step-by-step solution

1. 1
   $\cos \theta =\frac{1}{\sqrt{5}}\Rightarrow {\sin }^2\theta =1-\frac{1}{5}=\frac{4}{5},\sin \theta =\frac{2}{\sqrt{5}}$

   Why: $\theta$ is in the first quadrant, so the sine is positive. Equivalently: a right triangle with adjacent side $1$ and hypotenuse $\sqrt{5}$ has opposite side $2$.

   Pythagorean trigonometric identities
2. 2
   $\tan \theta =\frac{\sin \theta }{\cos \theta }=\frac{2/\sqrt{5}}{1/\sqrt{5}}=2$

   Why: The $\sqrt{5}$ cancels, so the tangent is a whole number.

   Reciprocal and quotient relations between the ratios
3. 3
   $\frac{2\tan \theta }{1-{\tan }^2\theta }=\frac{2(2)}{1-4}=\frac{4}{-3}=-\frac{4}{3}$

   Why: Substitute. The denominator is negative because $\tan \theta >1$, and that negative sign is the whole point of the question.

   Sine, cosine and tangent in a right triangle

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Step-by-step solution

1. 1
   $\cos \theta \text{decreases strictly as}\theta \text{runs from}{0}^{\circ }\text{to}90^{\circ }$

   Why: On the first quadrant the cosine starts at $1$ and falls to $0$, taking each value once.

   How the ratios move across the first quadrant
2. 2
   $\cos \theta <\cos \varphi \Rightarrow \theta >\varphi$

   Why: A strictly decreasing function reverses the inequality, and both angles are inside the range where that holds.

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Step-by-step solution

1. 1
   $\text{centre}O\text{on the ground, dome of radius}r,\text{flag top at height}r+h\text{above}O$

   Why: The hemisphere's highest point is at height $r$ directly above its centre, and the flag stands on it.

   Angles of elevation and depression
2. 2
   $\text{at the nearer point}Q,\text{the}45^{\circ }\text{ray touches the dome}$

   Why: "Just visible" is the boundary case: a step closer and the dome's surface cuts the line of sight off. At the boundary the ray is tangent, and a tangent is at distance $r$ from the centre.

   The two tangents from an external point
3. 3
   $\text{distance from}O\text{to a}45^{\circ }\text{line through}Q=(-y,0)\text{is}\frac{y}{\sqrt{2}}=r\Rightarrow y=r\sqrt{2}$

   Why: The ray is $z=x+y$; its perpendicular distance from the origin is $\frac{y}{\sqrt{1+1}}$. Setting that equal to $r$ fixes how far back $Q$ stands.

4. 4
   $\text{the same ray at}x=0\text{has height}y=r\sqrt{2},\text{and it passes through the flag top}$

   Why: The flag top is the point the observer is looking at, and it sits directly above $O$, i.e. at $x=0$. A $45^{\circ }$ ray rises one unit for every unit travelled, so its height above $O$ equals the horizontal distance $y$.

   Exact trigonometric ratios of the standard angles
5. 5
   $r+h=r\sqrt{2}\Rightarrow \frac{h}{r}=\sqrt{2}-1$

   Why: Equate the two expressions for the flag top's height and divide by $r$. The distance $d$ and the $30^{\circ }$ reading are not needed: they only fix the scale.

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Step-by-step solution

1. 1
   $\cos B=\frac{\sqrt{3}}{2},B\text{acute}\Rightarrow B=30^{\circ }$

   Why: Among the standard angles only $30^{\circ }$ has cosine $\frac{\sqrt{3}}{2}$, and the cosine is strictly decreasing on the first quadrant so there is no second acute possibility.

   Exact trigonometric ratios of the standard angles
2. 2
   $\sin (A+B)=\frac{\sqrt{3}}{2}\Rightarrow A+B=60^{\circ }\text{or}120^{\circ }$

   Why: Sine is not monotonic over $0^{\circ }$ to $180^{\circ }$, so it takes the value $\frac{\sqrt{3}}{2}$ twice. Both cases must be written down.

3. 3
   $A+B=120^{\circ }\Rightarrow A=90^{\circ },\text{not acute}\Rightarrow \text{rejected}$

   Why: The question states that $A$ is acute, which kills the second case. This is the step the distractors are built on.

   How the ratios move across the first quadrant
4. 4
   $A+B=60^{\circ }\Rightarrow A=30^{\circ }\Rightarrow 2A-B=60^{\circ }-30^{\circ }=30^{\circ }$

   Why: Substitute the two angles.

5. 5
   $\tan 30^{\circ }=\frac{1}{\sqrt{3}}$

   Why: Read off the standard value.

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Step-by-step solution

1. 1
   $\frac{\cos \theta }{1-\sin \theta }+\frac{\cos \theta }{1+\sin \theta }=\frac{\cos \theta \left[(1+\sin \theta )+(1-\sin \theta )\right]}{1-{\sin }^2\theta }$

   Why: Add over the common denominator $(1-\sin \theta )(1+\sin \theta )$, which is a difference of two squares.

   Difference of two squares
2. 2
   $=\frac{2\cos \theta }{{\cos }^2\theta }=\frac{2}{\cos \theta }=2\sec \theta$

   Why: $1-{\sin }^2\theta ={\cos }^2\theta$, and one factor of $\cos \theta$ cancels. Cancelling both — leaving $\frac{2}{{\cos }^2\theta }$ — is the standard slip here.

   Pythagorean trigonometric identities
3. 3
   $2\sec \theta =4\Rightarrow \cos \theta =\frac{1}{2}\Rightarrow \theta =60^{\circ }:\text{Statement 1 is correct}$

   Why: The cosine is strictly decreasing on $0^{\circ }<\theta <90^{\circ }$, so $60^{\circ }$ is the only acute solution.

   Reciprocal and quotient relations between the ratios
4. 4
   $3\tan \theta +\cot \theta =5\mathrm{cosec}\theta \Rightarrow 3{\sin }^2\theta +{\cos }^2\theta =5\cos \theta$

   Why: Write everything in sines and cosines and multiply through by $\sin \theta \cos \theta$, which is non-zero on the given range.

5. 5
   $3-2{\cos }^2\theta =5\cos \theta \Rightarrow 2{\cos }^2\theta +5\cos \theta -3=0$

   Why: Replace ${\sin }^2\theta$ by $1-{\cos }^2\theta$ to get a quadratic in $\cos \theta$ alone.

6. 6
   $(2\cos \theta -1)(\cos \theta +3)=0\Rightarrow \cos \theta =\frac{1}{2}\Rightarrow \theta =60^{\circ }:\text{Statement 2 is correct}$

   Why: Split the middle term: $-1\times 6=-6$ and $-1+6=5$. The root $\cos \theta =-3$ is impossible, since a cosine lies between $-1$ and $1$.

   Factorise a quadratic by splitting the middle term

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Step-by-step solution

1. 1
   $0\le {\cos }^2\theta \le 1$, so Statement 1 needs $0\le \frac{p^2+q^2}{2pq}\le 1$.

   Why: A squared cosine cannot leave $[0,1]$, so everything turns on where that fraction is allowed to sit.

   Sine and cosine lie between $-1$ and $1$
2. 2
   $\frac{p^2+q^2}{2pq}=\frac{1}{2}\left(\frac{p}{q}+\frac{q}{p}\right)$, which is $\ge 1$ when $pq>0$ and $\le -1$ when $pq<0$.

   Why: A quantity plus its reciprocal is at least $2$ for positives and at most $-2$ for negatives; halving gives $\pm 1$.

   A positive number plus its reciprocal is at least $2$
3. 3
   The only value in $[0,1]$ it can take is $1$, and $\frac{p^2+q^2}{2pq}=1$ means $(p-q{)}^2=0$, so $p=q$ and ${\cos }^2\theta =0$. Statement 1 is correct.

   Why: $p^2+q^2-2pq$ is $(p-q{)}^2$, so equality in the bound is exactly the condition $p=q$.

   Square of a sum and of a difference
4. 4
   ${\tan }^2\theta \ge 0$, so Statement 2 needs $\frac{4pq}{(p+q{)}^2}\ge 1$; but $(p+q{)}^2-4pq=(p-q{)}^2\ge 0$ forces $\frac{4pq}{(p+q{)}^2}\le 1$.

   Why: The same identity read the other way caps the second fraction at $1$ whenever $(p+q{)}^2>0$.

5. 5
   Equality again needs $p=q$, giving ${\tan }^2\theta =0$. Statement 2 is correct, so both statements hold.

   Why: Both statements therefore say the same thing: the expression is pinned at its only admissible value, and that pins $p=q$.

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Step-by-step solution

1. 1
   $\cos \theta +\sec \theta =x+\frac{1}{x}$ with $x=\cos \theta$ and $x\ne 0$.

   Why: Writing $\sec \theta$ as $\frac{1}{\cos \theta }$ turns Statement 1 into a question about one number and its reciprocal.

   Reciprocal and quotient relations between the ratios
2. 2
   For $x>0$, $x+\frac{1}{x}\ge 2$; for $x<0$, $x+\frac{1}{x}\le -2$. So the expression never lands in $(-2,2)$, and $1.5$ is impossible. Statement 1 is correct.

   Why: The whole interval between $-2$ and $2$ is unreachable, so no case work on $\theta$ is needed.

   A positive number plus its reciprocal is at least $2$
3. 3
   ${\sec }^2\theta +{\mathrm{cosec}}^2\theta =\frac{1}{{\cos }^2\theta }+\frac{1}{{\sin }^2\theta }=\frac{{\sin }^2\theta +{\cos }^2\theta }{{\sin }^2\theta {\cos }^2\theta }=\frac{1}{{\sin }^2\theta {\cos }^2\theta }$

   Why: Putting the two reciprocals over a common denominator lets the Pythagorean identity collapse the numerator to $1$.

   Pythagorean trigonometric identities
4. 4
   Write $s={\sin }^2\theta$, so the denominator is $s(1-s)$. The two factors have fixed sum $1$, so their product is largest at $s=\frac{1}{2}$, where it is $\frac{1}{4}$.

   Why: Maximising a product with a fixed sum is a standard move and avoids any calculus.

   A product with a fixed sum is largest when the factors are equal
5. 5
   Hence ${\sin }^2\theta {\cos }^2\theta \le \frac{1}{4}$ and ${\sec }^2\theta +{\mathrm{cosec}}^2\theta \ge 4$. Statement 2 is correct, so both statements hold.

   Why: A smaller denominator makes a bigger fraction, so the largest denominator gives the least value of the whole expression.

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Step-by-step solution

1. 1
   ${\sin }^{2}x+\sin x=1\Rightarrow \sin x=1-{\sin }^{2}x={\cos }^{2}x$

   Why: Isolating $\sin x$ and using ${\sin }^{2}x+{\cos }^{2}x=1$ replaces the condition by one clean relation between $\sin x$ and $\cos x$.

   Pythagorean trigonometric identities
2. 2
   ${\cos }^{12}x+3{\cos }^{10}x+3{\cos }^{8}x+{\cos }^{6}x={\cos }^{6}x\left({\cos }^{6}x+3{\cos }^{4}x+3{\cos }^{2}x+1\right)={\cos }^{6}x({\cos }^{2}x+1{)}^3$

   Why: The coefficients $1,3,3,1$ are the giveaway: with $t={\cos }^{2}x$ the bracket is $(t+1{)}^3$.

   Cube of a sum and of a difference
3. 3
   ${\cos }^{6}x({\cos }^{2}x+1{)}^3={\left[{\cos }^{2}x({\cos }^{2}x+1)\right]}^3={\left[\sin x(\sin x+1)\right]}^3$

   Why: Both factors are cubes, so they can be gathered under one cube before substituting ${\cos }^{2}x=\sin x$.

4. 4
   $\sin x(\sin x+1)={\sin }^{2}x+\sin x=1$, so the value is $1^3=1$.

   Why: The bracket is exactly the left-hand side of the given equation, which is why the expression was built this way.

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Step-by-step solution

1. 1
   $(3\sin \theta +5\cos \theta {)}^2+(3\cos \theta -5\sin \theta {)}^2=9{\sin }^2\theta +30\sin \theta \cos \theta +25{\cos }^2\theta +9{\cos }^2\theta -30\sin \theta \cos \theta +25{\sin }^2\theta$

   Why: The two cross terms are $+30\sin \theta \cos \theta$ and $-30\sin \theta \cos \theta$, which is exactly why pairing these two expressions is worth trying.

   Square of a sum and of a difference
2. 2
   $=9({\sin }^2\theta +{\cos }^2\theta )+25({\sin }^2\theta +{\cos }^2\theta )=9+25=34$

   Why: Whatever $\theta$ is, the sum of the two squares is the fixed number $3^2+5^2$.

   Pythagorean trigonometric identities
3. 3
   $(3\cos \theta -5\sin \theta {)}^2=34-(3\sin \theta +5\cos \theta {)}^2=34-4^2=18$

   Why: The given value $4$ is substituted only at the end, so no angle ever has to be found.

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Step-by-step solution

1. 1
   $16mn={\cot }^2\theta (1+\sin \theta )(1-\sin \theta )={\cot }^2\theta {\cos }^2\theta =\frac{{\cos }^4\theta }{{\sin }^2\theta }$

   Why: Multiplying is the natural first move because $(1+\sin \theta )(1-\sin \theta )=1-{\sin }^2\theta ={\cos }^2\theta$ clears the sine entirely.

   Difference of two squaresPythagorean trigonometric identitiesReciprocal and quotient relations between the ratios
2. 2
   $16(m^2-n^2)=(4m{)}^2-(4n{)}^2={\cot }^2\theta \left[(1+\sin \theta {)}^2-(1-\sin \theta {)}^2\right]=4\sin \theta {\cot }^2\theta$

   Why: The options are built from $m^2\pm n^2$, so squaring each given and taking the difference is the move that produces them.

3. 3
   $m^2-n^2=\frac{\sin \theta {\cot }^2\theta }{4}=\frac{{\cos }^2\theta }{4\sin \theta }$, so $(m^2-n^2{)}^2=\frac{{\cos }^4\theta }{16{\sin }^2\theta }$

   Why: Writing ${\cot }^2\theta$ as $\frac{{\cos }^2\theta }{{\sin }^2\theta }$ puts this in the same form as Step 1.

4. 4
   Step 1 gives $mn=\frac{{\cos }^4\theta }{16{\sin }^2\theta }$ as well, so $(m^2-n^2{)}^2=mn$.

   Why: The two sides are literally the same expression in $\theta$, so the relation holds for every $\theta$ where the givens make sense.

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Step-by-step solution

1. 1
   Perpendicular $=\sqrt{(u^2+v^2{)}^2-(u^2-v^2{)}^2}=\sqrt{4u^2{v}^2}=2uv$

   Why: The pair $(u^2-v^2,u^2+v^2)$ is the standard Pythagorean parametrisation, and the difference of two squares collapses at once.

   Pythagoras' theoremDifference of two squares
2. 2
   Area $=\frac{1}{2}(u^2-v^2)(2uv)=uv(u-v)(u+v)=2016$

   Why: The two legs are the base and the perpendicular, so half their product is the area.

   Area of a triangle from base and height
3. 3
   $2016=2^5\times 3^2\times 7$. Trying $u=9$ gives $9v(81-v^2)=2016$, i.e. $v(81-v^2)=224$, met by $v=7$ because $7\times 32=224$.

   Why: Factorising $2016$ first keeps the search short: $u$ must divide it, and $u=9$ is the first value that leaves an integer for $v$.

   Prime factorisation
4. 4
   Perimeter $=(u^2-v^2)+2uv+(u^2+v^2)=2u^2+2uv=2u(u+v)=2\times 9\times 16=288$ units

   Why: The three sides add in a form that factorises, so the perimeter needs no further arithmetic once $u$ and $v$ are known.

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Step-by-step solution

1. 1
   An equilateral triangle of side $l$ inscribed in a circle of radius $R$ has $l=R\sqrt{3}$, so its circumradius is $R=\frac{l}{\sqrt{3}}$.

   Why: The incircle and circumcircle of an equilateral triangle share a centre, so the circumradius is the quickest way in.

   Side of a regular polygon inscribed in a circle
2. 2
   That centre lies at distance $\frac{R}{2}$ from each side, so the inscribed circle has radius $r=\frac{R}{2}=\frac{l}{2\sqrt{3}}$.

   Why: The distance from the centre to a side is exactly what 'inscribed circle' means, and for an equilateral triangle it is half the circumradius.

3. 3
   A square inscribed in a circle of radius $r$ has its diagonal equal to the diameter $2r$, so its side is $\frac{2r}{\sqrt{2}}$ and its area is $\frac{(2r{)}^2}{2}=2r^2$.

   Why: Every square inscribed in that circle is the same square up to rotation, which is why the question can say 'any square'.

   Perimeter and area of a square
4. 4
   $2r^2=2\times \frac{l^2}{12}=\frac{l^2}{6}$

   Why: $r^2=\frac{l^2}{4\times 3}=\frac{l^2}{12}$, so the factor of $2$ leaves $\frac{l^2}{6}$.

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Step-by-step solution

1. 1
   Walls of one room $=2(l+b)h=2(6+4)(2.5)=50$ m${}^2$

   Why: Roofs and floors are excluded, so this is the four-walls figure and not the total surface.

   Volume and wall area of a cuboid
2. 2
   Five identical rooms: $5\times 50=250$ m${}^2$

   Why: The rooms are identical, so one calculation serves all five.

3. 3
   Two square windows of side $2.5$ m: $2\times (2.5{)}^2=12.5$ m${}^2$ is not painted, leaving $250-12.5=237.5$ m${}^2$.

   Why: Only two of the five rooms have a window, so the deduction is two windows and not five.

   Perimeter and area of a square
4. 4
   $\frac{237.5}{20}=11.875$ litres, and paint is sold only in whole $1$-litre cans, so $12$ cans are required.

   Why: The fraction has to be rounded up: $11$ cans would leave part of the wall bare.

   Unitary method: find one unit, then scale

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Step-by-step solution

1. 1
   Let $T=ABCD$ with $P,Q,R,S$ the midpoints of $AB,BC,CD,DA$. In $△ABC$ the midpoint theorem gives $PQ=\frac{1}{2}AC$, and in $△BCD$ it gives $QR=\frac{1}{2}BD$.

   Why: Each side of $S$ joins the midpoints of two sides of $T$, so each is half of one diagonal of $T$.

   Midpoint theorem
2. 2
   Corner triangle $△APS$ has sides $\frac{AB}{2}$ and $\frac{AD}{2}$ about $\angle A$, so its area is $\frac{1}{4}$ of $\frac{1}{2}AB\cdot AD\sin A$, which is $\frac{1}{8}$ of the area of $T$.

   Why: Halving both sides about the same angle quarters the triangle, and the full triangle $ABD$ is half the parallelogram.

   Area of a triangle from two sides and the included angle
3. 3
   The four corner triangles take $4\times \frac{1}{8}=\frac{1}{2}$ of $T$, so $S$ keeps the other half and $[T]:[S]=2:1$. Statement 1 is correct.

   Why: Opposite corners are congruent in pairs, so all four eighths are genuinely there.

4. 4
   Perimeter of $S=2\left(\frac{AC}{2}\right)+2\left(\frac{BD}{2}\right)=AC+BD$, the full sum of the diagonals of $T$, not half of it. Statement 2 is wrong.

   Why: Each diagonal is halved once but then counted twice, because $S$ has two sides of each length.

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Step-by-step solution

1. 1
   $s=\frac{5+6+7}{2}=9$

   Why: Heron's formula needs the semi-perimeter first, and here it is a whole number.

   Heron's formula
2. 2
   Area $=\sqrt{9(9-5)(9-6)(9-7)}=\sqrt{9\times 4\times 3\times 2}=\sqrt{216}$

   Why: The four factors are small, so the product can be formed before taking any root.

3. 3
   $\sqrt{216}=\sqrt{36\times 6}=6\sqrt{6}\approx 6\times 2.449=14.70$ cm${}^2$

   Why: Pulling out the perfect square $36$ leaves only $\sqrt{6}$ to estimate.

   $\sqrt{ab}=\sqrt{a}\sqrt{b}$ and $\sqrt{a/b}=\sqrt{a}/\sqrt{b}$

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Step-by-step solution

1. 1
   $\pi r^2=144\pi \Rightarrow r=12$ m

   Why: The $\pi$ cancels, so the radius comes straight out of $144$.

   Area and circumference of a circle
2. 2
   The path is $5$ m wide all round, so the outer radius is $12+5=17$ m.

   Why: A uniform path adds its width to the radius, not to the area.

3. 3
   Total area $=\pi (17{)}^2=289\pi$ m${}^2$

   Why: This is the plot plus the path, which is what the question asks for.

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