Geometry Solutions

Part of the Oxford MAT Livestream.

 

Revision Questions

1. Something like

Three vectors pointing out from the origin. One is pointing roughly West and is large in magnitude than the other two, which are pointing roughly South-East and North-East.

2. We add the components separately, so $\displaystyle \binom{3}{2}+\binom{-4}{1}=\binom{-1}{3}$.

My diagram now looks like this.

The same image, with another vector marked; this one is roughly North-West. There's a dotted line connecting the tips of the northerly vectors; this dotted line segment has the same length and direction as the original West-pointing vector.

3. You multiply a vector by a scalar by multiplying each component, so $\displaystyle 3 \binom{-4}{1}=\binom{-12}{3}$ and $\displaystyle 2\binom{1}{-2}=\binom{2}{-4}$. Then add them together $\displaystyle \binom{-12}{3}+\binom{2}{-4}=\binom{-10}{-1}$.

My diagram now looks like this.

Same diagram again, but with one extra-long vector pointing West. There's a pair of dotted line segments to explain how to reach the very westerly tip of this vector  by traveling in the direction of the West-pointing vector  some distance and then travelling in the direction of the roughly-South-East-pointing vector for some distance.


4. This line has gradient $(-1-5)/(3-1)=-3$ and goes through $(1,5)$ so it's $y-5=-3(x-1)$ which can also be written as $y=8-3x$.

5. This must be $y=2x+c$ for some constant $c$, and the line goes through $(3,5)$ so $5=6+c$ and so the line is $y=2x-1$.

6. I might try to show that all the sides are the same length, and that all the corners are right angles. First I need to draw a diagram to get the points in the right order.

Four points arranged in a square. The square is at an angle; not enough that you'd call it a diamond, but it's close.

Now I can check that the distances from $(1,3)$ to $(3,4)$, from $(3,4)$ to $(2,6)$, from $(2,6)$ to $(0,5)$, and from $(0,5)$ to $(1,3)$ are all $\sqrt{5}$.

To check the corners are right angles, I could check that the gradients of the lines for each side multiply to $-1$. Those gradients are all either $\frac{1}{2}$ or $-2$, so all the corners are right angles.

7. There are lots of examples that work! I decided to use the $x$-axis as one of my lines (that's $y=0$), and then use something like $y=\sqrt{3}-ax$ and $y=\sqrt{3}+bx$ for some $a$ and $b$; I've chosen those $y$-intercepts so that $(0,\sqrt{3})$ is a corner of the triangle.

I need those two lines to go through $(\pm 1,0)$. I can do that by choosing $a$ and $b$ carefully, and I end up with the three lines $y=0$ and $y=\sqrt{3}(1-x)$ and $y=\sqrt{3}(1+x)$.

8. $(x+1)^2+(y-2)^2=3^2$

The area is $\pi r^2$ and $r=3$ so the area is $9\pi$.

The circle meets the $x$-axis where $(x+1)^2+(0-2)^2=3^2$. That's $x=-1\pm \sqrt{5}$.

The circle meets the $y$-axis where $(0+1)^2+(y-2)^2=3^2$. That's $y=2\pm\sqrt{8}$.

9. $x^2+9x+y^2-3y=\left(x+\frac{9}{2}\right)^2+\left(y-\frac{3}{2}\right)^2-\frac{81}{4}-\frac{9}{4}$. The equation of the circle is $\left(x+\frac{9}{2}\right)^2+\left(y-\frac{3}{2}\right)^2=10+\frac{90}{4}$. So the centre is $\left(-\frac{9}{2},\frac{3}{2}\right)$ and the radius is $\sqrt{\frac{65}{2}}$.

10. Draw a diagram.

A sector of a circle is marked out between two radii, from the origin to points A and B.

Since $120^\circ$ is one-third of $360^\circ$, the length of the arc is one-third of the length of the circumference $2\pi r$ with $r=2$. So the length of the arc is $\frac{4}{3}\pi$.    The area is one-third of $\pi r^2$, which works out to be $\frac{4}{3}\pi$.

11. Draw a diagram.

Two overlapping circles. A line connects the points of intersection. Two more line segments (radii) connect the  points of intersection to the centre of the left circle, forming an isosceles triangle if you include the line connecting the two points of intersection.


Find the points of intersection. Taking the difference between the two equations gives $x^2=(x-2)^2$, so $x=2-x$ or $x=x-2$, which only has $x=1$ as a solution. The $y$-coordinates are $\pm \sqrt{3}$, and the angle at the centre is $120^\circ$. Let's aim to find the area to the right of $x=1$ that's inside both circles. That's the area of the sector from the previous question, minus the area of a triangle. We can use $\frac{1}{2}ab \sin\theta$ to work out the area of the triangle, $\sqrt{3}$.

Then we'll need to double the area to get our final answer of $\frac{8}{3}\pi -2\sqrt{3}$.

12. Draw a diagram.

A circle with two chords marked, which meet at a point on one side of the circle. The angle where they meet is marked with a right angle.

We could write down equations for the distance of a general point $(x,y)$ to each of these points and set them equal to each other, but that's a lot of work.

Instead, note that the gradient of the line from $(0,0)$ to $(1,a)$ is $a$ and the gradient of the line from $(1,a)$ to $(0,a+a^{-1})$ is $-a^{-1}$. These gradients multiply to $-1$, so the lines are at right-angles.

The angle in a semi-circle is a right-angle, so the line from the first point to the third point is the diameter of the circle.

The centre is at the midpoint of the diameter, so it's at $(0,\frac{1}{2}\left(a+a^{-1}\right))$.

13. The area $A(c)$ is zero if $c<-1$ and it's $\pi$ if $c>1$. In between, the area rises from 0 to $\pi$ in a nice symmetric manner; slow then fast then slow.

Graph of a function is zero for a long time, then increases to pi (slowly then quickly then slowly), then at some point it's pi forever after. The graph has rotational symmetry of order 2 about the point (0,pi/2).

 

MAT Questions

MAT 2016 Q1C

  • The equation can be written as $$\left(x+\frac{a}{2}\right)+\left(y+\frac{a}{2}\right)=c+\frac{a^2}{4}+\frac{b^2}{4}.$$
  • The centre is at $\left(-\frac{a}{2},-\frac{b}{2}\right)$ and the radius is $\sqrt{c+\frac{a^2}{4}+\frac{b^2}{4}}$.
  • The origin is in the circle if the distance to the centre is less than the radius
  • This gives the inequality $\sqrt{\frac{a^2}{4}+\frac{b^2}{4}}<\sqrt{c+\frac{a^2}{4}+\frac{b^2}{4}}$. Both sides are positive, so we want $\frac{a^2}{4}+\frac{b^2}{4}<c+\frac{a^2}{4}+\frac{b^2}{4}$, which happens only if $c>0$.
  • Alternatively, substitute in $x=0$ and $y=0$ to discover that the origin lies on the boundary of the circle when $c=0$. The origin is outside the circle if $c$ is very small, so the answer must be $c>0$.
  • The answer is (a).

MAT 2017 Q1G

  • No matter the value of $\theta$, the point $(-1,1)$ is always on the line. As $\theta$ changes, the line rotates around, such that the line makes an angle of $\theta$ with the positive $x$-axis.
  • The "larger" region is maximised if the line is at right angles to the radius between $(0,0)$ and $(-1,1)$.
  • This happens for two values of $\theta$, $135^\circ$ and $315^\circ$. (there are two values of $\theta$ corresponding to the same line, because replacing $\theta$ with $\theta+180^\circ$ introduces a minus sign on both sides of the equation).
  • The answer is (b).

MAT 2014 Q1D

  • The line perpendicular to $y=mx$ that passes through $(1,0)$ is $y=-\frac{1}{m}(x-1)$.
  • This line meets $y=mx$ when when $\displaystyle x=\frac{1}{1+m^2}$ and so $\displaystyle y=\frac{m}{1+m^2}$.
  • We want the point on the "other side" of the line. Moving an equal distance will change the $y$-coordinate by the same amount again, that is, doubling it. So the point we're looking for has $\displaystyle y=\frac{2m}{1+m^2}$.
  • The answer is (d).

MAT 2008 Q4

(i) Let's write down $(x-a)^2+(y-b)^2=r^2$ for the equation of the circle. The three points $(0,0)$ and $(p,0)$ and $(0,q)$ all lie on the circle, which gives three equations;
\begin{equation*}
(0-a)^2+(0-b)^2=r^2, \qquad (p-a)^2+(0-b)^2=r^2, \qquad (0-a)^2+(q-b)^2=r^2.
\end{equation*}
We can simplify these a bit, and multiply out some squares, and take the difference between pairs of equations, to get
\begin{equation*}
p^2-2pa=0,\qquad q^2-2qb=0,\qquad a^2+b^2=r^2
\end{equation*}
Now $p\neq 0$ and $q\neq 0$ so $a=\frac{p}{2}$ and $b=\frac{q}{2}$, and $\displaystyle r=\sqrt{\frac{p^2+q^2}{4}}$.

If we substitute those numbers into the equation $(x-a)^2+(y-b)^2=r^2$ and multiply out the squares, we get the equation in the question.

Along the way, we found the centre $(a,b)=\left(\frac{p}{2},\frac{q}{2}\right)$ and we found the radius. The area of $C$ is $\displaystyle \pi\frac{p^2+q^2}{4}$.

(ii) We just found the area of the circle. The area of the triangle is $\frac{1}{2}pq$. If we write out the inequality in the question, we see that we're being asked to prove that
$$
\pi\frac{p^2+q^2}{2pq}\geq \pi
$$
for all positive real numbers $p$ and $q$. This is true because $(p-q)^2\geq 0$, and that rearranges to the inequality above (we can divide by $pq$ because $p$ and $q$ are not zero). If I'm honest, I rearranged the equation first, factorised it as $(p-q)^2$, realised that was positive or zero, then presented all of that to you in the opposite order. Sometimes it's a good idea to work backwards as well as forwards... provided that your final argument makes sense, of course.

(iii) Now we're asked to solve
$$
\pi\frac{p^2+q^2}{2pq}=2\pi
$$
which rearranges to $p^2+q^2=4pq$. This is one equation for two variables, so we can't really solve it for $p$ and $q$. But we just want expressions for the angles. From trigonometry, we know that $q/p$ is $\tan\angle OPQ$, and something similar is true for $\tan \angle OQP$. That inspires me to divide the equation by $p^2$ and solve
\begin{equation*}
1+\left(\frac{q}{p}\right)^2=4\left(\frac{q}{p}\right)
\end{equation*}
which is just a quadratic equation. The roots are $2\pm\sqrt{3}$ so the angles are $\tan^{-1} \left(2\pm \sqrt{3}\right)$.

 

Extension

  • If $\tan\theta=2-\sqrt{3}$ then $$\tan 2\theta = \frac{4-2\sqrt{3}}{1-\left(4-4\sqrt{3}+3\right)}=\frac{4-2\sqrt{3}}{4\sqrt{3}-6}=\frac{1}{\sqrt{3}}$$
    so $2\theta$ is $30^\circ$ (restricting to the range $0\leq \theta\leq 180^\circ$). So $\theta$ must be $15^\circ$.

    If $\tan\theta=2+\sqrt{3}$ then $$\tan 2\theta = \frac{4+2\sqrt{3}}{1-\left(4+4\sqrt{3}+3\right)}=\frac{4+2\sqrt{3}}{-4\sqrt{3}-6}=-\frac{1}{\sqrt{3}}$$
    so $2\theta$ is $150^\circ$ (restricting to the range $0\leq \theta\leq 180^\circ$). So $\theta$ must be $75^\circ$.
  • Differentiate to get $1-x^{-2}$ which is zero at $x=\pm 1$. The minimum value of the function occurs when $x=1$, and the value is $2$.

    Alternatively, use the fact that $a^2+b^2\geq 2ab$ with $a=x^{1/2}$ and $b=x^{-1/2}$. Then $x+x^{-1}\geq 2x^{1/2}x^{-1/2}=2$.

 

MAT 2010 Q4

(i) If I drop a perpendicular from $(1,2h)$ to $(1,0)$ then I have a right-angled triangle with angle $\theta$, opposite a side of length $2h$ and adjacent to a side of length $2h$. So $\tan\theta=2h$.

(ii) $(1,2h)$ lies in $x^2+y^2<4$ if and only if $1+4h^2<4$, which rearranges to $h^2<\frac{3}{4}$. Since $h>0$, this condition is equivalent to $h<\sqrt{3}/2$.

(iii) The gradient of the line is $-h$ because the $y$-value changes by $-2h$ between $x=1$ and $x=3$. The line goes through $(3,0)$ and has equation $y=-h(x-3)$.

We could look for repeated roots between $x^2+y^2=4$ and $y=-h(x-3)$. In general if I substitute one into the other I get $x^2+h^2(x-3)^2=4$. If $h=2/\sqrt{5}$ then this is $5x^2+4(x-3)^2=20$. If we multiply this out, rearrange, and factorise, we get $(3x-4)^2=0$ indicating that there is a double root at $x=4/3$, so the line is tangent to the circle.

(iv) In this case, the point $(1,2h)$ is outside the circle, because $\frac{2}{\sqrt{5}}>\frac{\sqrt{3}}{2}$ (I know this because $\frac{4}{5}>\frac{3}{4}$). The diagram is like the first picture in the question. The area inside both is the area of the sector with angle $\theta$. So the area is $\displaystyle 4\pi \frac{\theta}{360^\circ}$ where $\tan\theta=2h$.

(v) In this case, the point $(1,2h)$ is inside the circle, because $\frac{6}{7}<\frac{\sqrt{3}}{2}$ (I know this because $\frac{36}{49}<\frac{3}{4}$ (I know this because $144<147$)). The diagram is like the second picture in the question.

Check that $(8/5,6/5)$ lies on the circle; $(8/5)^2+(6/5)^2=(64/25)+(36/25)=4$. Check that $(8/5,6/5)$ lies on the line; $-\frac{6}{7}\left(\frac{8}{5} - 3\right)=\frac{6}{5}$.

The area is made up of a triangle with corners $(0,0)$ and  $(1,12/7)$ and $(8/5,6/5)$, plus a sector of a circle from the $x$-axis up to the line from the origin to $(8/5,6/5)$.

Using the formula in the question, the area of the triangle is $\displaystyle \frac{27}{35}$. The area of the sector is $\displaystyle 4\pi\frac{\alpha}{360^\circ}$ where $\displaystyle \tan\alpha=\frac{6}{8}$. Add these.

 

Extension

  • Let $A$ be $(a,b)$ and let $C$ be $(c,d)$ and let $M$ be $(a,0)$ and let $N$ be $(c,0)$. The area we want is the triangle $OCN$ plus the trapezium $NCAM$ minus the triangle $OAM$. If we find all of these and simplify, we get $\frac{1}{2}\left(ad-bc\right)$. The absolute value signs come in because I haven't been very careful; I've assumed that $(a,b)$ and $(c,d)$ are a particular way around.
  • Consider the cross product of the vectors $(a,b,0)\times (c,d,0)=(0,0,ad-bc)$. We also know that the magnitude of $\mathbf{a}\times \mathbf{b}$ is $|\mathbf{a}| |\mathbf{b}| \sin\theta$, which is (almost) exactly the formula for the area of the triangle. So we just need to take the magnitude of $\mathbf{a}\times \mathbf{b}$ and divide by 2 to get the area of the triangle.
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