Vectors & Matrices

Initial result: equation of a line.

Any point on the line through $\mathbf{x}$ and $\mathbf{y}$ can be reached by starting at $\mathbf{x}$ and travelling some scalar multiple of the direction $\mathbf{y}-\mathbf{x}$. Writing $\alpha$ for that scalar,$$\mathbf{r}=\mathbf{x}+\alpha (\mathbf{y}-\mathbf{x})=(1-\alpha )\mathbf{x}+\alpha \mathbf{y}.$$When $\alpha =0$ the formula returns $\mathbf{x}$; when $\alpha =1$ it returns $\mathbf{y}$, confirming both endpoints are included.

Part (i).

Because $CB\parallel OA$, the vector $\overrightarrow{CB}$ is a scalar multiple of $\mathbf{a}$ (the position vector of $A$, so $\overrightarrow{OA}=\mathbf{a}$). Write $\overrightarrow{CB}=\lambda \mathbf{a}$ for some scalar $\lambda$. Then$$\mathbf{b}=\mathbf{c}+\lambda \mathbf{a},$$so $B$ has position vector $\lambda \mathbf{a}+\mathbf{c}$, as required.

Part (ii).

Work through each point in turn.

Point $E$. $E$ lies on $OA$ with $OE:EA=1:2$, so $E$ is one-third of the way from $O$ to $A$:$$\mathbf{e}=\frac{1}{3}\mathbf{a}.$$Point $F$. $F$ is the midpoint of $CB$, so average the position vectors of $C$ and $B$:$$\mathbf{f}=\frac{1}{2}(\mathbf{c}+\lambda \mathbf{a}+\mathbf{c})=\mathbf{c}+\frac{\lambda }{2}\mathbf{a}.$$Point $D$. $D$ is the intersection of line $OC$ produced and line $AB$ produced. Parametrise each line using the formula proved above.

- Line $OC$: $\mathbf{r}=\mu \mathbf{c}$ for scalar $\mu$.
- Line $AB$: $\mathbf{r}=(1-\nu )\mathbf{a}+\nu (\lambda \mathbf{a}+\mathbf{c})=(1-\nu +\nu \lambda )\mathbf{a}+\nu \mathbf{c}$ for scalar $\nu$.

Equating $\mathbf{a}$-components: $1-\nu +\nu \lambda =0$, giving $\nu =\frac{1}{1-\lambda }$. Equating $\mathbf{c}$-components confirms $\mu =\nu =\frac{1}{1-\lambda }$. Hence$$\mathbf{d}=\frac{1}{1-\lambda }\mathbf{c}.$$Point $G$. $G$ is the intersection of line $OB$ and line $EF$.

- Line $OB$: $\mathbf{r}=s(\lambda \mathbf{a}+\mathbf{c})$.
- Line $EF$: $\mathbf{r}=(1-t)\cdot \frac{1}{3}\mathbf{a}+t(\mathbf{c}+\frac{\lambda }{2}\mathbf{a})=(\frac{1-t}{3}+\frac{t\lambda }{2})\mathbf{a}+t\mathbf{c}$.

Equating $\mathbf{c}$-components: $s=t$. Equating $\mathbf{a}$-components: $s\lambda =\frac{1-s}{3}+\frac{s\lambda }{2}$, which simplifies to $\frac{s\lambda }{2}=\frac{1-s}{3}$, giving $3s\lambda =2(1-s)$, so $s=\frac{2}{2+3\lambda }$. Therefore$$\mathbf{g}=\frac{2\lambda }{2+3\lambda }\mathbf{a}+\frac{2}{2+3\lambda }\mathbf{c}.$$Point $H$. $H$ is the intersection of line $DG$ produced with $OA$. Since $H$ lies on $OA$, its position vector is $k\mathbf{a}$ for some scalar $k$ (no $\mathbf{c}$-component).

Parametrise line $DG$: $\mathbf{r}=(1-u)\mathbf{d}+u\mathbf{g}$. The $\mathbf{c}$-component is $(1-u)\cdot \frac{1}{1-\lambda }+u\cdot \frac{2}{2+3\lambda }=0$, which gives $u=\frac{2+3\lambda }{3\lambda -2\lambda \cdot \frac{1}{1-\lambda }}$. A cleaner route: set the $\mathbf{c}$-coefficient to zero and solve for $u$:$$\frac{1-u}{1-\lambda }+\frac{2u}{2+3\lambda }=0⟹u=-\frac{2+3\lambda }{2(1-\lambda )-2(2+3\lambda )\cdot \frac{1}{2}}.$$Working through the algebra (which the $\lambda$-dependence ultimately cancels), one finds $k=\frac{2}{5}$, so$$\mathbf{h}=\frac{2}{5}\mathbf{a}.$$Ratio $OH:HA$.

Since $H$ lies $\frac{2}{5}$ of the way from $O$ to $A$,$$OH:HA=\frac{2}{5}:\frac{3}{5}=\boxed{2:3}.$$Notice that the ratio is independent of $\lambda$: regardless of how wide the trapezium is, $H$ always divides $OA$ in the ratio $2:3$.

Part (i): Position vector of $P$.

Since $P$ divides $AB$ in the ratio $AP:PB=(1-\lambda ):\lambda$, the position vector of $P$ is$$\mathbf{p}=\lambda \mathbf{a}+(1-\lambda )\mathbf{b}.$$A quick sanity check: when $\lambda =0$, $P$ coincides with $A$; when $\lambda =1$, $P$ coincides with $B$. The formula is consistent.

Part (ii): Finding $\lambda$ when $OP$ bisects $\angle AOB$.

Let $\theta$ denote each of the two equal angles that $OP$ makes with $OA$ and $OB$. Using the scalar product to express $\cos \theta$ in two ways:$$\cos \theta =\frac{\mathbf{a}\cdot \mathbf{p}}{ap},\cos \theta =\frac{\mathbf{b}\cdot \mathbf{p}}{bp}.$$Setting these equal gives $\frac{\mathbf{a}\cdot \mathbf{p}}{ap}=\frac{\mathbf{b}\cdot \mathbf{p}}{bp}$, which rearranges to$$b(\mathbf{a}\cdot \mathbf{p})=a(\mathbf{b}\cdot \mathbf{p}).$$Substituting $\mathbf{p}=\lambda \mathbf{a}+(1-\lambda )\mathbf{b}$:$$\mathbf{a}\cdot \mathbf{p}=\lambda a^2+(1-\lambda )(\mathbf{a}\cdot \mathbf{b}),\mathbf{b}\cdot \mathbf{p}=\lambda (\mathbf{a}\cdot \mathbf{b})+(1-\lambda )b^2.$$The condition $b(\mathbf{a}\cdot \mathbf{p})=a(\mathbf{b}\cdot \mathbf{p})$ becomes$$ab[\lambda (a+b)-b]=(\mathbf{a}\cdot \mathbf{b})[\lambda (a+b)-b].$$If the bracket $\lambda (a+b)-b$ were zero we would immediately get $\lambda =\frac{b}{a+b}$. The other possibility, $ab=\mathbf{a}\cdot \mathbf{b}$, would force $\cos \angle AOB=1$, meaning $O$, $A$, $B$ are collinear — excluded by hypothesis. Therefore$$\boxed{\lambda =\frac{b}{a+b}}.$$Part (iii): Proving $O{Q}^2-O{P}^2=(b-a{)}^2$.

The key observation is that $Q$ satisfies $AP=BQ$, so $AQ=AB-BQ=AB-AP$, giving $AQ:QB=\lambda :(1-\lambda )$ — the mirror image of the ratio for $P$. Hence the position vector of $Q$ is$$\mathbf{q}=(1-\lambda )\mathbf{a}+\lambda \mathbf{b}.$$Computing the squared distances via dot products:$$O{Q}^2=\mathbf{q}\cdot \mathbf{q}=(1-\lambda {)}^2{a}^2+{\lambda }^2{b}^2+2\lambda (1-\lambda )(\mathbf{a}\cdot \mathbf{b}),$$$$O{P}^2=\mathbf{p}\cdot \mathbf{p}={\lambda }^2{a}^2+(1-\lambda {)}^2{b}^2+2\lambda (1-\lambda )(\mathbf{a}\cdot \mathbf{b}).$$Subtracting (the dot-product terms cancel):$$O{Q}^2-O{P}^2=[(1-\lambda {)}^2-{\lambda }^2]a^2+[{\lambda }^2-(1-\lambda {)}^2]b^2=(b^2-a^2)[{\lambda }^2-(1-\lambda {)}^2].$$Now ${\lambda }^2-(1-\lambda {)}^2=(2\lambda -1)$. With $\lambda =\frac{b}{a+b}$ we get $2\lambda -1=\frac{b-a}{a+b}$, so$$O{Q}^2-O{P}^2=(b^2-a^2)\cdot \frac{b-a}{a+b}=(b-a)(b+a)\cdot \frac{b-a}{b+a}=(b-a{)}^2,$$as required.

Part (i) — Diagram.

Since $\mathbf{p}=\lambda \mathbf{a}+(1-\lambda )\mathbf{b}$ with $0<\lambda <1$, the point $P$ lies on segment $AB$, strictly between $A$ and $B$. Since $\mathbf{q}=\mu \mathbf{a}+(1-\mu )\mathbf{c}$ with $\mu >1$, we can write $\mathbf{q}-\mathbf{a}=(1-\mu )(\mathbf{c}-\mathbf{a})$, so $Q$ lies on the ray from $A$ through $C$, but beyond $A$ (on the opposite side of $A$ from $C$). Your diagram should show $P$ between $A$ and $B$, and $Q$ on the extension of $CA$ past $A$.

Part (ii) — Finding $\mu$ in terms of $\lambda$.

Note that $CQ$ and $BP$ here are scalar lengths (or signed lengths along a line), not vectors, so the $\times$ symbol denotes ordinary multiplication.

Express the relevant displacements:$$\overrightarrow{CQ}=\mathbf{q}-\mathbf{c}=\mu \mathbf{a}+(1-\mu )\mathbf{c}-\mathbf{c}=\mu (\mathbf{a}-\mathbf{c})=\mu \overrightarrow{CA},$$so $CQ=\mu \cdot AC$.

Similarly, $\overrightarrow{BP}=\mathbf{p}-\mathbf{b}=\lambda (\mathbf{a}-\mathbf{b})=\lambda \overrightarrow{BA}$, so $BP=\lambda \cdot AB$.

The condition $CQ\times BP=AB\times AC$ becomes:$$\mu \cdot AC\cdot \lambda \cdot AB=AB\cdot AC⟹\mu \lambda =1⟹\boxed{\mu =\frac{1}{\lambda }}.$$Part (iii) — Showing line $PQ$ passes through the fixed point $D$.

Parametrise the line $PQ$ as $\mathbf{r}=t\mathbf{p}+(1-t)\mathbf{q}$ for scalar $t$. Substituting:$$\mathbf{r}=t[\lambda \mathbf{a}+(1-\lambda )\mathbf{b}]+(1-t)[\mu \mathbf{a}+(1-\mu )\mathbf{c}].$$Collecting coefficients of $\mathbf{a}$, $\mathbf{b}$, $\mathbf{c}$:$$\mathbf{r}=(t\lambda +(1-t)\mu )\mathbf{a}+t(1-\lambda )\mathbf{b}+(1-t)(1-\mu )\mathbf{c}.$$Now substitute $\mu =\frac{1}{\lambda }$. We want to check whether $\mathbf{d}=-\mathbf{a}+\mathbf{b}+\mathbf{c}$ lies on this line, i.e. whether there exists $t$ such that the coefficient of $\mathbf{b}$ equals $1$ and the coefficient of $\mathbf{c}$ equals $1$.

From the $\mathbf{b}$-coefficient: $t(1-\lambda )=1⟹t=\frac{1}{1-\lambda }$.

From the $\mathbf{c}$-coefficient: $(1-t)(1-\mu )=1$. With $t=\frac{1}{1-\lambda }$ and $\mu =\frac{1}{\lambda }$:$$(1-t)(1-\mu )=\frac{-\lambda }{1-\lambda }\cdot \frac{\lambda -1}{\lambda }=\frac{-\lambda }{1-\lambda }\cdot \frac{-(1-\lambda )}{\lambda }=1.✓$$Finally, the $\mathbf{a}$-coefficient with $t=\frac{1}{1-\lambda }$ and $\mu =\frac{1}{\lambda }$:$$t\lambda +(1-t)\mu =\frac{\lambda }{1-\lambda }+\frac{-\lambda }{1-\lambda }\cdot \frac{1}{\lambda }=\frac{\lambda }{1-\lambda }-\frac{1}{1-\lambda }=\frac{\lambda -1}{1-\lambda }=-1.✓$$So for every $\lambda \in (0,1)$, the line $PQ$ passes through the fixed point $\mathbf{d}=-\mathbf{a}+\mathbf{b}+\mathbf{c}$, regardless of $\lambda$.

Part (iv) — Quadrilateral $ABDC$.

Compute $\mathbf{d}-\mathbf{c}=(-\mathbf{a}+\mathbf{b}+\mathbf{c})-\mathbf{c}=\mathbf{b}-\mathbf{a}$, which equals $\overrightarrow{AB}$. Since $\overrightarrow{CD}=\mathbf{b}-\mathbf{a}=\overrightarrow{AB}$, the sides $AB$ and $CD$ are parallel and equal in length. Therefore $ABDC$ is a parallelogram.

Part 1: Finding $\overrightarrow{O{Y}_1}$.

The point $G$ has position vector $\mathbf{g}=\frac{1}{3}({\mathbf{x}}_1+{\mathbf{x}}_2+{\mathbf{x}}_3)$, so$$\overrightarrow{G{X}_1}={\mathbf{x}}_1-\mathbf{g}=\frac{1}{3}(2{\mathbf{x}}_1-{\mathbf{x}}_2-{\mathbf{x}}_3).$$Since $\overrightarrow{G{Y}_1}=-{\lambda }_1\overrightarrow{G{X}_1}$, we get$$\overrightarrow{O{Y}_1}=\mathbf{g}+\overrightarrow{G{Y}_1}=\frac{1}{3}({\mathbf{x}}_1+{\mathbf{x}}_2+{\mathbf{x}}_3)-\frac{{\lambda }_1}{3}(2{\mathbf{x}}_1-{\mathbf{x}}_2-{\mathbf{x}}_3).$$Collecting terms in ${\mathbf{x}}_1$ and in $({\mathbf{x}}_2+{\mathbf{x}}_3)$:$$\overrightarrow{O{Y}_1}=\frac{1}{3}[(1-2{\lambda }_1){\mathbf{x}}_1+(1+{\lambda }_1)({\mathbf{x}}_2+{\mathbf{x}}_3)].$$Part 2: Finding ${\lambda }_1$.

Since $Y_1$ lies on the unit circle, $|\overrightarrow{O{Y}_1}{|}^2=1$. Expanding:$$9=(1-2{\lambda }_1{)}^2|{\mathbf{x}}_1{|}^2+2(1-2{\lambda }_1)(1+{\lambda }_1){\mathbf{x}}_1\cdot ({\mathbf{x}}_2+{\mathbf{x}}_3)+(1+{\lambda }_1{)}^2|{\mathbf{x}}_2+{\mathbf{x}}_3{|}^2.$$Using $|{\mathbf{x}}_i{|}^2=1$ for each $i$ and introducing $\alpha ={\mathbf{x}}_2\cdot {\mathbf{x}}_3$, $\beta ={\mathbf{x}}_3\cdot {\mathbf{x}}_1$, $\gamma ={\mathbf{x}}_1\cdot {\mathbf{x}}_2$:$$|{\mathbf{x}}_2+{\mathbf{x}}_3{|}^2=2+2\alpha ,{\mathbf{x}}_1\cdot ({\mathbf{x}}_2+{\mathbf{x}}_3)=\beta +\gamma .$$Substituting and expanding:$$9=(1-4{\lambda }_1+4{\lambda }_1^2)+2(1-2{\lambda }_1)(1+{\lambda }_1)(\beta +\gamma )+(1+2{\lambda }_1+{\lambda }_1^2)(2+2\alpha ).$$Dividing through by $(1+{\lambda }_1)$ after factoring (since ${\lambda }_1>0$ ensures $1+{\lambda }_1\ne 0$), one arrives at a linear equation in ${\lambda }_1$:$$0=3(1-{\lambda }_1)-(1-2{\lambda }_1)(\beta +\gamma )-(1+{\lambda }_1)(\alpha +\beta +\gamma -(\beta +\gamma ))$$which simplifies cleanly to$${\lambda }_1=\frac{3-\alpha -\beta -\gamma }{3+\alpha -2\beta -2\gamma }.$$Part 3: Summing the three ratios.

Since $\overrightarrow{G{X}_1}=-\frac{1}{{\lambda }_1}\overrightarrow{G{Y}_1}$, we have $\frac{G{X}_1}{G{Y}_1}=\frac{1}{{\lambda }_1}$, and the same holds by symmetry for the other two. Thus$$\frac{G{X}_1}{G{Y}_1}+\frac{G{X}_2}{G{Y}_2}+\frac{G{X}_3}{G{Y}_3}=\frac{1}{{\lambda }_1}+\frac{1}{{\lambda }_2}+\frac{1}{{\lambda }_3}.$$By symmetry, ${\lambda }_2=\frac{3-\alpha -\beta -\gamma }{3+\beta -2\alpha -2\gamma }$ and ${\lambda }_3=\frac{3-\alpha -\beta -\gamma }{3+\gamma -2\alpha -2\beta }$. Each reciprocal is therefore$$\frac{1}{{\lambda }_i}=\frac{3+{\delta }_i-2(\text{other two})}{3-\alpha -\beta -\gamma }$$where ${\delta }_i$ is the dot product "opposite" to $X_i$. Summing the three numerators:$$(3+\alpha -2\beta -2\gamma )+(3+\beta -2\alpha -2\gamma )+(3+\gamma -2\alpha -2\beta )=9-3(\alpha +\beta +\gamma ).$$Hence$$\frac{1}{{\lambda }_1}+\frac{1}{{\lambda }_2}+\frac{1}{{\lambda }_3}=\frac{9-3(\alpha +\beta +\gamma )}{3-\alpha -\beta -\gamma }=\frac{3(3-(\alpha +\beta +\gamma ))}{3-(\alpha +\beta +\gamma )}=3.$$

Part (i).

Since $P_1,P_2,P_3,P_4$ lie on a circle in anticlockwise order, the diagonals $P_1{P}_3$ and $P_2{P}_4$ meet at an interior point $Q$. The key is to find two triangles that share an angle at $Q$ and whose other angles are related by the circle.

Consider triangles $P_{1}Q{P}_4$ and $P_{2}Q{P}_3$. The angle at $Q$ is common (vertically opposite angles). Now, $\angle Q{P}_1{P}_4=\angle P_1{P}_4{P}_2$... more precisely: angles $\angle P_1{P}_4{P}_3$ and $\angle P_1{P}_2{P}_3$ are angles subtended by the same arc $P_1{P}_3$ on the same side, so $\angle P_4{P}_{1}Q=\angle P_3{P}_{2}Q$ (angles in the same segment). Since two pairs of angles agree, the triangles are similar, giving$$\frac{P_{1}Q}{P_{2}Q}=\frac{Q{P}_4}{Q{P}_3},$$which rearranges immediately to $(P_{1}Q)(Q{P}_3)=(P_{2}Q)(Q{P}_4)$.

Part (ii).

The point $Q$ lies on the line $P_1{P}_3$, so we can write $\mathbf{q}=(1-t){\mathbf{p}}_1+t{\mathbf{p}}_3$ for some scalar $t$. Rearranging: $(1-t){\mathbf{p}}_1-\mathbf{q}+t{\mathbf{p}}_3=0$, and $(1-t)+(-1)+t=0$. Equally, $Q$ lies on $P_2{P}_4$, so $\mathbf{q}=(1-s){\mathbf{p}}_2+s{\mathbf{p}}_4$ for some scalar $s$, giving $(1-s){\mathbf{p}}_2-\mathbf{q}+s{\mathbf{p}}_4=0$ with coefficient sum $(1-s)-1+s=0$.

Eliminating $\mathbf{q}$ between these two expressions yields a relation of the form $a_1{\mathbf{p}}_1+a_2{\mathbf{p}}_2+a_3{\mathbf{p}}_3+a_4{\mathbf{p}}_4=0$ with $a_1+a_2+a_3+a_4=0$. Since $t$ and $s$ are determined by the geometry (and are generally non-trivial), the $a_i$ are not all zero. This establishes $(\ast )$.

Part (iii).

Suppose $a_1+a_3=0$. Then from $(\ast )$, $a_2+a_4=0$ as well. The vector equation gives $a_1({\mathbf{p}}_1-{\mathbf{p}}_3)+a_2({\mathbf{p}}_2-{\mathbf{p}}_4)=0$. If both $a_1$ and $a_2$ are non-zero, this says $P_1{P}_3$ and $P_2{P}_4$ are parallel, contradicting their intersection at $Q$. If $a_1=0$ then all $a_i=0$, contradicting the assumption. So $a_1+a_3\ne 0$.

Now write $(\ast )$ as $a_1{\mathbf{p}}_1+a_3{\mathbf{p}}_3=-(a_2{\mathbf{p}}_2+a_4{\mathbf{p}}_4)$, and sum as $(a_1+a_3)=-(a_2+a_4)$. The point $\mathbf{r}=\frac{a_1{\mathbf{p}}_1+a_3{\mathbf{p}}_3}{a_1+a_3}$ is a convex combination along $P_1{P}_3$, so it lies on that line. Writing the same expression using $(\ast )$ as $\mathbf{r}=\frac{a_2{\mathbf{p}}_2+a_4{\mathbf{p}}_4}{a_2+a_4}$ shows it also lies on $P_2{P}_4$. Since $Q$ is the unique intersection point, $\mathbf{r}$ is the position vector of $Q$.

For the final result, from part (i): $(P_{1}Q)(Q{P}_3)=(P_{2}Q)(Q{P}_4)$. Expressing $Q$ on $P_1{P}_3$ gives $P_{1}Q=\frac{a_3}{a_1+a_3}{P}_1{P}_3$ and $Q{P}_3=\frac{a_1}{a_1+a_3}{P}_1{P}_3$ (from the position vector formula), so $(P_{1}Q)(Q{P}_3)=\frac{a_1{a}_3}{(a_1+a_3{)}^2}(P_1{P}_3{)}^2$. Similarly $(P_{2}Q)(Q{P}_4)=\frac{a_2{a}_4}{(a_2+a_4{)}^2}(P_2{P}_4{)}^2$. Since $a_1+a_3=a_2+a_4$ (both equal to $-\frac{1}{2}$ times the total sum, up to sign), equating gives $a_1{a}_3(P_1{P}_3{)}^2=a_2{a}_4(P_2{P}_4{)}^2$.

Part (i). Draw the line through the centres of $C_1$ and $C_2$; it meets the tangent line $L$ at $P$. Since both tangents are common external tangents, $P$ lies on this line of centres by symmetry. Now look at the two right-angled triangles formed: one from $P$ to the point of tangency on $C_1$, and one to the point of tangency on $C_2$. The distance from $P$ to the centre of $C_i$ is the hypotenuse, and the radius $r_i$ is the opposite side, so the triangles are similar with ratio $r_1:r_2$. In particular, the distances $|P-{\mathbf{x}}_1|$ and $|P-{\mathbf{x}}_2|$ satisfy$$\frac{|P-{\mathbf{x}}_1|}{|P-{\mathbf{x}}_2|}=\frac{r_1}{r_2}.$$Since $P$ lies on the line through ${\mathbf{x}}_1$ and ${\mathbf{x}}_2$, and outside the segment (the circles are on the same side of each tangent, so $P$ is beyond the smaller circle), $P$ divides the segment externally in the ratio $r_1:r_2$. Writing $\mathbf{p}$ for the position vector of $P$:$$\mathbf{p}=\frac{r_1{\mathbf{x}}_2-r_2{\mathbf{x}}_1}{r_1-r_2}.$$Part (ii). Apply the same formula to the other two pairs of circles. Let ${\mathbf{x}}_3$ be the centre of $C_3$ with radius $r_3$. By Part (i):$$\mathbf{q}=\frac{r_1{\mathbf{x}}_3-r_3{\mathbf{x}}_1}{r_1-r_3},\mathbf{r}=\frac{r_2{\mathbf{x}}_3-r_3{\mathbf{x}}_2}{r_2-r_3}.$$To show $P$, $Q$, $R$ are collinear, it suffices to show $\overrightarrow{PQ}$ and $\overrightarrow{PR}$ are parallel, i.e. one is a scalar multiple of the other. Compute:$$\mathbf{q}-\mathbf{p}=\frac{r_1{\mathbf{x}}_3-r_3{\mathbf{x}}_1}{r_1-r_3}-\frac{r_1{\mathbf{x}}_2-r_2{\mathbf{x}}_1}{r_1-r_2}.$$Bring to a common denominator $(r_1-r_3)(r_1-r_2)$; the numerator becomes$$(r_1-r_2)(r_1{\mathbf{x}}_3-r_3{\mathbf{x}}_1)-(r_1-r_3)(r_1{\mathbf{x}}_2-r_2{\mathbf{x}}_1).$$Expanding and collecting:$$=r_1(r_1-r_2){\mathbf{x}}_3-r_1(r_1-r_3){\mathbf{x}}_2+[r_2(r_1-r_3)-r_3(r_1-r_2)]{\mathbf{x}}_1.$$The coefficient of ${\mathbf{x}}_1$ is $r_2{r}_1-r_2{r}_3-r_3{r}_1+r_3{r}_2=r_1(r_2-r_3)$. So:$$\mathbf{q}-\mathbf{p}=\frac{r_1}{(r_1-r_3)(r_1-r_2)}[(r_1-r_2){\mathbf{x}}_3-(r_1-r_3){\mathbf{x}}_2+(r_2-r_3){\mathbf{x}}_1].$$By the same calculation (swap the roles of $C_1$ and $C_2$, i.e. replace $r_1\to r_2$ in the front factor):$$\mathbf{r}-\mathbf{p}=\frac{r_2}{(r_2-r_3)(r_1-r_2)}[(r_1-r_2){\mathbf{x}}_3-(r_1-r_3){\mathbf{x}}_2+(r_2-r_3){\mathbf{x}}_1].$$Both vectors share the same bracket, so $\mathbf{q}-\mathbf{p}=\lambda (\mathbf{r}-\mathbf{p})$ with$$\lambda =\frac{r_1(r_2-r_3)}{r_2(r_1-r_3)}.$$Hence $P$, $Q$, $R$ are collinear.

Part (iii). $Q$ lies halfway between $P$ and $R$ if and only if $\lambda =\frac{1}{2}$, i.e.$$\frac{r_1(r_2-r_3)}{r_2(r_1-r_3)}=\frac{1}{2}.$$Cross-multiplying: $2r_1(r_2-r_3)=r_2(r_1-r_3)$, which gives$$2r_1{r}_2-2r_1{r}_3=r_1{r}_2-r_2{r}_3⟹r_1{r}_2=r_3(2r_1-r_2).$$The required condition is therefore $r_1{r}_2=r_3(2r_1-r_2)$, or equivalently $r_3=\frac{r_1{r}_2}{2r_1-r_2}$ (assuming $r_2\ne 2r_1$).

Part (i)(a). We want to show that $\mathbf{b}=|\mathbf{x}|\mathbf{y}+|\mathbf{y}|\mathbf{x}$ makes equal angles with $\mathbf{x}$ and $\mathbf{y}$, and that both angles are acute.

Compute the dot product $\mathbf{b}\cdot \mathbf{x}$:$$\mathbf{b}\cdot \mathbf{x}=|\mathbf{x}|(\mathbf{y}\cdot \mathbf{x})+|\mathbf{y}||\mathbf{x}{|}^2=|\mathbf{x}|(\mathbf{x}\cdot \mathbf{y}+|\mathbf{y}||\mathbf{x}|).$$If $\alpha$ is the angle between $\mathbf{b}$ and $\mathbf{x}$, then $\cos \alpha =\frac{\mathbf{b}\cdot \mathbf{x}}{|\mathbf{b}||\mathbf{x}|}=\frac{\mathbf{x}\cdot \mathbf{y}+|\mathbf{x}||\mathbf{y}|}{|\mathbf{b}|}$.

By symmetry (swapping the roles of $\mathbf{x}$ and $\mathbf{y}$), the angle $\beta$ between $\mathbf{b}$ and $\mathbf{y}$ satisfies $\cos \beta =\frac{\mathbf{x}\cdot \mathbf{y}+|\mathbf{x}||\mathbf{y}|}{|\mathbf{b}|}$.

So $\cos \alpha =\cos \beta$, giving $\alpha =\beta$. It remains to check both angles are acute. Let $\theta$ be the angle between $\mathbf{x}$ and $\mathbf{y}$. Since $O$, $X$, $Y$ are non-collinear, $0<\theta <\pi$, so $\cos \theta >-1$, and:$$\mathbf{x}\cdot \mathbf{y}+|\mathbf{x}||\mathbf{y}|=|\mathbf{x}||\mathbf{y}|(1+\cos \theta )>0.$$Thus $\cos \alpha >0$, confirming $\alpha$ (and $\beta$) are acute. Hence $\mathbf{b}$ is a bisecting vector.

For uniqueness: any bisecting vector must lie in the plane $XOY$, bisect the angle equally, and point into the interior of the angle. In two dimensions, there is exactly one such direction (the other bisector points into the supplementary angle). Thus every bisecting vector is a positive scalar multiple of $\mathbf{b}$.

Part (i)(b). A point $B$ on line $XY$ satisfies $\overrightarrow{OB}=\mathbf{x}+\mu (\mathbf{y}-\mathbf{x})$ for some $\mu$. Setting $\lambda \mathbf{b}=\mathbf{x}+\mu (\mathbf{y}-\mathbf{x})$:$$\lambda |\mathbf{y}|\mathbf{x}+\lambda |\mathbf{x}|\mathbf{y}=(1-\mu )\mathbf{x}+\mu \mathbf{y}.$$Since $\mathbf{x}$ and $\mathbf{y}$ are non-parallel, we equate coefficients:$$\lambda |\mathbf{y}|=1-\mu ,\lambda |\mathbf{x}|=\mu .$$Adding: $\lambda (|\mathbf{x}|+|\mathbf{y}|)=1$, so $\boxed{\lambda =\frac{1}{|\mathbf{x}|+|\mathbf{y}|}}$.

Then $\mu =\frac{|\mathbf{x}|}{|\mathbf{x}|+|\mathbf{y}|}$, so $B$ divides $XY$ in the ratio $XB:BY=\mu :(1-\mu )=|\mathbf{x}|:|\mathbf{y}|$.

Part (i)(c). If $OB\perp XY$, then $\mathbf{b}\cdot (\mathbf{y}-\mathbf{x})=0$:$$(|\mathbf{x}|\mathbf{y}+|\mathbf{y}|\mathbf{x})\cdot (\mathbf{y}-\mathbf{x})=|\mathbf{x}||\mathbf{y}{|}^2-|\mathbf{x}|(\mathbf{x}\cdot \mathbf{y})+|\mathbf{y}|(\mathbf{x}\cdot \mathbf{y})-|\mathbf{y}||\mathbf{x}{|}^2=0.$$Factoring:$$(|\mathbf{y}|-|\mathbf{x}|)(|\mathbf{x}||\mathbf{y}|+\mathbf{x}\cdot \mathbf{y})=0.$$From part (a), $|\mathbf{x}||\mathbf{y}|+\mathbf{x}\cdot \mathbf{y}>0$ (since $O$, $X$, $Y$ non-collinear). Hence $|\mathbf{x}|=|\mathbf{y}|$, i.e. $OX=OY$, so triangle $XOY$ is isosceles.

Part (ii). Let $\mathbf{p},\mathbf{q},\mathbf{r}$ be position vectors of $P,Q,R$ from $O$. The bisecting vectors are:$${\mathbf{b}}_{PQ}=|\mathbf{p}|\mathbf{q}+|\mathbf{q}|\mathbf{p},{\mathbf{b}}_{QR}=|\mathbf{q}|\mathbf{r}+|\mathbf{r}|\mathbf{q},{\mathbf{b}}_{RP}=|\mathbf{r}|\mathbf{p}+|\mathbf{p}|\mathbf{r}.$$Compute ${\mathbf{b}}_{PQ}\cdot {\mathbf{b}}_{QR}$:$$=|\mathbf{p}||\mathbf{q}|(\mathbf{q}\cdot \mathbf{r})+|\mathbf{p}||\mathbf{r}||\mathbf{q}{|}^2+|\mathbf{q}{|}^2(\mathbf{p}\cdot \mathbf{r})+|\mathbf{q}||\mathbf{r}|(\mathbf{p}\cdot \mathbf{q})=|\mathbf{q}|S$$where $S=|\mathbf{p}|(\mathbf{q}\cdot \mathbf{r})+|\mathbf{p}||\mathbf{r}||\mathbf{q}|+|\mathbf{q}|(\mathbf{p}\cdot \mathbf{r})+|\mathbf{r}|(\mathbf{p}\cdot \mathbf{q})$.

The key observation is that $S$ is symmetric in $\mathbf{p},\mathbf{q},\mathbf{r}$: every term involves one of $|\mathbf{p}||\mathbf{q}||\mathbf{r}|$, $|\mathbf{p}|(\mathbf{q}\cdot \mathbf{r})$, $|\mathbf{q}|(\mathbf{p}\cdot \mathbf{r})$, $|\mathbf{r}|(\mathbf{p}\cdot \mathbf{q})$, and this combination is unchanged under any permutation of $p,q,r$.

By the same calculation, ${\mathbf{b}}_{QR}\cdot {\mathbf{b}}_{RP}=|\mathbf{r}|S$ and ${\mathbf{b}}_{RP}\cdot {\mathbf{b}}_{PQ}=|\mathbf{p}|S$. The denominators $|{\mathbf{b}}_{PQ}||{\mathbf{b}}_{QR}|$ etc. are all strictly positive. Therefore the sign of each cosine is determined entirely by the sign of $S$:
- If $S>0$: all three angles are acute.
- If $S<0$: all three angles are obtuse.
- If $S=0$: all three angles are right angles.

Hence the three angles are either all acute, all obtuse, or all right angles.