2019 Canadian Senior
Mathematics Contest
Solutions

Wednesday, November 20, 2019
(in North America and South America)

Thursday, November 21, 2019
(outside of North American and South America)

©2019 University of Waterloo

PART A

1. Since Zipporah is 7 years old and the sum of Zipporah’s age and Dina’s age is 51, then Dina is $51-7=44$ years old.
   Since Dina is 44 years old and the sum of Julio’s age and Dina’s age is 54, then Julio is $54-44=10$ years old.

   Answer: 10

2. Since the circular track has radius 60 m, its circumference is $2\pi \cdot 60\text{ m}$ which equals $120\pi \text{ m}$.
   Since Ali runs around this track at a constant speed of 6 m/s, then it takes Ali ${\displaystyle \frac{120\pi \text{ m}}{6\text{ m/s}}}=20\pi \text{ s}$ to complete one lap.
   Since Ali and Darius each complete one lap in the same period of time, then Darius also takes $20\pi \text{ s}$ to complete one lap.
   Since Darius runs at a constant speed of 5 m/s, then the length of his track is $20\pi \text{ s}\cdot 5\text{ m/s}$ or $100\pi \text{ m}$.
   Since Darius’s track is in the shape of an equilateral triangle with side length $x$ m, then its perimeter is $3x\text{ m}$ and so $3x\text{ m}=100\pi \text{ m}$ and so $x=\frac{100\pi }{3}$.

   Answer: $x=\frac{100\pi }{3}$

3. Since $2^a\cdot 2^b=2^{a+b}$, then $$\begin{array}{rl}{32}^n& =2^{200}\cdot 2^{203}+2^{163}\cdot 2^{241}+2^{126}\cdot 2^{277}\\ & =2^{200+203}+2^{163+241}+2^{126+277}\\ & =2^{403}+2^{404}+2^{403}\\ & =2^{403}+2^{403}+2^{404}\end{array}$$ Since $2^c+2^c=2(2^c)=2^1\cdot 2^c=2^{c+1}$, then $$\begin{array}{rl}{32}^n& =2^{403+1}+2^{404}\\ & =2^{404}+2^{404}\\ & =2^{404+1}\\ & =2^{405}\end{array}$$ Since $(2^d{)}^e=2^{de}$, then ${32}^n=(2^5{)}^n=2^{5n}$.
   Since ${32}^n=2^{405}$, then $2^{5n}=2^{405}$ which means that $5n=405$ and so $n=81$.

   Answer: $n=81$

4. For there to exist a pair of integers $(x,y)$ with $x^2\le y\le x+6$, it must be the case that $x^2\le x+6$ and so $x^2-x-6\le 0$.
   Now $x^2-x-6=(x-3)(x+2)$, so $x^2-x-6\le 0$ exactly when $-2\le x\le 3$. (If we consider the function $f(x)=(x-3)(x+2)$, whose graph is a parabola opening upwards, its values are less than or equal to $0$ between its roots.)
   Therefore, any pair of integers $(x,y)$ with $x^2\le y\le x+6$ must have $x$ equal to one of $-2,-1,0,1,2,3$.
   When $x=-2$, the original inequality becomes $4\le y\le 4$ and so $y$ must equal 4. There is 1 pair in this case, namely $(-2,4)$.
   When $x=-1$, we obtain $1\le y\le 5$ and so $y$ must equal one of $1,2,3,4,5$. There are 5 pairs in this case.
   When $x=0$, we obtain $0\le y\le 6$ and so $y$ must equal one of $0,1,2,3,4,5,6$. There are 7 pairs in this case.
   When $x=1$, we obtain $1\le y\le 7$. There are 7 pairs in this case.
   When $x=2$, we obtain $4\le y\le 8$. There are 5 pairs in this case.
   When $x=3$, we obtain $9\le y\le 9$ and so $y$ must equal $9$. There is 1 pair in this case.
   In total, there are $1+5+7+7+5+1=26$ pairs of integers that satisfy the inequality.

   Answer: 26

5. Since 605 is the middle side length of the right-angled triangle, we suppose that the side lengths of the triangle are $a,605,c$ for integers $a<605<c$. (Why do we not need to consider the cases $a=605$ or $605=c$?)
   By the Pythagorean Theorem, knowing that $c$ (the longest side length) must be the length of the hypotenuse, we obtain $a^2+{605}^2=c^2$ and so $c^2-a^2={605}^2$.
   We want to determine the maximum possible length of the shortest side of the triangle.
   In other words, we want to try to determine the maximum possible length of $a$ which is less than 605.
   We note that $c^2-a^2={605}^2$ exactly when $(c+a)(c-a)={605}^2$.
   We note also that $605=5\cdot 121=5\cdot {11}^2$ and so ${605}^2=5^2\cdot {11}^4$.
   Therefore, we have $(c+a)(c-a)=5^2\cdot {11}^4$. This means that $c+a$ and $c-a$ are a divisor pair of $5^2\cdot {11}^4$.
   Since $a$ and $c$ are positive integers, then $c+a>c-a$. Note that $c>a$ and so $c+a>c-a>0$.
   We make a table of the possible values for $c+a$ and $c-a$, and use these to determine the possible values of $c$ and $a$.

   $c+a$	$c-a$	$2c=(c+a)+(c-a)$	$c$	$a=(c+a)-c$
   $5^2\cdot {11}^4=366025$	$1$	$366026$	$183013$	$103012$
   $5\cdot {11}^4=73205$	$5$	$73210$	$36605$	$36600$
   $5^2\cdot {11}^3=33275$	$11$	$33286$	$16643$	$16632$
   ${11}^4=14641$	$5^2=25$	$14666$	$7333$	$7308$
   $5\cdot {11}^3=6655$	$5\cdot 11=55$	$6710$	$3355$	$3300$
   $5^2\cdot {11}^2=3025$	${11}^2=121$	$3146$	$1573$	$1452$
   ${11}^3=1331$	$5^2\cdot 11=275$	$1606$	$803$	$528$
   $5\cdot {11}^2=605$	$5\cdot {11}^2=605$	$1210$	$605$	$0$

   These are all of the possible factorizations of ${605}^2$, and so give all of the possible pairs $(a,c)$ that satisfy the equation.
   Therefore, the maximum possible value of $a$ that is less than $605$ is 528.

   Answer: 528

6. Since square $ABCD$ has side length 4, then its area is $4^2$, which equals $16$.
   The area of quadrilateral $PQRS$, which we expect to be a function of $k$, equals the area of square $ABCD$ minus the combined areas of $\mathrm{△}ABP$, $\mathrm{△}PCQ$, $\mathrm{△}QDR$, and $\mathrm{△}ARS$.
   Since ${\displaystyle \frac{BP}{PC}}={\displaystyle \frac{k}{4-k}}$, then there is a real number $x$ with $BP=kx$ and $PC=(4-k)x$.
   Since $BP+PC=BC=4$, then $kx+(4-k)x=4$ and so $4x=4$ or $x=1$.
   Thus, $BP=k$ and $PC=4-k$.
   Similarly, $CQ=DR=k$ and $QD=RA=4-k$.
   $\mathrm{△}ABP$ is right-angled at $B$ and so its area is $\frac{1}{2}(AB)(BP)=\frac{1}{2}(4k)=2k$.
   $\mathrm{△}PCQ$ is right-angled at $C$ and so its area is $\frac{1}{2}(PC)(CQ)=\frac{1}{2}(4-k)k$.
   $\mathrm{△}QDR$ is right-angled at $D$ and so its area is $\frac{1}{2}(QD)(DR)=\frac{1}{2}(4-k)k$.
   To find the area of $\mathrm{△}ARS$, we first join $R$ to $P$.

   

   Now $\mathrm{△}ARP$ can be seen as having base $RA=4-k$ and perpendicular height equal to the distance between the parallel lines $CB$ and $DA$, which equals $4$.
   Thus, the area of $\mathrm{△}ARP$ is $\frac{1}{2}(4-k)(4)$.
   Now we consider $\mathrm{△}ARP$ as having base $AP$ divided by point $S$ in the ratio $k:(4-k)$.
   This means that the ratio of $AS:AP$ equals $k:((4-k)+k)$ which equals $k:4$.
   This means that the area of $\mathrm{△}ARS$ is equal to ${\displaystyle \frac{k}{4}}$ times the area of $\mathrm{△}ARP$(The two triangles have the same height – the distance from $R$ to $AP$ – and so the ratio of their areas equals the ratio of their bases.)
   Thus, the area of $\mathrm{△}ARS$ equals ${\displaystyle \frac{\frac{1}{2}(4-k)(4)\cdot k}{4}}=\frac{1}{2}k(4-k)$.
   Thus, the area of quadrilateral $PQRS$ is $$\begin{array}{rl}16-2k-3\cdot \frac{1}{2}k(4-k)& =16-2k-\frac{3}{2}\cdot 4k+\frac{3}{2}{k}^2\\ & =\frac{3}{2}{k}^2-2k-6k+16\\ & =\frac{3}{2}{k}^2-8k+16\end{array}$$ The minimum value of the quadratic function $f(t)=a{t}^2+bt+c$ with $a>0$ occurs when $t=-{\displaystyle \frac{b}{2a}}$ and so the minimum value of $\frac{3}{2}{k}^2-8k+16$ occurs when $k=-\frac{-8}{2(3/2)}=\frac{8}{3}$.
   Therefore, the area of quadrilateral $PQRS$ is minimized when $k=\frac{8}{3}$.

   Answer: $k=\frac{8}{3}$

PART B

1. 1. Since each of Rachel’s jumps is 168 cm long, then when Rachel completes 5 jumps, she jumps $5\times 168\text{ cm}=840\text{ cm}$.
      Since each of Joel’s jumps is 120 cm long, then when Joel completes $n$ jumps, he jumps $120n\text{ cm}$.
      Since Rachel and Joel jump the same total distance, then $120n=840$ and so $n=7$.

   2. Since each of Joel’s jumps is 120 cm long, then when Joel completes $r$ jumps, he jumps $120r\text{ cm}$.
      Since each of Mark’s jumps is 72 cm long, then when Mark completes $t$ jumps, he jumps $72t\text{ cm}$.
      Since Joel and Mark jump the same total distance, then $120r=72t$ and so dividing by 24, $5r=3t$.
      Since $5r$ is a multiple of 5, then $3t$ must also be a multiple of 5, which means that $t$ is a multiple of 5.
      Since $11\le t\le 19$ and $t$ is a multiple of 5, then $t=15$.
      Since $t=15$, then $5r=3\cdot 15=45$ and so $r=9$.
      Therefore, $r=9$ and $t=15$.

   3. When Rachel completes $a$ jumps, she jumps $168a$ cm.
      When Joel completes $b$ jumps, he jumps $120b$ cm.
      When Mark completes $c$ jumps, he jumps $72c$ cm.
      Since Rachel, Joel and Mark all jump the same total distance, then $168a=120b=72c$.
      Dividing by 24, we obtain $7a=5b=3c$.
      Since $7a$ is divisible by 7, then $3c$ is divisible by 7, which means that $c$ is divisible by 7.
      Since $5b$ is divisible by 5, then $3c$ is divisible by 5, which means that $c$ is divisible by 5.
      Since $c$ is divisible by 5 and by 7 and because 5 and 7 have no common divisor larger than 1, then $c$ must be divisible by $5\cdot 7$ which equals 35.
      Since $c$ is divisible by 35 and $c$ is a positive integer, then $c\ge 35$.
      We note that if $c=35$, then $3c=105$ and since $7a=5b=105$, we obtain $a=15$ and $b=21$. In other words, $c=35$ is possible.
      Therefore, the minimum possible value of $c$ is $c=35$.
      

2. 1. For the sequence ${\displaystyle \frac{1}{w}},{\displaystyle \frac{1}{2}},{\displaystyle \frac{1}{3}},{\displaystyle \frac{1}{6}}$ to be an arithmetic sequence, it must be the case that $${\displaystyle \frac{1}{2}}-{\displaystyle \frac{1}{w}}={\displaystyle \frac{1}{3}}-{\displaystyle \frac{1}{2}}={\displaystyle \frac{1}{6}}-{\displaystyle \frac{1}{3}}$$ Since ${\displaystyle \frac{1}{3}}-{\displaystyle \frac{1}{2}}={\displaystyle \frac{1}{6}}-{\displaystyle \frac{1}{3}}=-{\displaystyle \frac{1}{6}}$, then ${\displaystyle \frac{1}{2}}-{\displaystyle \frac{1}{w}}=-{\displaystyle \frac{1}{6}}$ and so ${\displaystyle \frac{1}{w}}={\displaystyle \frac{1}{2}}+{\displaystyle \frac{1}{6}}={\displaystyle \frac{2}{3}}$, which gives $w={\displaystyle \frac{3}{2}}$.

   2. The sequence ${\displaystyle \frac{1}{y+1}},x,{\displaystyle \frac{1}{z+1}}$ is arithmetic exactly when $x-{\displaystyle \frac{1}{y+1}}={\displaystyle \frac{1}{z+1}}-x$ or $2x={\displaystyle \frac{1}{y+1}}+{\displaystyle \frac{1}{z+1}}$.
      Since $y,1,z$ is a geometric sequence, then ${\displaystyle \frac{1}{y}}={\displaystyle \frac{z}{1}}$ and so $z={\displaystyle \frac{1}{y}}$. Since $y$ and $z$ are positive, then $y\ne -1$ and $z\ne -1$.
      In this case, ${\displaystyle \frac{1}{y+1}}+{\displaystyle \frac{1}{z+1}}={\displaystyle \frac{1}{y+1}}+{\displaystyle \frac{1}{{\displaystyle \frac{1}{y}}+1}}={\displaystyle \frac{1}{y+1}}+{\displaystyle \frac{y}{1+y}}={\displaystyle \frac{y+1}{y+1}}=1$.
      Since ${\displaystyle \frac{1}{y+1}}+{\displaystyle \frac{1}{z+1}}=1$, then the sequence ${\displaystyle \frac{1}{y+1}},x,{\displaystyle \frac{1}{z+1}}$ is arithmetic exactly when $2x=1$ or $x={\displaystyle \frac{1}{2}}$.

   3. Since $a,b,c,d$ is a geometric sequence, then $b=ar$, $c=a{r}^2$ and $d=a{r}^3$ for some real number $r$. Since $a\ne b$, then $a\ne 0$. (If $a=0$, then $b=0$.)
      Since $a\ne b$, then $r\ne 1$. Note that ${\displaystyle \frac{b}{a}}={\displaystyle \frac{ar}{a}}=r$ and so we want to determine all possible values of $r$.
      Since $a$ and $b$ are both positive, then $r>0$.
      Since ${\displaystyle \frac{1}{a}},{\displaystyle \frac{1}{b}},{\displaystyle \frac{1}{d}}$ is an arithmetic sequence, then $$\begin{array}{rl}{\displaystyle \frac{1}{b}}-{\displaystyle \frac{1}{a}}& ={\displaystyle \frac{1}{d}}-{\displaystyle \frac{1}{b}}\\ {\displaystyle \frac{1}{ar}}-{\displaystyle \frac{1}{a}}& ={\displaystyle \frac{1}{a{r}^3}}-{\displaystyle \frac{1}{ar}}\\ {\displaystyle \frac{1}{r}}-1& ={\displaystyle \frac{1}{r^3}}-{\displaystyle \frac{1}{r}}\text{(since }a\ne 0\text{)}\\ r^2-r^3& =1-r^2\\ 0& =r^3-2r^2+1\\ 0& =(r-1)(r^2-r-1)\end{array}$$ Since $r\ne 1$, then $r^2-r-1=0$.
      By the quadratic formula, $r={\displaystyle \frac{1\pm \sqrt{(-1{)}^2-4(1)(-1)}}{2}}={\displaystyle \frac{1\pm \sqrt{5}}{2}}$.
      Since $a$ and $b$ are both positive, then $r>0$ and so $r={\displaystyle \frac{1+\sqrt{5}}{2}}$.
      This is the only possible value of $r$.
      We can check that $r$ satisfies the conditions by verifying that when $a=1$ (for example) and $r={\displaystyle \frac{1+\sqrt{5}}{2}}$, giving $b={\displaystyle \frac{1+\sqrt{5}}{2}}$, $c={\left({\displaystyle \frac{1+\sqrt{5}}{2}}\right)}^2$, and $d={\left({\displaystyle \frac{1+\sqrt{5}}{2}}\right)}^3$, then we do indeed obtain ${\displaystyle \frac{1}{b}}-{\displaystyle \frac{1}{a}}={\displaystyle \frac{1}{d}}-{\displaystyle \frac{1}{b}}$.
      

3. 1. Since $AS=ST=AT$, then $\mathrm{△}AST$ is equilateral.
      This means that $\mathrm{\angle }TAS=\mathrm{\angle }AST=\mathrm{\angle }ATS={60}^{\circ }$.
      Join $B$ to $P$, $B$ to $S$, $D$ to $Q$ and $D$ to $S$.

      

      Since $AS$ is tangent to the circle with centre $B$ at $P$, then $BP$ is perpendicular to $PS$.
      Since $BP$ and $BC$ are radii of the circle with centre $B$, then $BP=BC=1$.
      Consider $\mathrm{△}SBP$ and $\mathrm{△}SBC$.
      Each is right-angled (at $P$ and $C$), they have a common hypotenuse $BS$, and equal side lengths ($BP=BC$).
      This means that $\mathrm{△}SBP$ and $\mathrm{△}SBC$ are congruent.
      Thus, $\mathrm{\angle }PSB=\mathrm{\angle }CSB=\frac{1}{2}\mathrm{\angle }AST={30}^{\circ }$.
      This means that $\mathrm{△}SBC$ is a ${30}^{\circ }$-${60}^{\circ }$-${90}^{\circ }$ triangle, and so $SC=\sqrt{3}BC=\sqrt{3}$.
      Since $\mathrm{\angle }CSQ={180}^{\circ }-\mathrm{\angle }CSP={180}^{\circ }-{60}^{\circ }={120}^{\circ }$, then using a similar argument we can see that $\mathrm{△}DSC$ is also a ${30}^{\circ }$-${60}^{\circ }$-${90}^{\circ }$ triangle.
      This means that $CD=\sqrt{3}SC=\sqrt{3}\cdot \sqrt{3}=3$.
      Since $CD$ is a radius of the circle with centre $D$, then $r=CD=3$.

   2. Solution 1
      From the given information, $DQ=QP=r$.
      Again, join $B$ to $P$, $B$ to $S$, $D$ to $Q$, and $D$ to $S$.
      As in (a), $\mathrm{△}SBP$ and $\mathrm{△}SBC$ are congruent which means that $SP=SC$.
      Using a similar argument, $\mathrm{△}SDC$ is congruent to $\mathrm{△}SDQ$.
      This means that $SC=SQ$.
      Since $SP=SC$ and $SC=SQ$, then $SP=SQ$.
      Since $QP=r$, then $SP=SQ=\frac{1}{2}r$.
      Suppose that $\mathrm{\angle }PSC=2\theta$.
      Since $\mathrm{△}SBP$ and $\mathrm{△}SBC$ are congruent, then $\mathrm{\angle }PSB=\mathrm{\angle }CSB=\frac{1}{2}\mathrm{\angle }PSC=\theta$.
      Since $\mathrm{\angle }QSC={180}^{\circ }-\mathrm{\angle }PSC={180}^{\circ }-2\theta$, then $\mathrm{\angle }QSD=\mathrm{\angle }CSD=\frac{1}{2}\mathrm{\angle }QSC={90}^{\circ }-\theta$.
      Since $\mathrm{△}SDQ$ is right-angled at $Q$, then $\mathrm{\angle }SDQ={90}^{\circ }-\mathrm{\angle }QSD=\theta$.
      This means that $\mathrm{△}SBP$ is similar to $\mathrm{△}DSQ$.
      Therefore, ${\displaystyle \frac{SP}{BP}}={\displaystyle \frac{DQ}{SQ}}$ and so ${\displaystyle \frac{\frac{1}{2}r}{1}}={\displaystyle \frac{r}{\frac{1}{2}r}}=2$, which gives $\frac{1}{2}r=2$ and so $r=4$.
      
      Solution 2
      From the given information, $DQ=QP=r$.
      Join $B$ to $P$ and $D$ to $Q$. As in (a), $BP$ and $DQ$ are perpendicular to $PQ$.
      Join $B$ to $F$ on $QD$ so that $BF$ is perpendicular to $QD$.

      

      This means that $\mathrm{△}BFD$ is right-angled at $F$.
      Also, since $BPQF$ has three right angles, then it must have four right angles and so is a rectangle.
      Thus, $BF=PQ=r$ and $QF=PB=1$.
      Since $QD=r$, then $FD=r-1$.
      Also, $BD=BC+CD=1+r$.
      Using the Pythagorean Theorem in $\mathrm{△}BFD$, we obtain the following equivalent equations: $$\begin{array}{rl}B{F}^2+F{D}^2& =B{D}^2\\ r^2+(r-1{)}^2& =(r+1{)}^2\\ r^2+r^2-2r+1& =r^2+2r+1\\ r^2& =4r\end{array}$$ Since $r\ne 0$, then it must be the case that $r=4$.

   3. As in Solution 1 to (b), $\mathrm{△}SBP$ is similar to $\mathrm{△}DSQ$ and $SP=SQ$. Therefore, ${\displaystyle \frac{SP}{BP}}={\displaystyle \frac{DQ}{SQ}}$ or ${\displaystyle \frac{SP}{1}}={\displaystyle \frac{r}{SP}}$ which gives $S{P}^2=r$ and so $SP=\sqrt{r}$.
      Thus, $SP=SQ=SC=\sqrt{r}$.
      Next, $\mathrm{△}APB$ is similar to $\mathrm{△}AQD$ (common angle at $A$, right angle).
      Therefore, ${\displaystyle \frac{AB}{BP}}={\displaystyle \frac{AD}{DQ}}$ and so ${\displaystyle \frac{AB}{1}}={\displaystyle \frac{AB+BD}{r}}$ and so $AB={\displaystyle \frac{AB+1+r}{r}}$.
      Re-arranging gives $rAB=AB+1+r$ and so $(r-1)AB=r+1$ and so $AB={\displaystyle \frac{r+1}{r-1}}$.
      This means that $AC=AB+BC=AB+1={\displaystyle \frac{r+1}{r-1}}+1={\displaystyle \frac{(r+1)+(r-1)}{r-1}}={\displaystyle \frac{2r}{r-1}}$.
      Next, draw the circle with centre $O$ that passes through $A$, $S$ and $T$ and through point $V$ on the circle with centre $D$ so that $OV$ is perpendicular to $DV$.

      

      Let the radius of this circle be $R$. Note that $OS=AO=R$.
      Consider $\mathrm{△}OSC$.
      This triangle is right-angled at $C$.
      Using the Pythagorean Theorem, we obtain the following equivalent equations: $$\begin{array}{rl}O{S}^2& =O{C}^2+S{C}^2\\ R^2& =(AC-AO{)}^2+S{C}^2\\ R^2& =(AC-R{)}^2+S{C}^2\\ R^2& =A{C}^2-2R\cdot AC+R^2+S{C}^2\\ 2R\cdot AC& =A{C}^2+S{C}^2\\ R& ={\displaystyle \frac{AC}{2}}+{\displaystyle \frac{S{C}^2}{2AC}}\\ R& ={\displaystyle \frac{2r}{2(r-1)}}+{\displaystyle \frac{(\sqrt{r}{)}^2}{4r/(r-1)}}\\ R& ={\displaystyle \frac{r}{r-1}}+{\displaystyle \frac{r-1}{4}}\end{array}$$ Since $OV$ is perpendicular to $DV$, then $\mathrm{△}OVD$ is right-angled at $V$.
      Using the Pythagorean Theorem, noting that $OV=R$ and $DV=r$, we obtain the following equivalent equations: $$\begin{array}{rl}O{V}^2+D{V}^2& =O{D}^2\\ R^2+r^2& =(OC+CD{)}^2\\ R^2+r^2& =(AC-AO+CD{)}^2\\ R^2+r^2& ={({\displaystyle \frac{2r}{r-1}}-R+r)}^2\\ R^2+r^2& ={({\displaystyle \frac{2r+r(r-1)}{r-1}}-R)}^2\\ R^2+r^2& ={({\displaystyle \frac{r^2+r}{r-1}}-R)}^2\\ R^2+r^2& ={\left({\displaystyle \frac{r^2+r}{r-1}}\right)}^2-2R\left({\displaystyle \frac{r^2+r}{r-1}}\right)+R^2\\ 2R\left({\displaystyle \frac{r^2+r}{r-1}}\right)& ={\left({\displaystyle \frac{r^2+r}{r-1}}\right)}^2-r^2\\ 2R\left({\displaystyle \frac{r(r+1)}{r-1}}\right)& ={\displaystyle \frac{r^2(r+1{)}^2}{(r-1{)}^2}}-r^2\\ 2R& ={\displaystyle \frac{r-1}{r(r+1)}}\cdot {\displaystyle \frac{r^2(r+1{)}^2}{(r-1{)}^2}}-{\displaystyle \frac{r-1}{r(r+1)}}\cdot r^2\\ 2R& ={\displaystyle \frac{r(r+1)}{r-1}}-{\displaystyle \frac{r(r-1)}{r+1}}\end{array}$$ Since $R={\displaystyle \frac{r}{r-1}}+{\displaystyle \frac{r-1}{4}}$, we obtain: $${\displaystyle \frac{2r}{r-1}}+{\displaystyle \frac{r-1}{2}}={\displaystyle \frac{r(r+1)}{r-1}}-{\displaystyle \frac{r(r-1)}{r+1}}$$ Multiplying both sides by $2(r+1)(r-1)$, expanding, simplifying, and factoring, we obtain the following equivalent equations: $$\begin{array}{rl}4r(r+1)+(r-1{)}^2(r+1)& =2r(r+1{)}^2-2r(r-1{)}^2\\ (4r^2+4r)+(r-1)(r^2-1)& =2r((r+1{)}^2-(r-1{)}^2)\\ (4r^2+4r)+(r^3-r^2-r+1)& =2r((r^2+2r+1)-(r^2-2r+1))\\ (4r^2+4r)+(r^3-r^2-r+1)& =2r(4r)\\ r^3-5r^2+3r+1& =0\\ (r-1)(r^2-4r-1)& =0\end{array}$$ Now $r\ne 1$. (If $r=1$, the circles would be the same size and the two common tangents would be parallel.)
      Therefore, $r\ne 1$ which means that $r^2-4r-1=0$.
      By the quadratic formula, $$r={\displaystyle \frac{4\pm \sqrt{(-4{)}^2-4(1)(-1)}}{2}}={\displaystyle \frac{4\pm \sqrt{20}}{2}}=2\pm \sqrt{5}$$ Since $r>1$, then $r=2+\sqrt{5}$.