Problems

---

Solution: Since \(A\) and \(B\) return to their starting points, and at their highest points there is no vertical component to their velocities, their horizontal must perfectly reverse, ie \begin{align*} && m u \cos \alpha - M v \cos \beta &= -m u \cos \alpha + M v \cos \beta \\ \Rightarrow && mu \cos \alpha &= Mv \cos \beta \end{align*} Since they reach their highest points at the same time, they must have the same initial vertical speed, ie \(u \sin \alpha = v \sin \beta\), so \begin{align*} && m v \frac{\sin \beta}{\sin \alpha} \cos \alpha &= M v \cos \beta \\ \Rightarrow && m \cot \alpha &= M \cot \beta \end{align*} The horizontal distance travelled by \(A\) & \(B\) will be: \begin{align*} && d_A &= u \cos \alpha t \\ && d_B &= v \cos \beta t \\ \Rightarrow && \frac{d_A}{d_A+d_B} &= \frac{u \cos \alpha}{u \cos \alpha + v \cos \beta} \\ &&&= \frac{\frac{M}{m}v \cos \beta}{\frac{M}{m}v \cos \beta + v \cos \beta} \\ &&&= \frac{M}{M+m} \\ \Rightarrow && d_A = b &= \frac{Md}{m+M} \end{align*} Applying \(v^2 = u^2 + 2as\) we see that \begin{align*} && 0 &= u \sin \alpha - gt \\ \Rightarrow && t &= \frac{u \sin \alpha}{g} \\ && b &=u \cos \alpha \frac{u \sin \alpha}{g} \\ \Rightarrow && u^2 &= \frac{2bg}{\sin 2 \alpha} \\ && 0 &= u^2 \sin^2 \alpha - 2g h \\ \Rightarrow && h &= \frac{u^2 \sin^2 \alpha}{2g} \\ &&&= \frac{2bg}{\sin 2 \alpha} \frac{ \sin^2 \alpha}{2g} \\ &&&= \frac12 b \tan \alpha \end{align*}

---

Solution:

1. Suppose \(t_n = x\) for all \(n\), then we must have \begin{align*} && x &= p x + q x \\ \Leftrightarrow && 1 &= p+q \end{align*} and this clearly works for any value of \(x\).
2. Suppose \(t_{2n} = x, t_{2n+1} = y\) for all \(n\), then \begin{align*} && x &= py + q x \\ && y &= px + q y \\ \Rightarrow && 0 &= py + (q-1) x \\ && 0 &= px + (q-1) y \\ \Rightarrow && p &= (q-1) \frac{x}{y} = (q-1) \frac{y}{x} \\ \Rightarrow && \frac{y}{x} = \pm 1 & \text{ or } q = 1, p = 0 \\ \Rightarrow && y = \pm x & \text{ or } (p,q) = (0,1) \\ \end{align*}
3. Suppose \(t_{3n} = x\), \(t_{3n+1}=y\) and \(t_{3n+2}=z\) , so \begin{align*} && x &= pz + qy \\ && y & = px + qz \\ && z &= py + qx \\ \\ && z &= p(px+qz) + q(pz + qy) \\ &&&= p^2x + 2pqz + q^2 y \\ &&&= p^2(pz+qy) + 2pqz + q^2(px+qz) \\ &&&= p^3 z + p^2qy + 2pqz + q^2p x + q^3 z \\ &&&= (p^3+q^3+2pq)z + pq(py+qx) \\ &&&= (p^3 + q^3 + 2pq)z + pq z \\ &&&= (p^3 + q^3 + 3pq)z \\ \Rightarrow && 0 &= p^3 + q^3 + 3pq- 1 \\ &&&= (p+q-1)(p^2+q^2+1+p+q-pq) \\ &&&= \tfrac12(p+q-1)((p-q)^2+(p+1)^2+(q+1)^2) \end{align*} Therefore \(p+q = 1\) or \((p-q)^2+(p+1)^2+(q+1)^2 = 0 \Rightarrow p = q = -1\). If \(p+q = 1\), then \(z = py + (1-p)x\) and \(x = p(py+(1-p)x) + (1-p)y \Rightarrow (1-p+p^2)x = (1-p+p^2)y \Rightarrow x = y \Rightarrow x= y = z\). If \(p = q = -1\) then adding all the equations we get \(x + y + z = -2(x+y+z) \Rightarrow x + y + z = 0\)

Note that what is actually going on here is that solutions must be of the form \(t_n = \lambda^n\) so the only way to be constant is for \(\lambda = 1\) to be a root, the only way for it to be \(2\)-periodic is for \(\lambda = -1\) to be a root, and the only way for it to be \(3\)-periodic is for \(\lambda = 1, \omega, \omega^2\) to be the roots (although we see this via the classic \(x^3 + y^3 + z^3 - 3xyz = (x+y+z)(x + \omega y + \omega^2 z)(x+\omega^2 y +\omega z)\) which is because of the real constraint in the question.

---

Solution: If \(\p(x)\) and \(\q(x)\) are polynomials of degree \(m\) and \(n\), \(\p(\q(x))\) has degree \(mn\).

1. Suppose \(\p(\p(\p(x)))- 3 \p(x)= -2x\), and suppose \(p(x)\) has degree \(n = \geq 2\), then \(\p(\p(\p(x)))\) has degree \(n^3\) and so the left hand side has degree higher than \(1\) and the right hand side is degree \(1\). Therefore \(\p(x)\) is degree \(1\) or \(0\). If \(p(x) = c\) then \(c^3-3c = -2x\) but the LHS doesn't depend on \(x\) which is also a contradiction. Therefore \(\p(x)\) is degree \(1\). Suppose \(\p(x) = ax+b\) then: \begin{align*} && -2x &= \p(\p(\p(x))) - 3\p(x) \\ &&&= \p(\p(ax+b)) - 3(ax+b) \\ &&&= \p(a(ax+b)+b) - 3ax -3b \\ &&&= a(a^2x+ab+b) + b - 3ax - 3b \\ &&&= (a^3-3a)x + b(a^2+a-2) \\ \Rightarrow &&& \begin{cases} a^3-3a&=-2 \\ b(a^2+a-2) &= 0\end{cases} \\ \Rightarrow &&& \begin{cases} a^3-3a+2 = 0 \\ b = 0, a = 1, a = -2\end{cases} \\ \Rightarrow &&& \begin{cases} (a-1)(a^2+a-2) = 0 \\ b = 0, a = 1, a = -2\end{cases} \\ \Rightarrow && (a,b) &= (1, b), (-2,b) \end{align*}
2. Suppose \(2\p(\p(x)) +3 [\p(x)]^2 -4\p(x) =x^4\) and let \(\deg \p(x) = n\), then LHS has degree \(\max(n^2,2n,n)\) and the right hand side has degree \(4\). Therefore \(\p(x)\) must have degree \(2\). Let \(\p(x) = ax^2 + bx + c\), then, considering the coefficient of \(x^4\) in \(2\p(\p(x)) + 3[\p(x)]^2 -4\p(x)\) we will have \(2a^3+3a^2=1 \Rightarrow 2a^3+3a^2-1 = (a+1)^2(2a-1) \Rightarrow a = -1, a=\frac12\). Consider the coefficient of \(x^3\) in \(2\p(\p(x)) + 3[\p(x)]^2 -4\p(x)\) we have \(4a^2b+6ab = 0 \Rightarrow 2ab(2a+3) = 0\) Since \(a = -1, \frac12\) this means \(b = 0\). Consider the constant coefficient in \(2\p(\p(x)) + 3[\p(x)]^2 -4\p(x)\) (using \(b = 0\)). \(2ac^2+c+3c^2-4c = 0 \Rightarrow c(2ac+3c-3) = 0\). Therefore \(c = 0\) or \(a = -1, c = 3, a = \frac12, c = \frac34\), so our possible polynomials are: \(\p(x) = -x^2, \frac12x^2, -x^2+3, \frac12x^2+\frac34\)

---

Solution:

1. 1. \(\,\)
      

      + - ↺
   2. \(\,\)
      

      + - ↺
   3. \(\,\)
      

      + - ↺
   4. \(\,\)
      

      + - ↺
2. Suppose \(m = 12\) \begin{align*} && y &= 4-x^2 \\ && 12x &= k-y^2 \\ \Rightarrow && 12 x&=k-(4-x^2)^2 \\ &&&= k-16+8x^2-x^4 \\ \Rightarrow && 0 &= x^4- 8x^2+12x+16-k \end{align*} When the curves touch, we will have repeated root, ie \(a\) is a root of \(4x^3-16x+12 \Rightarrow a^3-4a+3 =0\). \begin{align*} &&0 &= a^3-4a+3 \\ &&&= (a-1)(a^2+a-3) \\ \Rightarrow &&a &= 1, \frac{-1 \pm \sqrt{13}}{2} \end{align*} \begin{align*} && 0 &= a^4-8a^2+12a+16-k \\ \Rightarrow && k &= a(a^3-8a+12)+16 \\ &&&= a(4a-3-8a+12)+16 \\ &&&= -4a^2+9a+16 \\ \\ \Rightarrow && a = 1& \quad k = 21 \\ && k &= -4(3-a)+9a+16 = 13a+4\\ && a = \frac{-1-\sqrt{13}}2& \quad k = \frac{-5 - 13\sqrt{13}}{2} < 0 \\ && a = \frac{-1+\sqrt{13}}2& \quad k = \frac{-5 + 13\sqrt{13}}{2} \\ \end{align*} So we have type (a), and (d).

---

Solution:

1. \(\,\) \begin{align*} && 1 &= \frac{a}{x} + \frac{b}{y} \\ \frac{\d}{\d x}: && 0 &= -\frac{a}{x^2} - \frac{b}{y^2} \frac{\d y}{\d x} \\ \Rightarrow && \frac{\d y}{\d x} &= -\frac{ay^2}{bx^2} \\ \\ (p,q): && -\frac{aq^2}{bp^2} &= -\frac{a}{b} \\ \Rightarrow && p^2 &= q^2 \\ \Rightarrow && p &= \pm q \\ \\ \Rightarrow && ap \pm b p &= 1 \\ \Rightarrow && (a\pm b)p &= 1 \\ \Rightarrow && \frac{a}{p} \pm \frac{b}{p} &= 1 \\ \Rightarrow && (a \pm b)\frac{1}{p} &= 1 \\ \Rightarrow && (a \pm b)^2 &= 1 \end{align*}
2. \(\,\) \begin{align*} && 1 &= \frac{a}{x} - \frac{b}{y} \\ \Rightarrow && \frac{\d y}{\d x} &= \frac{ay^2}{bx^2} \\ \Rightarrow && \frac{aq^2}{bp^2} &= \frac{b}{a} \\ \Rightarrow && aq &= \pm bp \\ \Rightarrow && 1 &= \frac{a}{p} - \frac{b}{q} \\ &&&= \frac{aq-bp}{pq} \\ \Rightarrow && aq &= -bp \\ \Rightarrow && 1 &= \frac{2aq}{pq} \\ \Rightarrow && p &= 2a \\ \Rightarrow && q &= -2b \\ \Rightarrow && 1 &= 2a^2-2b^2 \\ \Rightarrow && \frac12 &= a^2-b^2 \end{align*}

---

Solution:

1. 

   + - ↺
   Clearly the only solution is \(x = 1\)
2. 

   + - ↺
   There is clearly only one solution, with \(x \approx -2\) \begin{align*} && 4(1-x) &= 6+2\sqrt{9-x^2} \\ && -2x-1 &=\sqrt{9-x^2} \\ \Rightarrow && 4x^2+4x+1 &= 9-x^2 \\ \Rightarrow && 0 &= 5x^2+4x-8 \\ &&x&= \frac{-2\pm 2\sqrt{11}}{5} \\ \Rightarrow && x &= -\left ( \frac{2+2\sqrt{11}}{5} \right) \end{align*}

---

Solution: \(a\left(x-y+2\right)^{2} +b\left(cx+y\right)^{2}+d\, \equiv (a + bc^2)x^2 + (a+b)y^2 + (-2a+2bc)xy + (4a)x+(-4ay) + 4a+d\) so we want to solve \[ \begin{cases} a + bc^2 &= 5 \\ a+b &= 2 \\ 2bc - 2a &= -6 \\ 4a &= 4 \\ -4a &= 4 \\ 4a+d &= -9 \end{cases} \Rightarrow a = 1, b = 1, c = -2, d = -13 \] Therefore we have: \((x-y+2)^2 + (2x+y)^2-13 = 0\) and our simultaneous equations will be: \[ \begin{cases} (x-y+2)^2 + (-2x+y)^2 &= 13 \\ 2(x-y+2)^2 + (-2x+y)^2 &= 22 \end{cases} \] which are simultaneous equations in \((x-y+2)^2\) and \((-2x+y)^2\) which solve to \((x-y+2)^2 = 9, (-2x+y)^2 = 4 \) so we need to solve \(4\) sets of simultaneous equations: \begin{align*} &\begin{cases} x - y + 2 &= 3 \\ -2x + y &= 2 \end{cases} &&\Rightarrow (x,y) = (-3, -4) \\ &\begin{cases} x - y + 2 &= -3 \\ -2x + y &= 2 \end{cases} &&\Rightarrow (x,y) = (3, 8) \\ &\begin{cases} x - y + 2 &= 3 \\ -2x + y &= -2 \end{cases} &&\Rightarrow (x,y) = (1, 0) \\ &\begin{cases} x - y + 2 &= -3 \\ -2x + y &= -2 \end{cases} &&\Rightarrow (x,y) = (7, 12) \\ \end{align*} So \((x,y) = (-3, -4), (3, 8), (1, 0), (7,12)\)

---

Solution:

1. \(\,\) \begin{align*} && 0 &= a+b \tag{1}\\ && 1 &= a\lambda -a\mu \tag{2} \\ && 1 &= a\lambda^2 -a\mu^2 \tag{3} \\ && 2 &= a\lambda^3 - a\mu^3 \tag{4} \\ (4) \div (2): && 2 & = \lambda^2+\lambda \mu + \mu^2 \\ (3) \div (2): && 1 &= \lambda + \mu \\ \Rightarrow && 2 &= \lambda^2 + \lambda(1-\lambda) + (1-\lambda)^2 \\ &&&= \lambda^2-\lambda+1\\ \Rightarrow && \lambda, \mu &= \frac{1 \pm \sqrt{5}}{2} \\ \Rightarrow && a &= \frac{1}{\lambda - \mu} = \frac{1}{\sqrt{5}} \\ \Rightarrow && b &= -\frac{1}{\sqrt{5}} \end{align*} (NB: This is Binet's formula)
2. \(\,\) \begin{align*} F_6 &= \frac{1}{\sqrt{5}} \left ( \left ( \frac{1 +\sqrt{5}}{2} \right)^6- \left ( \frac{1 -\sqrt{5}}{2} \right)^6 \right) \\ &= \frac{1}{2^6 \sqrt{5}} \left ( (1+\sqrt{5})^6-(1-\sqrt{5})^6 \right) \\ &= \frac{1}{2^5 \cdot \sqrt{5}} \left (6 \sqrt{5} +\binom{6}{3} (\sqrt{5})^3+\binom{6}{5}(\sqrt{5})^5 \right)\\ &= \frac{1}{2^5} \left (6 +20\cdot 5+6\cdot 5^2 \right)\\ &= \frac{1}{2^5} 256 = 2^3 = 8 \end{align*} (way more painful than just computing it by adding terms!)
3. \(\,\) \begin{align*} && \sum_{n=0}^{\infty} \frac{F_n}{2^{n+1}} &= \sum_{n=0}^{\infty} \frac{a\lambda^n + b\mu^n}{2^{n+1}} \\ &&&= \frac12 \left ( \frac{a}{1-\frac{\lambda}2} + \frac{b}{1-\frac{\mu}2} \right) \\ &&&= \frac12 \left ( \frac{2a}{2-\lambda} + \frac{2b}{2-\mu}\right) \\ &&&= \frac{2a}{4-2\lambda} + \frac{2b}{4-2\mu}\\ &&&= \frac{2a}{4-(1+\sqrt{5})} - \frac{2a}{4-(1-\sqrt{5})} \\ &&&= \frac{2}{3\sqrt{5}-5} - \frac{2}{3\sqrt{5}+5} \\ &&&= \frac{6\sqrt{5}+10-6\sqrt{5}+10}{45-25} \\ &&&= 1 \end{align*}

---

Solution: The curve \(x^4 + y^4 = u\) has lines of symmetry:

• \(y = 0\)
• \(x = 0\)
• \(y = x\)
• \(y = -x\)

The curve \(xy = v\) has lines of symmetry:

• \(y = x\)
• \(y = -x\)

The points are \(A = (\alpha, \beta), B = (\beta, \alpha), C = (-\alpha, -\beta), D = (-\beta, -\alpha)\) \(AD\) has gradient \(\frac{\beta+\alpha}{\alpha+\beta} = 1\), \(BC\) has the same gradient. \(AB\) has gradient \(\frac{\alpha-\beta}{\beta-\alpha} = -1\), as does \(CD\). Therefore it has two sets of perpendicular and parallel sides, hence a rectangle. The area is \(|AD||AB| = \sqrt{2(\alpha+\beta)^2}\sqrt{2(\alpha-\beta)^2} = 2(\alpha^2-\beta^2)\) The squared area is \(4(\alpha^4+\beta^4 - 2 \alpha^2\beta^2) = 4(u - 2v^2)\) ie the area is \(2\sqrt{u-2v^2}\) When \(u = 81, v = 4\) we have the area is \(2 \sqrt{81 - 2 \cdot 16} = 14\) as required.

---

Solution: This is a second order recurrence relation, so we need to find \(m\) and \(n\) such that; \begin{align*} &&\frac15 &= m\frac13 + n \\ &&\frac17 &= m \frac15 + n\frac13 \\ \Rightarrow && m,n &= \frac67, - \frac{3}{35} \end{align*} So we now need to solve the characteristic equation: \(\lambda^2 - \frac67 \lambda + \frac{3}{35} = 0\) So \(x,y = \frac{15 \pm 2 \sqrt{30}}{35}\). We need, \begin{align*} && 1 &= a+ b \\ && \frac13 &= a \frac{15 + 2 \sqrt{30}}{35} + b \frac{15 - 2 \sqrt{30}}{35} \\ && \frac13 &= \frac{15}{35} + \frac{2 \sqrt{30}}{35}(a-b) \\ \Rightarrow && -\frac{\sqrt{30}}{18} &= a-b \\ \Rightarrow && a &= \frac{18-\sqrt{30}}{36} \\ && b &= \frac{18+\sqrt{30}}{38} \end{align*} So our two answers are: \[ (a,b,x,y) = \left (\frac{18\pm\sqrt{30}}{36} ,\frac{18\mp\sqrt{30}}{36},\frac{15 \pm 2 \sqrt{30}}{35},\frac{15 \mp 2 \sqrt{30}}{35}, \right)\]

---

Solution:

1. Since \(AP = 2BP\) we also have \(|AP|^2 = 4|BP|^2\) ie \begin{align*} && (x-5)^2 + (y-16)^2 &= 4(x+4)^2 + 4(y-4)^2 \\ \Rightarrow && x^2 - 10x+25 + y^2 -32y + 256 &= 4x^2+32x+64+4y^2-32y+64 \\ \Rightarrow && 281 &= 3x^2+42x+3y^2+128\\ && 281 &= 3(x+7)^2-147+3y^2+128 \\ \Rightarrow && 300 &= 3(x+7)^2 + 3y^2 \\ && 100 &= (x+7)^2 + y^2 \end{align*}
2. Since \(|QC|^2 = k^2 |QD|^2\), \begin{align*} && (x-a)^2 + y^2 &= k^2 (x-b)^2 + k^2y^2 \\ \Rightarrow && x^2-2ax+a^2 &= k^2x^2-2k^2bx+k^2b^2 + (k^2-1)y^2 \\ && a^2-k^2b^2 &= (k^2-1)x^2-2(k^2b-a)x + (k^2-1)y^2 \\ && a^2-k^2b^2&= (k^2-1)\left(x-\frac{k^2b-a}{k^2-1}\right)^2-(k^2-1)\left(\frac{k^2b-a}{k^2-1}\right)^2+(k^2-1)y^2 \\ && \frac{a^2-k^2b^2}{k^2-1}+\left(\frac{k^2b-a}{k^2-1}\right)^2&= \left(x-\frac{k^2b-a}{k^2-1}\right)^2+y^2 \\ \Rightarrow && -7 &= \frac{k^2b-a}{k^2-1} \tag{*} \\ && 100 &= \frac{a^2-k^2b^2}{k^2-1}+\left(\frac{k^2b-a}{k^2-1}\right)^2 \\ &&&= \frac{a^2-k^2b^2}{k^2-1}+7^2 \\ \Rightarrow && 51 &= \frac{a^2-k^2b^2}{k^2-1} \tag{**} \\ (*) \Rightarrow && k^2(b+7)&= a+7 \\ (**) \Rightarrow && k^2(51+b^2)&= a^2+51 \\ \Rightarrow && \frac{a^2+51}{b^2+51} &= \frac{a+7}{b+7} \\ \\ \Rightarrow && a^2b+51b+7a^2 &= ab^2+51a+7b^2 \\ && 0 &= ab(b-a)-51(b-a)+7(b-a)(b+a) \\ &&&= (b-a)(ab+7(b+a)-51) \\ &&&= (b-a)((a+7)(b+7)-100) \\ \Rightarrow && 100 &= (a+7)(b+7) \end{align*} Since \(a \neq b\)

---

Solution: \begin{align*} && I &= \int_0^\pi x \sin x \d x \\ &&&= \left [ -x \cos x \right]_0^\pi + \int_0^{\pi} \cos x \d x \\ &&&= \pi \\ \\ && J &= \int_0^\pi x \cos x \d x \\ &&&= \left [ x \sin x \right]_0^\pi - \int_0^\pi \sin x \d x \\ &&&= -2 \end{align*} \begin{align*} && f(t) &= t + \int_0^\pi f(x) \sin (x+t) \d x \\ &&&= t + \int_0^\pi f(x) \left ( \sin t \cos x + \cos t \sin x \right) \d x \\ &&&= t + \sin t \int_0^{\pi} f(x) \cos x \d x + \cos t \int_0^{\pi} f(x) \sin x \d x \\ \\ && A &= \int_0^\pi (x + A \sin x + B \cos x) \cos x \d x \\ &&&= -2+ \frac{\pi}{2} B \\ && B &= \int_0^{\pi} (x + A \sin x + B \cos x ) \sin x \d x \\ &&&= \pi + \frac{\pi}{2} A \\ \Rightarrow && (A,B) &= (-2,0) \end{align*}