Problems

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Solution:

1. \(\,\)
   

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   Notice that \begin{align*} && (a+b)^2 &= PQ^2 + (a-b)^2 \\ \Rightarrow && PQ^2 &= 4ab \\ && (b+c)^2 &= QR^2 + (c-b)^2 \\ \Rightarrow && QR^2 &= 4bc \\ && (a+c)^2 &= PR^2 + (a-c)^2 \\ \Rightarrow && PR^2 &= 4ac \\ \Rightarrow && 2\sqrt{ac} &= 2\sqrt{ab}+2\sqrt{bc} \\ \Rightarrow && \frac{1}{\sqrt{b}} &= \frac{1}{\sqrt{c}} + \frac1{\sqrt{a}} \\ \end{align*} Let \(x, y, z = \frac{1}{\sqrt{a}}, \frac1{\sqrt{b}}, \frac{1}{\sqrt{z}}\) so we would like to prove that \(2(x^4+y^4+z^4) = (x^2+y^2+z^2)^2\) or \(x^4+y^4+z^4 = 2x^2y^2+2y^2z^2+2z^2x^2\). We also have \begin{align*} && y &= x+z \\ \Rightarrow &&y^2 &= x^2+z^2+2xz \\ \Rightarrow && (y^2-x^2-z^2)^2 &= 4x^2z^2 \\ \Rightarrow && y^4+x^4+z^4 - 2x^2y^2-2y^2z^2+2x^2z^2 &= 4x^2z^2\\ \Rightarrow && y^4+x^4+z^4 &= 2x^2y^2+2y^2z^2+2z^2x^2 \end{align*}
2. Notice that subject to \(y > z > x\) all these steps are reversible, so we must have the equality we desire

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Solution:

1. \(C\) has the equation \(x^2 + y^2 = a^2\). One suitable circle would ideally pass through \((0,a)\) and \((0,-a)\) have a centre on the positive \(x\)-axis, so we would need \(a^2+c^2 = r^2\) so \(c = \sqrt{r^2-a^2}\) and the equation would be \((x-\sqrt{r^2-a^2})^2 + y^2 = r^2\). Clearly a circle with radius \(r < a\) cannot pass through two diametrically opposed points of a circle radius \(a\), since the furthest two points can be on a circle is \(2r\), and diametrically opposed points are \(2a\) apart. Similarly if they are exactly the same radii, then if they pass through diametrically opposed points they must be the same circle.
2. Let the centre of \(D\) be at \((x,y)\), then it must be a distance of \(\sqrt{r^2-a_i}\) from each circle centre, ie \begin{align*} && (x-d)^2+y^2 &= r^2-a_2^2 \\ && (x+d)^2 + y^2 &= r^2-a_1^2 \\ \Rightarrow && 4dx &= a_2^2 - a_1^2 \\ \Rightarrow && x &= \frac{a_2^2-a_1^2}{4d} \\ \Rightarrow && y^2 &= r^2-a_1^2 - \left (\frac{a_2^2-a_1^2}{4d}+d \right)^2 \\ &&&= r^2 - a_1^2 - \frac{(a_2^2-a_1^2+4d^2)^2}{16d^2} \\ &&&= \frac{16d^2r^2-16d^2a_1^2 - a_2^4-a_1^4-16d^4+2a_1^2a_2^2+8a_1^2d^2-8a_2^2d^2}{16d^2} \\ &&&= \frac{16d^2r^2-8d^2a_1^2 - a_2^4-a_1^4-16d^4+2a_1^2a_2^2-8a_2^2d^2}{16d^2} \\ \Rightarrow && y &= \pm \sqrt{ \frac{16d^2r^2-8d^2a_1^2 - a_2^4-a_1^4-16d^4+2a_1^2a_2^2-8a_2^2d^2}{16d^2}} \end{align*} and we need \begin{align*} && 0 &\leq 16d^2r^2-8d^2a_1^2 - a_2^4-a_1^4-16d^4+2a_1^2a_2^2-8a_2^2d^2 \\ \Rightarrow && 16d^2 d^2 &\geq 8d^2a_1^2 + a_2^4+a_1^4+16d^4+2a_1^2a_2^2+8a_2^2d^2 \\ &&&= (4d^2+(a_2-a_1)^2)(4d^2+(a_2+a_1)^2) \end{align*}

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Solution:

1. Notice that \(P\) lies on \(C_1C_2\), and that the triangles formed from \(C_iPT_i\) where \(T_i\) are the tangent points are similar, with ratios \(\frac{r_1}{r_2}\). Therefore \(\frac{C_1P}{r_1} = \frac{C_2P}{r_2}\), and hence \(\frac{C_1P}{C_1C_2} = \frac{C_1P}{C_1P-C_2P} = \frac{1}{1-\frac{r_2}{r_1}} = \frac{r_1}{r_1-r_2}\) So we have \(\mathbf{p} = \mathbf{x_1} + (\mathbf{x}_2 - \mathbf{x}_1)\cdot\frac{r_1}{r_1-r_2} = \frac{r_1\mathbf{x}_2 - r_2\mathbf{x}_1}{r_1-r_2}\)
2. Suppose \(\mathbf{x}_3 = \binom{\alpha}{\beta}\) in the basis of \(\{ \mathbf{x}_1, \mathbf{x}_2 \}\), then we can see that \begin{align*} && \mathbf{p} &= \frac{1}{r_1-r_2}\binom{-r_2}{r_1} \\ && \mathbf{q} &= \frac{r_1(\alpha \mathbf{x}_1 +\beta \mathbf{x}_2) - r_3\mathbf{x}_1}{r_1-r_3} \\ &&&= \frac{1}{r_1-r_3} \binom{r_1\alpha -r_3}{r_1\beta} \\ && \mathbf{r} &=\frac{1}{r_2-r_3} \binom{r_2\alpha}{r_2\beta - r_3} \\ && \mathbf{p}-\mathbf{q} &= \frac{1}{r_1-r_2}\binom{-r_2}{r_1} - \frac{1}{r_1-r_3} \binom{r_1\alpha -r_3}{r_1\beta} \\ &&&= \frac{1}{(r_1-r_2)(r_1-r_3)} \binom{(r_1-r_3)(-r_2)-(r_1-r_2)(r_1\alpha-r_3)}{(r_1-r_3)r_1 - (r_1-r_2)r_1\beta} \\ &&&= \frac{r_1}{(r_1-r_2)(r_1-r_3)} \binom{(r_3-r_2)-\alpha(r_1-r_2)}{(r_1-r_3)-\beta(r_1-r_2)} \\ && \mathbf{q} - \mathbf{r} &= \frac{1}{r_1-r_3} \binom{r_1\alpha -r_3}{r_1\beta} - \frac{1}{r_2-r_3} \binom{r_2\alpha}{r_2\beta - r_3} \\ &&&= \frac{1}{(r_1-r_3)(r_2-r_3)}\binom{(r_2-r_3)(r_1\alpha-r_3) - (r_1-r_3)r_2\alpha)}{(r_2-r_3)r_1\beta - (r_1-r_3)(r_2\beta - r_3)} \\ &&&= \frac{1}{(r_1-r_3)(r_2-r_3)}\binom{(-r_2r_3+r_3^2) - \alpha(r_1r_3-r_3r_2)}{r_3(r_1-r_3)-\beta(r_1-r_2)} \\ &&&= \frac{r_3}{(r_1-r_3)(r_2-r_3)}\binom{(r_3-r_2)-\alpha(r_1-r_2)}{(r_1-r_3)-\beta(r_1-r_2)} \end{align*} Therefore they are clearly parallel, and hence lie on a line.
3. \(Q\) is halfway between \(P\) and \(R\) if \begin{align*} && \frac{r_1}{(r_1-r_2)(r_1-r_3)} &= \frac{r_3}{(r_1-r_3)(r_2-r_3)} \\ \Leftrightarrow && r_1(r_2-r_3) &= r_3(r_1-r_2) \\ \Leftrightarrow && r_1r_2 - r_1r_3 &= r_1r_3 - r_2r_3 \\ \Leftrightarrow && r_2 &= \frac{2r_1r_3}{r_1+r_3} \end{align*}

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Solution:

The shown isoceles triangle has base \(2r\), and the two side lengths are \(R\). The angle at the center of the circle is \(\frac{2\pi}{n}\). The height of the triangle (by Pythagoras) is \(\sqrt{R^2-r^2}\) and so the area enclosed in the triangle is \(\frac12 2r \sqrt{R^2-r^2}\). The area in the three sectors are: \(\frac{\pi}{n}(R-r)^2\), two sets of \(\frac12(\frac12 \l \pi - \frac{2\pi}{n}\r)r^2 = \l\frac{1}{2} - \frac{1}{n} \r \frac12 \pi r^2\). Therefore the remaining area \(Y/n\) is \(r \sqrt{R^2-r^2} -\frac{\pi}{n}(R-r)^2 - \l\frac{1}{2} - \frac{1}{n} \r \pi r^2\). Multiplying this by \(n\) we get the desired result. For \(X\) we can look at the larger sector, which we obtain using the extension of this triangle. This has area \(\frac12 \frac{2 \pi}{n} (R+r)^2\). The areas to be removed are the area of the triangle: \(r \sqrt{R^2-r^2}\) and the areas of the two sectors, which have radii \(r\) and angles \(\pi - \l \frac12 -\frac1{n} \r\pi = \l \frac{1}{2} + \frac{1}{n} \r \pi\). Therefore the area for \(X/n\) is \(\frac{\pi}{n} (R+r)^2 - r \sqrt{R^2-r^2} - \l \frac{1}{2} + \frac{1}{n} \r \pi r^2\) and so \(X\) has area: \[ X = \pi (R+r)^2 - nr \sqrt{R^2-r^2} - n\pi r^2\l \frac{1}{2} +\frac{1}{n} \r \] \(Z = n\pi r^2\) \begin{align*} \frac{(X+Y)}{Z} &= \frac{\pi (R+r)^2 - nr \sqrt{R^2-r^2} - n\pi r^2\l \frac{1}{2} +\frac{1}{n} \r}{n \pi r^2 }\\ &= \qquad \frac{nr\sqrt{R^{2}-r^{2}}-\pi(R-r)^{2}-n\pi r^{2}\left(\frac{1}{2}-\frac{1}{n}\right)}{n \pi r^2} \\ &= \frac{4\pi R r - n \pi r^2}{n\pi r^2} \\ &= \frac{4R-nr}{n r} \\ &= \frac{4R}{nr} - 1 \end{align*} Since \(\frac{r}{R} = \sin \frac{\pi}{n}\) we have: \begin{align*} \frac{(X+Y)}{Z} &= \frac{4R}{nr} - 1 \\ &= \frac{4}{n \sin \frac{\pi}n} - 1 \\ & \to \frac{4}{\pi} - 1 \end{align*}