Problems

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Solution:

The orange guard will always see \(2b+b-a = 3b-a\) The blue guard will see \(b + \frac{b(b-a)}{a} = \frac{b^2}{a}\) if \(b < 3a\) and \(3b + \frac{b(b-3a)}{(b-a)} = \frac{2b(2b-3a)}{b-a}\). Therefore the blue guard always sees more if \(b > 3a\). He sees more in the other case if \begin{align*} && \frac{b^2}{a} &> 3b - a \\ \Leftrightarrow && \frac{b^2}{a^2} &> 3\frac{b}{a} - 1 \\ \Leftrightarrow && x^2 - 3x + 1 &> 0\\ \Leftrightarrow && x > \frac{3 + \sqrt{5}}{2} \text{ or } x < \frac{3-\sqrt{5}}{2} \end{align*} Since \(b > a\) we must have \(b > \frac{3+\sqrt{5}}2 a\) There is an alternative interpretation which is that the orange guard is in the top left corner, ie In this case the green guard will always see \(2b + \frac{2b(b-a)}{b+a} = \frac{4b^2}{b+a}\) Comparing \(\frac{4b^2}{b+a}\) with \(\frac{b^2}{a}\) we can see the former is larger if \(3a > b\). Comparing \(\frac{4b^2}{b+a}\) with $$

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Solution: Let \(O\) be the origin, and let \(\mathbf{a}, \mathbf{b}, \mathbf{a}', \mathbf{b}'\) be the four points. The conditions give us \begin{align*} && \mathbf{a} \cdot \mathbf{b} &= 0 \\ && |\mathbf{a}| &= |\mathbf{b}| \\ && \mathbf{a}' \cdot \mathbf{b}' &= 0 \\ && |\mathbf{a}'| &= |\mathbf{b}'| \\ \end{align*} So \begin{align*} \text{midpoint }AB \text{ to midpoint } BA' &= (\tfrac12(\mathbf{a}+\mathbf{b}) - \tfrac12(\mathbf{b}+\mathbf{a}'))\cdot (\tfrac12(\mathbf{a}+\mathbf{b}) - \tfrac12(\mathbf{b}+\mathbf{a}')) \\ &= \frac12(\mathbf{a}-\mathbf{a}')\cdot \frac12(\mathbf{a} - \mathbf{a}') \\ \text{midpoint }BA' \text{ to midpoint } A'B' &= (\tfrac12(\mathbf{b}+\mathbf{a}') - \tfrac12(\mathbf{a}'+\mathbf{b}')) \cdot (\tfrac12(\mathbf{b}+\mathbf{a}') - \tfrac12(\mathbf{a}'+\mathbf{b}'))\\ &= \frac12(\mathbf{b}-\mathbf{b}')\cdot \frac12(\mathbf{b} - \mathbf{b}') \\ &= \frac14 (|\mathbf{b}|^2 + |\mathbf{b}'|^2 - 2\mathbf{b}\cdot\mathbf{b}')\\ &= \frac14(|\mathbf{a}|^2 + |\mathbf{a}'|^2 - 2\mathbf{b}\cdot\mathbf{b}') \\ \text{midpoint }A'B' \text{ to midpoint } B'A &= (\tfrac12(\mathbf{a}'+\mathbf{b}') - \tfrac12(\mathbf{b}'+\mathbf{a})) \cdot (\tfrac12(\mathbf{a}'+\mathbf{b}') - \tfrac12(\mathbf{b}'+\mathbf{a}))\\ &= \frac12(\mathbf{a}'-\mathbf{a})\cdot \frac12(\mathbf{a}' - \mathbf{a}) \\ \text{midpoint }B'A \text{ to midpoint } AB &= (\tfrac12(\mathbf{b}'+\mathbf{a}) - \tfrac12(\mathbf{a}+\mathbf{b})) \cdot (\tfrac12(\mathbf{b}'+\mathbf{a}) - \tfrac12(\mathbf{a}+\mathbf{b}))\\ &= \frac12(\mathbf{b}'-\mathbf{b})\cdot \frac12(\mathbf{b}' - \mathbf{b}) \\ \end{align*} So it's sufficient to prove \(\mathbf{a}\cdot \mathbf{a}' = \mathbf{b}\cdot \mathbf{b}'\) but this is clear from looking at a diagram for 1 second. Given the length of the square is what it is, we want to minimise \(\mathbf{b}\cdot \mathbf{b}'\) which is when they are vertically opposite each other, ie \(\angle BOA' = 90^\circ\)

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Solution: The vector representing the side \(AB\) is \(b - a\) and the vector representing the side \(DC\) is \(c - d\). \(ABCD\) is a parallelogram if and only if these opposite sides are parallel and equal in length, which is given by \(b - a = c - d\), or equivalently \(a + c = b + d\). Similarly, if \(a + c = b + d\), then \(c - b = d - a\), so the side \(BC\) is parallel and equal in length to the side \(AD\). Thus, \(a + c = b + d\) is the necessary and sufficient condition for \(ABCD\) to be a parallelogram. In a parallelogram, the shape is a square if and only if the diagonals are equal in length and perpendicular to each other. The diagonals are represented by the vectors \(c - a\) and \(d - b\). For these to be equal in length and perpendicular, one must be a \(90^\circ\) rotation of the other. Since \(A, B, C, D\) are labeled anticlockwise, a \(90^\circ\) anticlockwise rotation of the vector \(\vec{AC}\) (which is \(c-a\)) would point in the direction of \(\vec{DB}\) (which is \(b-d\) if we consider the relative orientation). Specifically: \(i(c - a) = d - b \implies -i(a - c) = d - b \implies i(a - c) = b - d\). Thus, \(ABCD\) is a square if and only if \(i(a - c) = b - d\).

1. The midpoint of the side \(PQ\) is \(\frac{1}{2}(p + q)\). To find the centre \(X\) of the square built externally on \(PQ\), we start at the midpoint and move a distance equal to half the side length in a direction perpendicular to \(PQ\). Since \(P, Q, R, S\) are anticlockwise, the outward direction is a \(90^\circ\) clockwise rotation of the vector \(\vec{PQ}\). A clockwise rotation of \(90^\circ\) corresponds to multiplication by \(-i\). \[ x = \frac{p+q}{2} + (-i)\left(\frac{q-p}{2}\right) = \frac{p + q - iq + ip}{2} = \frac{1}{2} \big( p(1+i) + q(1-i) \big) \]
2. From part (i), we have the representations for the centres: \begin{align*} x &= \tfrac{1}{2}(p(1+i) + q(1-i)) \\ y &= \tfrac{1}{2}(q(1+i) + r(1-i)) \\ z &= \tfrac{1}{2}(r(1+i) + s(1-i)) \\ t &= \tfrac{1}{2}(s(1+i) + p(1-i)) \end{align*} As shown in the first part of the problem, \(XYZT\) is a square if and only if: (1) \(x+z = y+t\) (it is a parallelogram) (2) \(i(x-z) = y-t\) (it is a square) First, examine condition (1): \begin{align*} x+z - (y+t) &= \tfrac{1}{2} \big[ (p+r)(1+i) + (q+s)(1-i) - (q+s)(1+i) - (r+p)(1-i) \big] \\ &= \tfrac{1}{2} \big[ (p+r)(1+i - (1-i)) - (q+s)(1+i - (1-i)) \big] \\ &= \tfrac{1}{2} \big[ (p+r)(2i) - (q+s)(2i) \big] \\ &= i(p+r - (q+s)) \end{align*} Thus, \(x+z = y+t\) if and only if \(p+r = q+s\), which is the condition that \(PQRS\) is a parallelogram. Next, examine condition (2): \begin{align*} i(x-z) &= \tfrac{1}{2} i \big[ p(1+i) + q(1-i) - r(1+i) - s(1-i) \big] \\ &= \tfrac{1}{2} \big[ p(i-1) + q(i+1) - r(i-1) - s(i+1) \big] \\ y-t &= \tfrac{1}{2} \big[ q(1+i) + r(1-i) - s(1+i) - p(1-i) \big] \\ \text{So, } i(x-z) - (y-t) &= \tfrac{1}{2} \big[ p(i-1 + 1-i) + q(i+1 - 1-i) + r(-i+1 - 1+i) + s(-i-1 + 1+i) \big] \\ &= 0 \end{align*} Since \(i(x-z) = y-t\) is an identity (always true for any \(PQRS\)), \(XYZT\) is a square if and only if it is a parallelogram. As established above, this occurs if and only if \(PQRS\) is a parallelogram.

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Solution:

Note that \(CP\) has equation \(\frac{y-0}{x-an} = \frac{a-0}{a-an} = \frac{1}{1-n} \Rightarrow y = \frac{x-an}{1-n}\) Therefore \(R = (0, -\frac{an}{1-n})\) Note that \(CQ\) has equation \(\frac{y-am}{x} = \frac{a-am}{a} = 1-m \Rightarrow y = (1-m)x + am\) Therefore \(S = (-\frac{am}{1-m}, 0)\) \(PQ\) has equation \(\frac{y}{x-an} = \frac{am-0}{0-an} \Rightarrow y = -\frac{m}{n}x +am\) \(SR\) has equation \(\frac{y}{x+\frac{am}{1-m}} = \frac{-\frac{an}{1-n}}{\frac{am}{1-m}} = -\frac{n(1-m)}{m(1-n)} \Rightarrow y =-\frac{n(1-m)}{m(1-n)} x -a\frac{n}{1-n}\) So \(PQ \cap SR\) has \begin{align*} && -\frac{m}{n}x +am &= -\frac{n(1-m)}{m(1-n)} x -a\frac{n}{1-n} \\ && x \left (\frac{n(1-m)}{m(1-n)} - \frac{m}{n} \right) &= -am - \frac{an}{1-n} \\ \Rightarrow && x \left ( \frac{n^2(1-m)-m^2(1-n)}{nm(1-n)} \right) &= -\frac{a(m(1-n)+n)}{1-n} \\ \Rightarrow && x \left ( \frac{(m-n)(mn-n-m)}{mn(1-n)} \right) &= \frac{a(mn-m-n)}{1-n} \\ \Rightarrow && x &= \frac{amn}{m-n} \\ && y &= -\frac{amn}{m-n} \end{align*} Therefore clearly \(TA\) is perpendicular to \(AC\) since they are the lines \(y = -x\) and \(y = x\) Given this method we can construct the perpendicular to the diagonal through the vertex. Doing this at \(A\) we can construct \(C'\) the reflection of \(C\) in \(AB\). We can do the same to find the reflection of \(A\) and so we have a square with sidelengths \(\sqrt{2}a\) and hence area \(2a^2\)

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Solution:

In the coordinate system where \(A\) is \((0,0)\) and \(AD\) is the \(x\)-axis and \(AB\) the \(y\)-axis and all side lengths are \(2\), we find the centre of mass of each of the sides are: \begin{align*} AB :& (0,1) \\ BC :& (1,2) \\ CD :& (2,1) \\ \\ ABCD:& \l 1, \frac{4}{3} \r \end{align*} When hung from \(A\), the angle \(AB\) makes to the vertical is \(\alpha\) and the angle \(BC\) makes to the vertical will be \(90^{\circ} + \alpha\). When hung from \(B\) the angle \(BC\) makes to the vertical will be \(\beta\). The value we are interested in therefore is \(\beta + 90^{\circ} + \alpha\) \begin{align*} && \tan \alpha &= \frac{1}{\frac{4}{3}} \\ &&& = \frac{4}{3} \\ \\ && \tan \beta &= \frac{\frac{2}{3}}{1} \\ &&&= \frac{2}{3} \\ \\ && \tan \l \beta + (90^{\circ} + \alpha) \r &= \frac{\tan \beta + \tan\l 90^{\circ} + \alpha \r}{1 - \tan \beta \tan\l 90^{\circ} + \alpha \r} \\ &&&= \frac{\frac23 + \frac43}{1- \frac23 \frac43} \\ &&&= \frac{2}{1 - \frac89} \\ &&&= 18 \end{align*}

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Solution: First note that \((\sqrt{6} - \sqrt{2})^2 = 8 - 2\sqrt{12} = 8 - 4\sqrt{3}\) and since \(49 > 16 \times 3\) \(\sqrt{6}-\sqrt{2} > 1\). Therefore we can assume without loss of generality that \(P\) and \(Q\) both do not lie on the same side as each other, a side containing \(O\), otherwise one of those lengths would be \(1 \text{ cm} < (\sqrt{6}-\sqrt{2}) \text{ cm} \). Let \(O = (0,0)\), \(P = (1,x)\), \(Q = (y,1)\), then our lengths squared are: \(1 + x^2, 1 + y^2, (1-x)^2+(1-y)^2\). To maximise the length of the smallest side, each side should be equal in length (otherwise we could increase the length of the smallest side by moving the point between the shortest side and the longest side (without affecting the other side). Therefore \(x = y\) and \(1+x^2 = 2(1-x)^2 \Rightarrow x^2-4x+1 = 0 \Rightarrow x = 2 - \sqrt{3} \). Therefore the distances are all \(\sqrt{1+7-4\sqrt{3}} = \sqrt{8-4\sqrt{3}} = (\sqrt{6}-\sqrt{2}) \text{ cm}\)