From all corners

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(A) Apply the sine rule to triangle $OMP$:

(B) Consider just the square in the bottom left-hand quarter of the diagram. Assume that its sides are 2 units long. The diagonal lines coming from the corners will still go through the midpoints of the sides. Line $AO$ has been added, and it is a line of symmetry.

(C) Angle $M\hat{O}P = 45^\circ$, and so angle $O\hat{P}M = 180^\circ-45^\circ-63.43...=71.56...$

(D) Triangle $OBM$ is right-angled with sides of length 1 and 2, and contains angle $O\hat{M}B$, so $O\hat{M}B=\text{tan}^{-1}\left(\frac21\right) = 63.43...^\circ$

(E) The area of triangle $OMP$ can be found using $\frac12ab\sin{C}$:

(F) Since the area of the whole square is $4$, and the area of each half of the part of the octagon is $\frac1 3$, the octagon occupies $\dfrac{\frac23}4 = \dfrac{\frac13}2=\frac16$ of the area of the square.

(J) $\dfrac {MP}{\sin{45}} = \dfrac{1}{\sin{71.56...}} \Rightarrow MP=\dfrac {\sin{45}}{\sin{71.56}} = 0.74... $ or $\frac{\sqrt5}{3}$

(K) $\frac12\times 1 \times \frac {\sqrt 5}{3} \times \sin{63.43...} = \frac 1 3$

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(A) P lies on $L$, so $b=-2a+1$

(B) Line $M$ has gradient $-\frac12$

(C) Put the origin (0,0) at the centre of the diagram, and let the top right-hand corner of the square be at (1,1). Consider only the part of the diagram that lies in the first quadrant.

(D) $\left(\frac13\right)^2 + \frac12\times\left(\frac13\right)\left(\frac12-\frac13\right)\times2$

      $=\frac19 + \left(\frac13\right)\left(\frac12-\frac13\right)$

      $=\frac19 + \frac13 \times \frac16$

      $=\frac19+\frac1{18}$

      $=\frac3{18}=\frac16$

(E) Let the lines drawn to the midpoints (0,1) and (1,0) be $L$ and $M$ respectively. $L$ meets the vertex at (1,-1), so it also goes through the point ($\frac12$,0) and similarly $M$ goes through the point (0,$\frac12$). $L$ and $M$ meet at P, with coordinates ($a$, $b$).

(F) Line $L$ has equation $y=-2x+1$

(G) $b = - 2\times\frac13 + 1 =\frac13$

(H) Since the total area of this quarter of the diagram is $1$, the fraction of the area occupied by the octagon is $\frac16$.

(I) Line $L$ has gradient $-2$

(J) $-2a+1=-\frac12a+\frac12$

    $\Rightarrow -4a + 2 = -a+1$

    $\Rightarrow 1 = 3a$

    $\Rightarrow a=\frac13$

(K) Line $M$ has $y$ intercept $\frac12$

(L) P lies on $M$, so $b=-\frac12a+\frac12$

(M) The area inside the octagon is equal to the area of the two small right-angled triangles plus the area of the square. That is equal to:

(N) Line $L$ has $y$ intercept $1$

(O) Line $M$ has equation $y=-\frac12x+\frac12$

 

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(A) Since the angles in triangle $ACM$ add up to $180^\circ$, angle $A\hat{M}C = 90 - 26.56... = 63.43...^\circ$

(B) $\Rightarrow \frac35\times MP = \frac{1}{\sqrt5}$

      $\Rightarrow MP = \frac {\sqrt5}{3}$ or $0.745...$

(C) Angle $M\hat{A}C = \text{tan}^{-1}\left(\frac12\right)=26.56...^\circ$

(D) Consider just the square in the bottom left-hand quarter of the diagram. Assume that its sides are 2 units long. The diagonal lines coming from the corners will still go through the midpoints of the sides. Points $W$ and $X$ have been added where vertical and horizontal lines from $P$ meet the edges of the diagram.

(E) The angles in triangle $CMQ$ are equal to the angles in triangle $ACM$, so triangle $CQM$ is similar to triangle $ACM$, and $AM$ and $CN$ are perpendicular

(F) The area of square $OWPX$ is $\left(\frac23\right)^2 = \frac49$, so the area of the part of the diagram that is in the octagon is $\frac69=\frac23$

(G) Triangle $ACM$ is congruent to triangle $CON$, so angles $N\hat{C}O = Q\hat{C}M$ and $M\hat{A}C$ are equal

(H) From triangle $MPQ$, $\cos{53.13...}=\frac{MQ}{MP}$

(I) So $MW = \frac13$ and $WP = \frac23$

(J) So the hypotenuse of triangle $MPW$ is $\frac{\sqrt 5}{3}$ or $0.745...$

(K) The angles on the line $CO$ at point $M$ add up to $180^\circ$, so angle $A\hat{M}B = 180^\circ - 2\times 63.43...^\circ = 53.13...^\circ$

(L) The total area of this diagram is $2\times2=4$, so the fraction that the octagon occupies is $\dfrac{\frac23}{4} = \dfrac16$

(M) So the scale factor from triangle $ACM$ to triangle $MPW$ is $\frac13$

(N) From triangle $CMQ$, $\sin{26.56...}=\frac{MQ}{1} \Rightarrow MQ = \frac{1}{\sqrt5}$ or $0.447...$

(O) The area of triangle $MPW$ is $\frac12\times\frac13\times\frac23 = \frac19$, so the areas of triangles $MPW$ and $NPX$ add up to $\frac29$

(P) Triangle $MPW$ is similar to triangle $ACM$, and the hypotenuse of triangle $ACM$ can be found using Pythagoras' Theorem : $1^2 + 2^2 = 5$ so $AM=\sqrt5$

(Q) Triangle $MBO$ is congruent to triangle $ACM$, so angle $B\hat{M}O = P\hat{M}W =63.43...^\circ$

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(A) Consider just the square in the bottom left-hand quarter of the diagram. Assume that its sides are 2 units long. The diagonal lines coming from the corners will still go through the midpoints of the sides. Line $AO$ has been added, and it is a line of symmetry.

(B) Applying Pythagoras' Theorem to triangles $ACM$ and $CON$ gives $1^2+2^2 = 5$, so $AM=CN=\sqrt{5}$

(C) Quadrilateral $ACMP$ has perpendicular diagonals, so its area can be found by multiplying its diagonals and dividing by $2$ (since it contains $4$ right-angled triangles). So the area of quadrilateral $ACMP$ is $AM\times CP\div2$

(D) Triangle $ACM$ is right-angled, and so angles $A\hat{M}C$ and $M\hat{A}C$ add up to $90^\circ$

(E) Triangle $ACM$ is congruent to triangle $CON$, so angles $N\hat{C}O = Q\hat{C}M$ and $M\hat{A}C$ are equal

(F) $PM=PN$ by symmetry

(G) Triangle $CQM$ has hypotenuse $1$, so the scale factor from triangle $CQM$ to triangle $ACM$ is $\sqrt5$. So $CQ = \frac2{\sqrt5}$ and $MQ=\frac1{\sqrt5}$ (or $0.894$ and $0.447$)

(H) So the total area of quadrilateral $ACMP$ and quadrilateral $ABNP$ is $\frac{10}3$, and the total area of the diagram is $2\times2 = 4$. So $\dfrac{\frac{10}3}{4} = \dfrac{\frac{20}{6}}{4}=\dfrac{5}{6}$ of the diagram is outside of the octagon, which means the octagon occupies $\frac16$ of the diagram.

(I) So $AM$ and $CN$ are perpendicular, and triangle $CQM$ is similar to triangle $ACM$

(J) $\Rightarrow \left(\frac1{\sqrt5}\right)^2 + QP^2 = PN^2$

      $\Rightarrow \frac15 + QP^2 = \left(\frac 3 {\sqrt5} - QP\right)^2$

      $\Rightarrow \frac15 + QP^2= \frac95 - \frac 6 {\sqrt5} QP + QP^2$

      $\Rightarrow \frac 15 = \frac 95 - \frac 6{\sqrt5} QP$

      $\Rightarrow \frac 6 {\sqrt5} QP = \frac 85$

      $\Rightarrow 6QP = \frac 85 \times \sqrt5 =\frac{8\sqrt5}5$

      $\Rightarrow QP =\frac{8\sqrt5}{5\times6} = \frac {4\sqrt5}{15}$ or $0.596$

(K) $CN = CQ + QP + PN \Rightarrow \sqrt5= \frac2 {\sqrt5} + QP + PN \Rightarrow QP + PN = \frac 3 {\sqrt5}$ or $1.342$, so $PN = \frac 3 {\sqrt5} - QP$ or $1.342-QP$

(L) $= \sqrt5 \times \left( \frac2{\sqrt5} + \frac {4\sqrt5}{15}\right)\div 2$

      $= \left( 2 + \frac {4\times5}{15} \right)\div2$

      $= \left( 2 + \frac 4 3\right)\div 2$

      $= \frac {10}3\div2$

      $= \frac 53$

(M) Therefore angles $Q\hat{C}M$ and $C\hat{M}Q$ add up to $90^\circ$, and so triangle $CQM$ is right-angled

(N)  Applying Pythagoras' Theorem to triangle $MPQ$ gives $MQ^2 + QP^2 = PM^2$