ABCDEFGH is a regular convex octagon, with $A (0, 1), B (\sqrt{2}, 1), E (\sqrt{2}, - 1 - \sqrt{2}) and F (0, - 1 - \sqrt{2})$ $A( 0, 1 ), B( sqrt2, 1 ), E( sqrt 2, -1-sqrt 2 ) and F( 0, -1-sqrt2 )$ . How do you find the coordinates of the remaining vertices? .

Question

ABCDEFGH is a regular convex octagon, with $A (0, 1), B (\sqrt{2}, 1), E (\sqrt{2}, - 1 - \sqrt{2}) and F (0, - 1 - \sqrt{2})$ $A( 0, 1 ), B( sqrt2, 1 ), E( sqrt 2, -1-sqrt 2 ) and F( 0, -1-sqrt2 )$ . How do you find the coordinates of the remaining vertices? .

2 Answers

Impact of this question

3024 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-02-08T19:51:50

Not so sure, but here is what I think one would do if they had solve this problem (or a similar one with different points).

Average all x values and y values

$\frac{0 + \sqrt{2} + \sqrt{2} + 0}{4}$ $(0+sqrt(2)+sqrt(2)+0)/4$ and $\frac{1 + 1 + (- 1 - \sqrt{2}) + (- 1 - \sqrt{2})}{4}$ $(1+1+(-1-sqrt(2))+(-1-sqrt(2)))/4$

So the center point is $(\frac{\sqrt{2}}{2}, - \frac{\sqrt{2}}{2})$ $((sqrt(2))/2,-sqrt(2)/2)$

Translate the center point to zero (change all x values by $- \frac{\sqrt{2}}{2}$ $-(sqrt(2))/2$ and increase all y values by $+ \frac{\sqrt{2}}{2}$ $+sqrt(2)/2$ )

Then, rotate each point by 90 degrees. ( $x = y and y = - x$ $x = y and y = -x$ )

Finally, translate each of these new points back to their original positions. (increase all x values by $+ \frac{\sqrt{2}}{2}$ $+(sqrt(2))/2$ and change all y values by $- \frac{\sqrt{2}}{2}$ $-sqrt(2)/2$ )

Depending on how the octagon was set up,
A' = G, B'=H, F'=D, E'=C

Answer 2 · 2017-02-09T03:42:15

$C (1 + \sqrt{2}, 0), D (1 + \sqrt{2}, - \sqrt{2}), G (- 1, - \sqrt{2}) and H (- 1, 0)$ $C(1+sqrt2, 0), D(1+sqrt2, -sqrt2), G(-1, -sqrt2) and H(-1, 0)$

Explanation:

As AB is in y-direction, side of the octagon L = $y_{B} - y_{A} = \sqrt{2}$ $y_B-y_A=sqrt2$

From the averages of the coordinates of A, B, E and F, the center M

has coordinates

#(x_M, y_M)=(1/sqrt2, -1/sqrt2).

C, D, G and H are equidistant from M. in the directions of the x and y

axes. These have the common distances

d_x = (side of octagon) $\times (\frac{1}{2} + {cos 45}^{o})$ $xx(1/2+cos45^o)$

$= \sqrt{2} (\frac{1}{2} + \frac{1}{\sqrt{2}}) = 1 + \frac{1}{\sqrt{2}}$ $=sqrt2(1/2+1/sqrt2)=1+1/sqrt2$

$d_{y} = \frac{1}{2} L = \frac{1}{\sqrt{2}}$ $d_y=1/2L=1/sqrt2$

From symmetry, the remaining vertices are

$C (x_{M} + d_{x}, y_{M} + d_{y}),$ $C(x_M+d_x, y_M+d_y),$ giving $C (1 + \sqrt{2}, 0)$ $C(1+sqrt2, 0)$

$D (x_{M} + d_{x}, y_{M} - d_{y}),$ $D(x_M+d_x, y_M-d_y),$ giving $D (1 + \sqrt{2}, - \sqrt{2})$ $D(1+sqrt2, -sqrt2)$

$G (x_{M} - d_{x}, y_{M} \pm d_{y}),$ $G(x_M-d_x, y_M+-d_y),$ giving G(-1, -sqrt2)#

$H (x_{M} + d_{x}, y_{M} + d_{y}),$ $H(x_M+d_x, y_M+d_y),$ giving H(-1, 0)#

Science

Math

Humanities

... and beyond

ABCDEFGH is a regular convex octagon, with $A (0, 1), B (\sqrt{2}, 1), E (\sqrt{2}, - 1 - \sqrt{2}) and F (0, - 1 - \sqrt{2})$ $A( 0, 1 ), B( sqrt2, 1 ), E( sqrt 2, -1-sqrt 2 ) and F( 0, -1-sqrt2 )$ . How do you find the coordinates of the remaining vertices? .

2 Answers

Explanation:

Related questions

Impact of this question

Science

Math

Humanities

... and beyond

ABCDEFGH is a regular convex octagon, with A(0,1),B(√2,1),E(√2,−1−√2)andF(0,−1−√2)A( 0, 1 ), B( sqrt2, 1 ), E( sqrt 2, -1-sqrt 2 ) and F( 0, -1-sqrt2 ). How do you find the coordinates of the remaining vertices? .

2 Answers

Explanation:

Related questions

Impact of this question

ABCDEFGH is a regular convex octagon, with $A (0, 1), B (\sqrt{2}, 1), E (\sqrt{2}, - 1 - \sqrt{2}) and F (0, - 1 - \sqrt{2})$ $A( 0, 1 ), B( sqrt2, 1 ), E( sqrt 2, -1-sqrt 2 ) and F( 0, -1-sqrt2 )$ . How do you find the coordinates of the remaining vertices? .