# pentagon

• Mar 10th 2011, 03:52 PM
Veronica1999
pentagon
Let A=(0,0), B=(7,2), C=(3,4), D=(3,7) and E=(-1,5). Cameron walks the polygonal path ABCDEA, writing down the number of degrees turned at each corner. What is the sum of these five numbers? Notice that ABCDE in not a convex pentagon.

I got 491.(using a protractor) Could someone take a look at my attached work?
oops, on my last line of attached work i meant if angle c was convex
• Mar 10th 2011, 04:00 PM
TKHunny
Why does convexity make any difference? Don't you have to end up where you started and facing the direction you started?
• Mar 11th 2011, 10:36 AM
bjhopper
Hi Veronica 1999,
I calculated the angle of the first turn at B to be 137.5. At C 116.6and 116.6 at D.Finish and let me know if you get the right answer

bjh
• Mar 11th 2011, 10:45 AM
TheChaz
TKHunny is suggesting a more theoretical approach, with which I agree. Are we explicitly calculating each angle as a means of verification?
• Mar 11th 2011, 01:08 PM
Soroban
Hello, Veronica1999!

Quote:

$\text{Let: }A=(0,0),\;B=(7,2),\;C=(3,4),\;D=(3,7),\;E=(\text{-}1,5).$

$\text{Cameron walks the polygonal path }ABCDEA,$

$\text{writing down the number of degrees turned at each corner.}$

$\text}What is the sum of these five numbers?}$

$\text{Note that }ABCDE\text{ is not a convex pentagon.}$

$\text{I got 491 (using a protractor).}$

While sketching, I found the problem more complex than anticipated.

Code:

      |        D       |        o       |  Y    *       |    . β *       |      . *       |        o                X       |        C  *          .       |            *  α  .       |              o       |          *    B       |      *       |  *     A o - - - - - - - - - - - - -       |

When he walks from $\,A$ to $\,B$, he is walking in the direction $\overrightarrow{AX}.$

To walk to $\,C$, he turns through $\text{angle }\alpha = \angle XBC$ . . . not $\angle ABC.$

Then he walks from $\,B$ to $\,C$, in the direction $\overrightarrow{BY}.$

To walk to $\,D$, he turns through $\text{angle }\beta = \angle YCD$

And so on . . .

• Mar 11th 2011, 04:15 PM
bjhopper
pentagon
I had the right idea but made a mistake in the turn at C I recalculated all the turns with the following results 137,5,63.5,116.5,74.5, 85.35.Is this correct?