Author Topic: Recreational Problems (SM level) (Read 96292 times) Tweet Share

dcc · « **Reply #195 on:** July 28, 2008, 08:03:28 pm »

Quote from: bigtick on July 19, 2008, 10:56:20 pm

Show $cos\frac{\pi }{5}+cos\frac{3\pi }{5}=\frac{1}{2}$ .

Since $\cos 5u = 5 \cos u - 20 \cos^3 u + 16 \cos^5 u$

$-1 = 5x - 20x^3 + 16x^5 \implies 0 = 16x^5 - 20x^3 + 5x + 1\ \mbox{(where $x = \cos \frac{\pi}{5}$)}$

$0 = \left(x + 1\right) \left(4x^2 - 2x - 1\right)^2$

$4x^2 - 2x - 1 = 0 \implies x^2 - \frac{1}{2} x - \frac{1}{4} = 0 \implies \left( x - \frac{1}{4}\right)^2 = \frac{5}{16}$

$\implies \cos \frac{\pi}{5} = \frac{\sqrt{5} + 1}{4}$

Since $\cos 3u = 4 \cos^3 u - 3 \cos u$

$\cos \frac{3\pi}{5} = 4 \left(\frac{\sqrt{5} + 1}{4}\right)^3 - \frac{3}{4} \left(\sqrt{5} + 1\right) \implies \cos \dfrac{\pi}{5}\ +\ \cos \frac{3\pi}{5} = 4 \left(\frac{\sqrt{5} + 1}{4}\right)^3 -\ \frac{1}{2} \left(\sqrt{5} + 1\right)$

$A = \frac{1}{16} \left( 5 \sqrt{5} + 15 + 3 \sqrt{5} + 1 \right) - \frac{8}{16} \left( \sqrt{5} + 1\right) \implies A = \frac{1}{2}$

bigtick · « **Reply #196 on:** July 28, 2008, 10:16:16 pm »

Think in a complex way since you like complexity.

Ahmad · « **Reply #197 on:** July 29, 2008, 02:08:25 pm »

Quote from: bigtick on July 25, 2008, 11:55:36 pm

Show that $cos\frac{\pi }{11}-cos\frac{2\pi }{11}+cos\frac{3\pi }{11}-cos\frac{4\pi }{11}+cos\frac{5\pi }{11}=\frac{1}{2}$ .

Equivalent to $\sum_{k=1}^5 \cos \frac{2\pi k}{11} = -\frac{1}{2}$

However, $z^{11} - 1 = 0 \implies \sum_{k=0}^{10} \cos \frac{2\pi k}{11} = 0 \implies \sum_{k=1}^{10} \cos \frac{2\pi k}{11} = -1$

By symmetry, $\sum_{k=1}^{5} \cos \frac{2\pi k}{11} = \sum_{k=6}^{10} \cos \frac{2\pi k}{11} \implies \sum_{k=1}^5 \cos \frac{2\pi k}{11} = -\frac{1}{2}$ as required.

I had a nice trigonometry problem but I forgot it

Ahmad · « **Reply #198 on:** July 29, 2008, 03:30:24 pm »

Here's one I made up:

Show that,

$\tan^2\left(\frac{\pi}{1337}\right) \,+\, \tan^2\left(\frac{3\pi}{1337}\right) \,+\, \tan^2\left(\frac{5\pi}{1337}\right) \,+\, \cdots \,+\, \tan^2\left(\frac{1333\pi}{1337}\right) \,+\, \tan^2\left(\frac{1335\pi}{1337}\right) = 893116$

bigtick · « **Reply #199 on:** July 29, 2008, 09:52:10 pm »

One step higher
${tan}^{2}\left(\frac{\pi }{1339} \right)+{tan}^{2}\left(\frac{3\pi }{1339} \right)+{tan}^{2}\left(\frac{5\pi }{1339} \right)+......+{tan}^{2}\left(\frac{1335\pi }{1339} \right)+{tan}^{2}\left(\frac{1337\pi }{1339} \right)=895791$

bigtick · « **Reply #200 on:** July 30, 2008, 01:14:26 pm »

Quote from: Ahmad on July 29, 2008, 02:08:25 pm

Quote from: bigtick on July 25, 2008, 11:55:36 pm
Show that $cos\frac{\pi }{11}-cos\frac{2\pi }{11}+cos\frac{3\pi }{11}-cos\frac{4\pi }{11}+cos\frac{5\pi }{11}=\frac{1}{2}$ .

Equivalent to $\sum_{k=1}^5 \cos \frac{2\pi k}{11} = -\frac{1}{2}$

However, $z^{11} - 1 = 0 \implies \sum_{k=0}^{10} \cos \frac{2\pi k}{11} = 0 \implies \sum_{k=1}^{10} \cos \frac{2\pi k}{11} = -1$

By symmetry, $\sum_{k=1}^{5} \cos \frac{2\pi k}{11} = \sum_{k=6}^{10} \cos \frac{2\pi k}{11} \implies \sum_{k=1}^5 \cos \frac{2\pi k}{11} = -\frac{1}{2}$ as required.

Easier $z^{11} + 1 = 0$

Ahmad · « **Reply #201 on:** July 30, 2008, 10:24:09 pm »

I think both approaches are the same (both viete's followed by symmetry, unless you had something else in mind), depending on how you initially want to transform the problem.

Care to discuss method of solution for the tan problem?

Mao · « **Reply #202 on:** July 31, 2008, 08:35:51 pm »

here's some that normal people might like to have a crack at

1) Express y in terms of x:

$\frac{dy}{dx}+\frac{y}{x+3}=5$ , given that it passes through $(0,7)$

2) Evaluate:

$\int e^{x}\cos (x)\; dx$
(MUEP people stay away

)

phagist_ · « **Reply #203 on:** July 31, 2008, 08:49:43 pm »

$\int e^x cos(x) dx$ = $\int e^x \times Re[e^{ix} dx]$

= $Re[\int e^x \times e^{ix} dx]$

=> $Re[\int e^{{(1+i)}x} dx]$

= $Re[ \frac{1}{1+i} e^{{(1+i)}x}]$

= $Re[ \frac{1-i}{2} e^{{(1+i)}x}]$

= $Re[ \frac{1-i}{2} e^x \times e^{ix}]$

= $Re[ \frac{1-i}{2} e^x \times (cos(x)+isin(x))]$

= $Re[ \frac{e^x}{2} cos(x) - \frac{ie^x}{2} cos(x) + \frac{ie^x}{2}sin(x) + \frac{e^x}{2} sin(x)]$

= $\frac{e^x}{2} cos(x) + \frac{e^x}{2} sin(x)$ + c

also could have used integration by parts.

dcc · « **Reply #204 on:** July 31, 2008, 11:06:40 pm »

Quote from: Mao on July 31, 2008, 08:35:51 pm

$\frac{dy}{dx}+\frac{y}{x+3}=5$ , given that it passes through $(0,7)$

Im never one to pass down a differential equation, so here we go

so: $\left(x + 3\right) \dfrac{dy}{dx} + y = 5 \left(x + 3\right)$

$\left[ \left(x + 3\right) y \right]' = 5x + 15 \implies \left(x + 3\right)y = \int 5x + 15\ dx$

$y = \dfrac{\dfrac{5}{2} x^2 + 15x + C}{x + 3}$ . Given the point $(0,\ 7)$ , we find $C = 21$ , so:

$y = \dfrac{\dfrac{5}{2} x^2 + 15x + 21}{x + 3} = \dfrac{5x^2 + 30x + 42}{2x + 6}$

p.s. what was that mao! weak!

Mao · « **Reply #205 on:** August 01, 2008, 05:17:26 pm »

dcc: I never said it was hard (at least not for you).

dcc · « **Reply #206 on:** August 01, 2008, 06:45:20 pm »

A circle is defined upon the complex plane as follows:

$A = \{ z\ :\ |z - (3 + 4i)| = 3,\ z \in \mathbb{C}\}$

Two points, $a_{1} = -4 - i$ and $a_{2} = -2 - 3i$ are also defined.

Let two points lie on $A$ , $a_{3}$ and $a_{4}$ .

Find the points $a_{3}$ and $a_{4}$ such that the area of the quadrilateral formed by connecting the points $a_{1},\ a_{2},\ a_{3}\ \& \ a_{4}$ is at a maximum.

bigtick · « **Reply #207 on:** August 02, 2008, 08:09:03 am »

a3=5.88+4.85i, a4=2.15+6.88i

dcc · « **Reply #208 on:** August 02, 2008, 04:08:50 pm »

Quote from: bigtick on August 02, 2008, 08:09:03 am

a3=5.88+4.85i, a4=2.15+6.88i

The answer is not important, how you got there is!

By all means, the points you have specified form a figure with an area of roughly $37\ units^2$ , which is not the maximum area possible.

AppleXY · « **Reply #209 on:** August 02, 2008, 05:39:42 pm »

My god. I feel (maybe 'cause I am) so stupid reading this thread. I can't do one

ROFL. Well at least Mao's ones were decent

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Author Topic: Recreational Problems (SM level) (Read 96292 times) Tweet Share

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