Hii !

Can You Help me For This Exercice :

http://img189.imageshack.us/img189/7995/problem003.gif

Question is : Find x !

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- Feb 2nd 2010, 02:44 PMPerelmanFind x !
Hii !

Can You Help me For This Exercice :

http://img189.imageshack.us/img189/7995/problem003.gif

Question is : Find x ! - Feb 2nd 2010, 02:48 PMe^(i*pi)
- Feb 3rd 2010, 03:22 AMPerelman
- Feb 3rd 2010, 05:07 AMAladdin
- Feb 4th 2010, 09:27 PMchosoi
first we have B = 180 - A - C = 180 - 3x - x

then we also find that B = B1+B2 (2 small triangle from left to right)

B1 = 180 - (A + D1) = 180 - (45 + 3x) = 135 - 3x

B2 = 180 - D2 - C = 180 - (180 - 45) - x = 45 - x

since B = B1 + B2 = 135 -3x + 45 - x = 180 - 3x - x

look up on the first line ( is this look familiar?)

result B = B1 + B2 = A + C

Now we got

B = A + C

and B = 180 - A - C

so A + C = 180 -A - C

=> A + C + A + C = 180

=> 2A + 2C = 180

plug in the real value A = 3x and C = x

=> 2(3x) + 2x = 180

=> 6x + 2x = 180

=> 8x = 180

=> x = 22.5

The End!(Hi) - Feb 5th 2010, 06:54 AMArchie Meade
From the attached sketch

$\displaystyle 3x+b=180^o-45^o=135^o=3(45^o)$

$\displaystyle x+a=180^o-135^o=45^o$

Therefore 3x+b = 3(x+a) = 3x+3a

Hence b = 3a.

When the third apex is on the circle, x=a and 3x=b=3a, so 2x=45 degrees, so x=22.5 degrees.

When it isn't...

If...

$\displaystyle (3x+z)=3(x+e)$

then, as

$\displaystyle (x+e)+c=45^o$

$\displaystyle (3x+z)+y=135^o=3(45^o)$

this means

$\displaystyle y=3c$

The third apex of triangles ABD and ABC may be__anywhere__on the red line,

__above the horizontal__.

Therefore x ranges from 0 to 45 degrees.

However, the interior triangles may no longer touch at B.

To find out if x can be any other angle than 22.5^o...

we can use the Law of Sines.

$\displaystyle \frac{Sin3x}{|BD|}=\frac{Sinb}{k},\ \frac{Sinx}{|BD|}=\frac{Sina}{k}$

$\displaystyle |BD|=\frac{kSin3x}{Sinb}=\frac{kSinx}{Sina}$

b=3a

$\displaystyle \frac{Sin3x}{Sin3a}=\frac{Sinx}{Sina}$

$\displaystyle Sin3x=3Sinx-4Sin^3x$

$\displaystyle \frac{3Sinx-4Sin^3x}{3Sina-4Sin^3a}=\frac{Sinx}{Sina}$

$\displaystyle \frac{Sinx\left(3-4Sin^2x\right)}{Sina\left(3-4Sin^2a\right)}=\frac{Sinx}{Sina}$

$\displaystyle 3-4Sin^2x=3-4Sin^2a$

$\displaystyle Sin^2x=Sin^2a$

$\displaystyle Sinx={\pm}Sina$

As x and "a" are acute, Sinx=Sina, x=a.

Thus x=22.5 degrees. - Feb 5th 2010, 12:17 PMOpalg
That argument assumes that the two green lines intersect at a point on the red line. But that is not the case. If z = 3e ≠ 0 then the two green lines will not intersect on the red line. Alternatively, if the two green lines do intersect on the red line then it is impossible to have z = 3e (or y = 3c) unless z = e = 0.

The only solution to the problem is x = 22.5º. - Feb 5th 2010, 12:38 PMArchie Meade
Yes,

i forgot to maintain the base dimensions equal, sorry!

Corrections are applied.