Hello everyone! I'm getting stuck at this problem and I dont know how to solve it. I'm just curious about it. Any help will be greatly appreciated

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- Nov 23rd 2012, 07:48 PMchubakuenoGeometry problems
Hello everyone! I'm getting stuck at this problem and I dont know how to solve it. I'm just curious about it. Any help will be greatly appreciated

Attachment 25877 - Nov 24th 2012, 03:59 AMBobPRe: Geometry problems
Could you resubmit, including a diagram, with labels for the various points so that we can refer to lines and angles ?

- Nov 24th 2012, 05:31 AMchubakuenoRe: Geometry problems
thanks for your attention, here is the diagram

Attachment 25900 - Nov 24th 2012, 11:08 AMearbothRe: Geometry problems
1. You are dealing with 2 right triangles: $\displaystyle \Delta(ABD)$ and $\displaystyle \Delta(ACE)$.

So the point A is located on a half circle (circle of Thales) over BD and point A is located on a half circle (circle of Thales) over CE.

2. You certainely have made an exact drawing(?). See attachment.

a) The point A is the intersection of the green circle and the circle of Thales over BD. The center of the green circle has the coordinates $\displaystyle \left(\frac32 , \frac32 \right)$ with the radius $\displaystyle \frac32 \sqrt{2}$ (why?)

b) Draw the line AC. Construct a right angle in A on AC. The leg of this right angle intersects the prolonged line BD in E.

3. I've done the construction in a coordinate system so you can read the length of x.

4. The line segment BD is divided harmonically by C (inner point) and E (outer point). According to the harmonic partition you'll get the proportion:

$\displaystyle \frac{BC}{CD}=\frac{BE}{DE}$

With your values:

$\displaystyle \frac{3}{2}=\frac{5+x}{x}$

Solve for x. - Nov 24th 2012, 04:00 PMchubakuenoRe: Geometry problems
No, it was just illustrative, but that's the way the problem was given to me. Thank you!

- Nov 25th 2012, 01:41 AMBobPRe: Geometry problems
My solution is rather more workmanlike !

Start with the right angled triangle ABD and let the side AB = a.

Then

$\displaystyle a^{2}+AD^{2}=5^{2}, $ so $\displaystyle AD = \sqrt{25-a^{2}}.$

Now let the angle ACD = $\displaystyle \theta,$ and use the sine rule in each of the triangles ABC and ACD.

From ACD,

$\displaystyle \frac{2}{\sin 45}=\frac{\sqrt{25-a^{2}}}{\sin \theta},$

so

$\displaystyle \sin \theta = \frac{\sin 45\sqrt{25-a^{2}}}{2}.$

From ABC,

$\displaystyle \frac{a}{\sin(180-\theta)}=\frac{3}{\sin 45},$

so

$\displaystyle \sin \theta=\frac{a\sin 45}{3}.$

Equate the two expessions for $\displaystyle \sin \theta, $ simplify, and you find that $\displaystyle a=\frac{15}{\sqrt{13}},$

and from which it follows that AD =$\displaystyle 10/\sqrt{13}.$

Call the angle ABD $\displaystyle \phi, $ then from the triangle ABD, $\displaystyle \tan \phi=\frac{10\sqrt{13}}{15\sqrt{13}}=\frac{2}{3},$ and from that you can deduce that $\displaystyle \sin \phi=2/\sqrt{13} $ and $\displaystyle \cos \phi=3/\sqrt{13}.$

Now, finally, turn your attention to the triangle at the other end, ADE.

The angle AED will equal $\displaystyle 45 - \phi, $ so making use of the sine rule again,

$\displaystyle \frac{AD}{\sin (45-\phi)}=\frac{x}{\sin 45},$

so

$\displaystyle x = \frac{AD\sin 45}{\sin (45 - \phi)}=\frac{10\sin 45}{\sqrt{13}(\sin 45 \cos \phi-\cos 45 \sin \phi)}. $

$\displaystyle \sin 45 = \cos 45, $ so after cancelling and then substituting for $\displaystyle \cos \phi $ and $\displaystyle \sin \phi,$

we finish up with

$\displaystyle x = \frac{10}{\sqrt{13}(3/\sqrt{13}-2/\sqrt{13})}=10.$