prove

**Archie Meade** · August 13th 2010, 08:14 AM

Originally Posted by yehoram

Here is a drawing of the problem !

Due to symmetry, if angle ACE is 30 degrees, then triangle AGC is equilateral.

If we use the corners of the square as circle centres, with radius "x",
and using the fact that the angle at the centre of a circle is twice the angle at the circumference
for angles standing on a common arc,

then it follows that if angle ACE is 30, then the angles ADE and AED are 15 degrees.

Triangle AGC is equilateral as all sides are R.
The angles of AGC are all $4\theta$ , so $\theta=15^o$

**Archie Meade** · August 13th 2010, 10:02 AM

Maybe simplest is to solve for $\beta$ in the diagram.

**Vlasev** · August 13th 2010, 12:49 PM

Originally Posted by sa-ri-ga-ma

From the picture it is clear that angle AEM = EMQ = 15 degrees.

The picture, however is not true, nor is it accurate. You cannot just make a statement like that.

EDIT: Your last post Archie, that's a great solution!

**bjhopper** · August 14th 2010, 03:25 AM

Hello yehoram,
Your problem was poorly worded,corrected by diagram but failing to redefine the. givens
At C draw an arc r=1 from Dto B and at B an arc fromCto E.These arcs meet at A along the perpendicular bisector of CB and DE
DA is a chord in a quarter circle. EDA is given 15 DE is tangent to circle at D
Arc DA = 30 Angle DCA =30 Angle ACB=60

bjh

**sa-ri-ga-ma** · August 14th 2010, 07:39 PM

Originally Posted by Vlasev

The picture, however is not true, nor is it accurate. You cannot just make a statement like that.

EDIT: Your last post Archie, that's a great solution!

Which picture is not correct?

What is wrong in my statement?

I have drawn PAQ parallel to DME. EMAQ is a rectangle. Angle AEM = 15 degrees is given in the problem. In the rectangle EMAQ angle MEA = Angle EMQ = 15 degrees. R is the point of intersection of the sector and MA. I have drawn SRT parallel to EMD.

**Vlasev** · August 14th 2010, 10:53 PM

Originally Posted by sa-ri-ga-ma

Which picture is not correct?

What is wrong in my statement?

I have drawn PAQ parallel to DME. EMAQ is a rectangle. Angle AEM = 15 degrees is given in the problem. In the rectangle EMAQ angle MEA = Angle EMQ = 15 degrees. R is the point of intersection of the sector and MA. I have drawn SRT parallel to EMD.

In the picture the angle is close to 30 degrees not 15. Look at Archie Meade's last post. That drawing is much more accurate. If your method works with it, I think it'll be fine!

**obidobi** · August 19th 2010, 09:08 PM

This reply might be a bit late. But why make it harder than it is

.

From your drawing

$\angle{DAE} = 180 - 15 - 15 = 150$

$\angle{ADB}=\angle{AEC} = 90 - 15 = 75$

$\angle{EAC}=\angle{DAB} = \beta$

If $\bigtriangleup{ABC}$ is equilateral set all its angles to $\alpha$

This gives us:
$150 + 2*\beta + \alpha = 360 \Rightarrow \alpha = 210 - 2*\beta$
$75 + \beta + (90-\alpha) = 180 \Rightarrow \beta = 15 - \alpha$
$\alpha = \frac{210-30}{3} = 60$

Which proves that $\bigtriangleup{ABC}$ is equilateral.

**Vlasev** · August 19th 2010, 11:16 PM

Originally Posted by obidobi

This reply might be a bit late. But why make it harder than it is

.

From your drawing

$\angle{DAE} = 180 - 15 - 15 = 150$

$\angle{ADB}=\angle{AEC} = 90 - 15 = 75$

$\angle{EAC}=\angle{DAB} = \beta$

If $\bigtriangleup{ABC}$ is equilateral set all its angles to $\alpha$

This gives us:
$150 + 2*\beta + \alpha = 360 \Rightarrow \alpha = 210 - 2*\beta$
$75 + \beta + (90-\alpha) = 180 \Rightarrow \beta = 15 - \alpha$
$\alpha = \frac{210-30}{3} = 60$

Which proves that $\bigtriangleup{ABC}$ is equilateral.

I'm afraid that is circular reasoning!

**umangarora** · August 20th 2010, 05:12 AM

If CAD is isoceles.. the lines are diagnols dude

**umangarora** · August 20th 2010, 05:31 AM

thats correct.. its circular reasoning.. uve taken a preusumption that the triangle is equ. and proved it.. doesnt amount to nething..

**obidobi** · August 20th 2010, 08:02 AM

Originally Posted by Vlasev

I'm afraid that is circular reasoning!

Originally Posted by umangarora

thats correct.. its circular reasoning.. uve taken a preusumption that the triangle is equ. and proved it.. doesnt amount to nething..

Yes. I should have turned on my brain before thinking

Oki lets do this the right way then:

Known:
$x = a + b$
$x = 2h$

Problem: Find a value for $a$ so that $r_0 = 2h = x$ , then left triangle is equilateral

Pythagoras gives:
$a^2 + h^2 = r_0^2$
$a^2 + (\frac{x}{2})^2 = x^2$
$a = \frac{\sqrt{3}}{2}x$
$b = \frac{2-\sqrt{3}}{2}x$

$\tan{\alpha} = \frac{b}{h}$
$\alpha = \arctan{(2-\sqrt{3})}$
$\alpha = 15^{\circ}$

**razemsoft21** · August 20th 2010, 03:59 PM

Originally Posted by yehoram

Hi ! I have some problem with this below !

Subjected to a square which has two isosceles triangles.
One of them is: a triangle whose base is one of the sides of the square root angles are 15.
The other is a triangle whose base is the side opposite the base of the triangle it first touches the apex of the triangle at the end of the first node.
Proved that the second triangle equilateral!

The prove must be geometric way , no trigonometric ! Please help me ! thanks all !

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