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Proof of a "Geometry"

Albert

Well-known member
Jan 25, 2013
1,225
AE is a diameter of a semicicle O,points B,C,D are three points on this semicicle
Given :AB=a ,BC=b ,CD=c ,DE=d ,and AE=2
please prove :
$a^2+b^2+c^2+d^2+abc+bcd < 4$

semicircle.jpg
 
Last edited:

MarkFL

Administrator
Staff member
Feb 24, 2012
13,775
Because of the symmetry between the pairs $(a,d)$ and $(b,c)$, we may reduce the problem to one of two variables:

Prove:

\(\displaystyle a^2+b^2+ab^2<2\)

where $a$ and $b$ are found from the following diagram:

albert01.jpg

Using the Law of Cosines, we find:

\(\displaystyle a^2=2-2\cos(\alpha)\)

\(\displaystyle b^2=2-2\cos(\beta)\)

Since $\alpha$ and $\beta$ are complementary, we now have the objective function in one variable (after dividing through by 2):

\(\displaystyle f(\alpha)=2-\left(\cos(\alpha)+\sin(\alpha) \right)+\sqrt{2-2\cos(\alpha)}\left(1-\sin(\alpha) \right)<1\)

Differentiating with respect to $\alpha$ and equating to zero, we may find the critical value(s) (using a numeric root-finding technique) on the relevant interval:

\(\displaystyle \alpha\approx1.06048703739259044748\)

Thus, checking the end-points of the interval, we find:

\(\displaystyle f(0)=1\)

\(\displaystyle f(1.06048703739259044748)\approx0.767829392509715815674068\)

\(\displaystyle f\left(\frac{\pi}{2} \right)=1\)

Thus, we may conclude that for \(\displaystyle 0<\alpha<\frac{\pi}{2}\) we must have:

\(\displaystyle f(\alpha)<1\)

which is what we needed to show.
 

Albert

Well-known member
Jan 25, 2013
1,225
AE is a diameter of a semicicle O,points B,C,D are three points on this semicicle
Given :AB=a ,BC=b ,CD=c ,DE=d ,and AE=2
please prove :
$a^2+b^2+c^2+d^2+abc+bcd < 4$

View attachment 968
connecting AC, and CE ,from law of cosine:

$AC^2=a^2+b^2-2ab\, cos B=a^2+b^2+2ab\,cos \angle AEC>a^2+b^2+abc(1)$

likewise :

$CE^2>c^2+d^2+bcd(2)$

now using pythagorean theorem (1)+(2) we have :

$AC^2+CE^2=AE^2=4>a^2+b^2+c^2+d^2+abc+bcd$