MHB Prove AB CD + AD BC < AC(AB+AD) for Convex Quadrilateral ABCD

  • Thread starter Thread starter anemone
  • Start date Start date
  • Tags Tags
    2017
AI Thread Summary
In a convex quadrilateral ABCD, the sum of angles A and C is less than 180 degrees. The goal is to prove that the inequality AB·CD + AD·BC < AC(AB + AD) holds true. The discussion includes a suggested solution that outlines the steps to demonstrate this inequality. Participants are encouraged to engage with the problem and provide their insights or alternative solutions. The thread emphasizes the importance of collaboration in solving mathematical challenges.
anemone
Gold Member
MHB
POTW Director
Messages
3,851
Reaction score
115
Here is this week's POTW:

-----

The sum of the angles $A$ and $C$ of a convex quadrilateral $ABCD$ is less than $180^\circ$.

Prove that $AB\cdot CD+AD \cdot BC <AC(AB+AD)$.

-----

Remember to read the http://www.mathhelpboards.com/showthread.php?772-Problem-of-the-Week-%28POTW%29-Procedure-and-Guidelines to find out how to http://www.mathhelpboards.com/forms.php?do=form&fid=2!
 
Physics news on Phys.org
No one answered last week's problem.(Sadface)

You can find the suggested solution below:

Let $O$ be the circumcircle $ABD$. Then the point $C$ is outside this circle, but inside the $\angle BAD$. Let us apply the inversion with the center $A$ and radius $1$. This inversion maps the circle $O$ to the line $O'=B'D'$, where $B'$ and $D'$ are images of $B$ and $D$. The point $C$ goes to the point $C'$ inside the triangle $AB'D'$. Therefore $B'C'+C'D'<AB'+AD'$. Now, due to inversion properties, we have

$B'C'=\dfrac{BC}{AB\cdot AC},\,C'D'=\dfrac{CD}{AC\cdot AD},\,AB'=\dfrac{1}{AB},\,AD'=\dfrac{1}{AD}$

Hence

$\dfrac{BC}{AB\cdot AC}+\dfrac{CD}{AC\cdot AD}<\dfrac{1}{AB}+\dfrac{1}{AD}$

Multiplying the above by $AB\cdot AC \cdot AD$ the result follows.
 
Back
Top