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April 22nd, 2018, 07:25 AM   #1
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Minimal value of the sum of two line segments

Quote:
 $\displaystyle ABCD$ is square with side $\displaystyle a$. There is a point E on side $\displaystyle AB$, so that $\displaystyle AE : EB = 2 : 1$. Let F be an arbitrary point on diagonal $\displaystyle BD$. Prove that $\displaystyle AF + EF \geqslant \frac{a \sqrt{10}}{3}$.
I proved that $\displaystyle AF + EF \geqslant \frac{a \sqrt{12}}{3}$.
It's the case when $\displaystyle AF \geqslant \frac{AC}{2}$, and $\displaystyle EF \geqslant \frac{a}{3\sqrt{2}}$ (when $\displaystyle EF \perp BD$).
Do you know how to prove the original inequality?
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Last edited by lua; April 22nd, 2018 at 07:27 AM.

April 24th, 2018, 03:13 AM   #2
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Reflect point $\displaystyle E$ around diagonal $\displaystyle BD$ to hit side $\displaystyle BC$ at point $\displaystyle G$. Connect $\displaystyle AG$ hitting the diagonal at point $\displaystyle F$.

Now $\displaystyle AF+EF=AF+FG=AG$. By the Pythagorean Theorem $\displaystyle AG = \dfrac{a}{3}\sqrt{10}$.

Sorry my picture is upside down from yours.
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Last edited by mrtwhs; April 24th, 2018 at 03:21 AM.

 April 24th, 2018, 04:51 AM #3 Member   Joined: Jan 2018 From: Belgrade Posts: 55 Thanks: 2 O, great. I see. If $\displaystyle F$ was anywhere else on $\displaystyle BD$, it would be: $\displaystyle AF+EF=AF+FG>AG$. So, $\displaystyle AG$ is minimal value of $\displaystyle AF+EF$. Thank you, very much.

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