Geometry Challenge: Prove $PT+PU\ge 2\sqrt{2}p$

[anemone]

Suppose that $PQRS$ is a square with side $p$. Let $A$ and $B$ be points on side $QR$ and $RS$ respectively, such that $\angle APB=45^{\circ}$. Let $T$ and $U$ be the intersections of $AB$ with $PQ$ and $PS$ respectively. Prove that $PT+PU\ge 2\sqrt{2}p$.

 

[Albert1]

Suppose that $PQRS$ is a square with side $p$. Let $A$ and $B$ be points on side $QR$ and $RS$ respectively, such that $\angle APB=45^{\circ}$. Let $T$ and $U$ be the intersections of $AB$ with $PQ$ and $PS$ respectively. Prove that $PT+PU\ge 2\sqrt{2}p$.

Spoiler

let $QA=QT=SU=SB=x$
where :$0<x<p$,then :$\angle APB=45^o$
by using $AM\geq GM$
if we can find $0<x<p$ and
$PT+PU=2(p+x)\geq 2\sqrt 2 p---(1)$ then the proof is done
the solution of (1) $x=(\sqrt 2-1)p$

 

[Greg]

Spoiler

Let \(\displaystyle \angle{QPA}=\angle{BPS}=22.5^{\circ}\). Let \(\displaystyle V\) be the intersection of \(\displaystyle \overline{PR}\) and \(\displaystyle \overline{AB}\). Note that \(\displaystyle \overline{PT}=\overline{PU}\).

\(\displaystyle \triangle{AQT}\cong\triangle{AVR}\cong\triangle{BVR}\cong\triangle{BSU}\) so \(\displaystyle \frac{\overline{PT}\cdot\overline{PU}}{2}=p^2\).

\(\displaystyle \tan(22.5)=\frac{\overline{QA}}{p}=\sqrt{2}-1\implies\overline{PT}=\sqrt2p\implies\overline{TU}=2p\).

\(\displaystyle \left(\overline{PT}+\overline{PU}\right)^2=\overline{PT}^2+4p^2+\overline{PU}^2=\overline{TU}^2+4p^2=8p^2\implies\overline{PT}+\overline{PU}=2\sqrt2p\)

Now, for any acceptable configuration of the problem:

\(\displaystyle \frac{\overline{PT}}{\overline{PU}}+\frac{\overline{PU}}{\overline{PT}}\ge2\) (I can provide a proof of this, if necessary).

\(\displaystyle \frac{\overline{PT}^2+\overline{PU}^2}{\overline{PT}\cdot\overline{PU}}\ge2\)

\(\displaystyle \left(\overline{PT}+\overline{PU}\right)^2\ge4\left(\overline{PT}\cdot\overline{PU}\right)\)

with equality when \(\displaystyle \overline{PT}=\overline{PU}\) hence

\(\displaystyle \overline{PT}+\overline{PU}\ge2\sqrt2p\).
\(\displaystyle \text{ }\)

 

[anemone]

Hi Albert and greg1313,

Very well done! And thanks for participating!

 

[Greg]

Spoiler

I made an error in my post above. It's corrected below.

Let \(\displaystyle \angle{QPA}=\angle{BPS}=22.5^{\circ}\). Let \(\displaystyle V\) be the intersection of \(\displaystyle \overline{PR}\) and \(\displaystyle \overline{AB}\). Note that \(\displaystyle \overline{PT}=\overline{PU}\).

\(\displaystyle \triangle{AQT}\cong\triangle{AVR}\cong\triangle{BVR}\cong\triangle{BSU}\) so \(\displaystyle \frac{\overline{PT}\cdot\overline{PU}}{2}=p^2\).

By the AM-GM inequality we have

\(\displaystyle \frac{\overline{PT}+\overline{PU}}{2}\ge\sqrt{\overline{PT}\cdot\overline{PU}}\implies\overline{PT}+\overline{PU}\ge2\sqrt2p\)

Hopefully that's correct. :)

 

[anemone]

Hi greg1313!

First off, I want to say sorry for I hastily said something that is purely wrong yesterday and I had deleted that nonsense reply. I have read your second submission thoroughly and yes, your second post is more easy to follow and your argument is complete.

Here is the solution of other that I want to share with MHB:

Spoiler

View attachment 4347
The circle centered at $P$ with radius $p$ has a tangent at $Q$, $S$. Let $AB'$ be the tangent at $H$ to the above circle. Then $PA$ is the bisector of $\angle QPH=\alpha$, $PB'$ is the bisector of $\angle HPS=\angle \beta$, and hence we have $\angle APB'=45^{\circ}$ because $\alpha+\beta=90^{\circ}$.

From this, we conclude that $B'=B$.

Thus,

$\begin{align*}PT+PU&=\dfrac{PH}{\cos \alpha}+\dfrac{PH}{\cos \beta}\\&=p\left(\dfrac{1}{\cos \alpha}+\dfrac{1}{\cos \beta}\right)\\& \ge p\left(\dfrac{2}{\sqrt{\cos \alpha \cos \beta}}\right)\\& \ge p\left(\dfrac{2}{\sqrt{\cos \alpha \cos (90^{\circ}-\alpha)}}\right)\\& \ge p\left(\dfrac{2}{\sqrt{\cos \alpha \sin \alpha}}\right)\\& \ge p\left(\dfrac{2\sqrt{2}}{\sqrt{\sin 2\alpha}}\right)\\& \ge 2\sqrt{2}p\end{align*}$

Equality holds when $\alpha=\beta$, i.e. $\angle QPA=45^{\circ}$