Nairi Sedrakyans new geometric ineqality

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Nairi Sedrakyans new geometric ineqality

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Source: Kazakhstan MO 2018 final round.Grade 11;Problem 6

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150 posts

#1 h May 4, 2018, 11:14 AM • 4 Y

Y by mathematicsy, Adventure10, Mango247, Rounak_iitr

Inside of convex quadrilateral $ABCD$ found a point $M$ such that $\angle AMB=\angle ADM+\angle BCM$ and $\angle AMD=\angle ABM+\angle DCM$ .Prove that $AM\cdot CM+BM\cdot DM\ge \sqrt{AB\cdot BC\cdot CD\cdot DA}.$

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150 posts

#2 h May 4, 2018, 11:51 AM • 3 Y

Y by Mprog., Adventure10, Mango247

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#3 h May 4, 2018, 11:53 AM • 2 Y

Y by Adventure10, Mango247

See here for picture

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5306 posts

soryn

#4 h Aug 15, 2018, 9:04 AM • 2 Y

Y by Adventure10, Mango247

Very nice!

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Fever

#5 h Sep 22, 2018, 5:49 PM • 1 Y

Y by Adventure10

Perform inversion in $M$ . Now we need to prove that for any point $M$ inside a parallelogram $ABCD$ next inequality holds:
$AM \cdot CM + BM \cdot DM \geq AB \cdot BC$ Draw line through $M$ parallel to $AB$ . Move $AM, DM$ to the top, $BM, CM$ to the bottom using parralel moving such that $AM \rightarrow XY, DM \rightarrow YZ, BM \rightarrow TX, CM \rightarrow ZT; X \in AB$ etc. Now for quarilateral $XYZT$ we need to prove that
$XY \cdot ZT + YZ \cdot TX \geq XZ \cdot YT$ which is Ptolemeus inequality

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yayups

#6 h Oct 16, 2019, 5:21 AM • 2 Y

Y by Adventure10, MS_asdfgzxcvb

It's not hard to see that the angle condition implies that $(AMD),(BMC)$ and $(AMB),(CMD)$ are pairs of tangent circles. Therefore, after inverting at $M$ with power $r^2$ we see that we get $A^*B^*C^*D^*$ a parallelogram. After using the inversion distance formula, we see that the inequality is equivalent to showing
$MA^*\cdot MC^*+MB^*\cdot MD^*\ge xy$ where $x=A^*B^*=C^*D^*$ and $y=A^*D^*B^*C^*$ . We'll drop the stars now.

Let $M'=M+\vec{CB}$ . By Ptolemy on $M'AMB$ , we see that
$M'M\cdot AB\le M'A\cdot MB+MA\cdot M'B,$ or
$xy\le MD\cdot MB+MA\cdot MC,$ as desired.

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pad

#7 h May 15, 2020, 8:18 PM

Y by

Invert about $M$ with arbitrary power $R^2$ . Briefly dropping stars, the new angle conditions are $\angle AMB=\angle MAD+\angle MBC$ and $\angle AMD=\angle MAB+\angle MDC$ . From the second condition, $\angle MAD+\angle MDA +\angle MAB+\angle MDC=180$ , so $\angle BAD+\angle CDA=180$ . Similarly, by the first condition, $\angle ABC+\angle BCD=180$ . Therefore, $A^*B^*C^*D^*$ is a parallelogram.

We have $MA=\tfrac{R^2}{MA^*}$ , and so on. We can verify that $AB = \frac{R^2}{MA^*\cdot MB^*} \cdot A^*B^*$ by using the Law of Sines. The LHS is
$AM\cdot CM + BM\cdot DM = \frac{R^4}{\prod_{\text{cyc}} MA^*} [MA^*\cdot MC^* + MB^*\cdot MD^*].$ And the RHS is
$\sqrt{\prod_{\text{cyc}} AB} = \frac{R^4}{\prod_{\text{cyc}} MA^*} \cdot \sqrt{\prod_{\text{cyc}} A^*B^*} = \frac{R^4}{\prod_{\text{cyc}} MA^*} \cdot A^*B^*\cdot B^*C^*$ since $A^*B^*C^*D^*$ is a parllelogram. So we want to show
$MA^*\cdot MC^* + MB^*\cdot MD^* \ge A^*B^*\cdot B^*C^*.$ The following finish is quite nice. Drop all stars. Construct $M'$ which is the point such that $MM'CD$ is a parallelogram; hence $MM'BA$ is also a parallelogram. Ptolemy on $BM'CM$ gives
$\begin{align*} AB\cdot BC &= MM'\cdot BC \\ &\le CM'\cdot MB + BM'\cdot MC \\ &= MD\cdot MB + MA\cdot MC. \end{align*}$ This completes the proof.

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soryn

#8 h Jan 13, 2021, 8:28 PM

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Please, I do not understand the "New angle conditions...

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8857 posts

#9 h Aug 15, 2023, 6:37 PM

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Invert about $M$ . Since I'm lazy we omit stars. Thus we have the angle conditions
$\begin{align*} \angle AMB &= \angle DAM + \angle CBM \\ \angle AMD &= \angle BAM + \angle CDM. \end{align*}$ The first condition implies $\overline{BC} \parallel \overline{AD}$ (just draw a diagram!) and similarly, $ABCD$ is a parallelogram.

Now consider $M'$ such that $\triangle BMC \cong \triangle AMD$ . By Ptolemy inequality, $AB \cdot BC = MM' \cdot BC \leq CM \cdot AM + BM \cdot DM.$ Using the inversion distance formula yields the desired result.

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1703 posts

#10 h Dec 7, 2023, 4:06 AM

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We consider inversion about $M$ . We can see that $\angle MA^*D^* + \angle MB^*C^* = \angle A^*MB^*$ . Thus, if we consider a line through $M$ parallel to $A^*D^*$ then we can see that $A^*D^* \parallel B^*C^*$ and similarly for the other pair of lines. Thus, $A^*B^*C^*D^*$ is a parallelogram.

Now we need to prove that for any point in a parallelogram $M$ then we have $AM \cdot CM + BM \cdot DM \geq AB \cdot BC$ Now consider a point $M'$ sucht that $MM'CD$ is a parallelogram. Then we can use inversion distance formula to yield the result. $\blacksquare$

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793 posts

#11 h Dec 28, 2023, 4:43 AM

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The angle conditions imply $(MAB)$ and $(MCD)$ are externally tangent, as are $(MBC)$ and $(MDA)$ . Inverting at $M$ with radius 1 makes $ABCD$ a parallelogram, and inversion distance formula rewrites the desired as
$AM \cdot CM + BM \cdot DM \ge \sqrt{AB \cdot BC \cdot CD \cdot DA} = AB \cdot BC.$
Translating $\triangle MCD$ to $\triangle M'C'D'$ such that $C' = B$ and $D' = A$ , this inequality is simply Ptolemy on $M'AMB$ . $\blacksquare$

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732 posts

#12 h Sep 27, 2024, 5:34 AM

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another problem where I missed the finish ;-;

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#13 h Feb 21, 2025, 11:57 PM

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First note that the conditions imply that $(MAB), (MCD)$ as well as $(MAD), (MBC)$ are externally tangent. By inverting about $M,$ we get that $ABCD$ becomes a parallelogram. Thus, the problem translates to $AM\cdot CM + BM\cdot DM \geq AB\cdot BC.$ Translate $MCD$ so that $CD$ matches with $AB.$ Then, Ptolemy finishes.

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#14 h Feb 22, 2025, 2:31 PM

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Inverting about $M$ , we get:

Let $ABCD$ be a parallelogram and $M$ a point in it, prove that $AM\cdot CM+BM\cdot DM\ge \sqrt{AB\cdot BC\cdot CD\cdot DA}.$ But translating triangle $MCD$ by vector $CB$ , this is equivalent to Ptolemy's inequality. $\square$

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1878 posts

#15 h Feb 28, 2025, 8:48 PM

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Note that $(ABM)$ and $(CDM)$ are tangent, and $(ADM)$ and $(BCM)$ are tangent. Invert around $M$ to get a problem involving parallelogram:

Parallelogram $ABCD$ has a point $M$ inside it. Prove that $AO\cdot CO+BO\cdot DO\ge AB\cdot BC$ .

Take $P=O+B-C$ , then apply Ptolemy inequality on $AMBP$ . $\blacksquare$

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