Kosovo MO 2021 Grade 10, Problem 4

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Kosovo MO 2021 Grade 10, Problem 4

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#1 h Feb 27, 2021, 7:11 PM

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Let $M$ be the midpoint of segment $BC$ of $\triangle ABC$ . Let $D$ be a point such that $AD=AB$ , $AD\perp AB$ and points $C$ and $D$ are on different sides of $AB$ . Prove that: $\sqrt{AB\cdot AC+BC\cdot AM}\geq\frac{\sqrt{2}}{2}CD.$

This post has been edited 1 time. Last edited by geometry6, Feb 27, 2021, 7:18 PM

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mihaig

#2 h Feb 27, 2021, 10:50 PM • 1 Y

Y by Mango247

Let $A=x+yi,\text{where} \left(y>0\right),B=-1,C=1\implies D=x-y+(x+y+1)i.$ We thus need to prove
$\sqrt{\left(x^2+y^2+1\right)^2-4x^2}+2\sqrt{x^2+y^2}\geq x^2+y^2+1+2y.$ After squaring, the latter writes as
$\sqrt{\left(x^2+y^2+1\right)^2-4x^2}\cdot\sqrt{x^2+y^2}\geq y(x^2+y^2+1).$ We square this too and get $\left(x^2+y^2+1\right)^2\geq4(x^2+y^2),$ which is $AM-GM.$ Equality at $x=0,$ i.e. $AB=AC.$
We also get equality at $x^2+y^2=1,$ which makes the angle from $A$ right. I have the strong belief we are talking about Tereshin here.

This post has been edited 1 time. Last edited by mihaig, Feb 27, 2021, 10:58 PM

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mihaig

#3 h Feb 28, 2021, 7:03 AM

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I am interested in another approaches, but especially I want to see the official solution.

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#4 h Feb 28, 2021, 3:01 PM

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I tried using Ptolemy's ineq but I couldn't do it, btw @above nice solution.

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mihaig

#5 h Feb 28, 2021, 6:54 PM • 1 Y

Y by Circumcircle

See here my second solution. I once created a problem with a part from this configuration. But it was a locus problem. I liked this one.

Let $A=x+yi,\text{where} \left(y>0\right),B=-1,C=1\implies D=x-y+(x+y+1)i.$ We thus need to prove
$\sqrt{\left(x^2+y^2+1\right)^2-4x^2}+\sqrt{4x^2+4y^2}\geq x^2+y^2+1+2y.$ If $x=0$ or if $x^2+y^2=1,$ then we clearly have equality. Otherwise, since
$\left(\left(x^2+y^2+1\right)^2,4y^2\right) \text{strictly majorize} \left(\left(x^2+y^2+1\right)^2-4x^2,4x^2+4y^2\right),\text{then by Karamata}$ $\sqrt{\left(x^2+y^2+1\right)^2-4x^2}+\sqrt{4x^2+4y^2}>\sqrt{\left(x^2+y^2+1\right)^2}+\sqrt{4y^2}= x^2+y^2+1+2y.$ The proof is complete. Equality if and only if $AB=AC$ or $AB\perp AC.$

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#6 h Feb 28, 2021, 7:03 PM • 1 Y

Y by Mango247

i heard that there exists a " geometrical
" solution

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#7 h Feb 28, 2021, 7:06 PM • 3 Y

Y by geometry6, mihaig, Leartia

Let $E$ be a point such that $EA=AC$ , $EA$ perpendicular to $AC$ , and $B$ and $E$ on different sides with respect to $AC$ .

$AD=AB$ , $AC=AE$ , $\angle DAC = 90 + \angle BAC = \angle BAE$ $\Rightarrow$ $\bigtriangleup DAC \cong \bigtriangleup BAE$ $\Rightarrow CD=BE$ ...(1).

Let $A'$ be the reflection of $A$ with respect to point $M$ . From this, we have that $ABA'C$ is parallelogram $\Rightarrow AC=BA'$ and $\angle ABA' = 180 - \angle BAC$ .

$AD=AB$ , $AE=AC=BA'$ , $\angle DAE = 360 - \angle DAB - \angle BAC - \angle CAE = 360 - 90 - \angle BAC - 90 = 180 - \angle BAC = \angle ABA'$ . $\Rightarrow \bigtriangleup DAE \cong \bigtriangleup ABA'$ $\Rightarrow DE= AA' = 2AM$ ...(2).

From Pythagorean Theorem in $\bigtriangleup DAB$ and $\bigtriangleup EAC$ , we can easily find that $DB=AB\cdot\sqrt{2}$ and $EC=AC\cdot\sqrt{2}$ ...(3).

From here, we use Ptolemy's inequality for quadrilateral $DECB$ and we have that $DB\cdot CE + BC\cdot DE \ge DC\cdot BE$ ...(4).

Substituting (1), (2), and (3) in (4), we have that $2AB\cdot AC + 2BC\cdot AM \ge CD^2$ .

The conclusion follows.

Motivation to this solution: The LHS is symmetrical wrt to $B$ and $C$ , so we try to take advantage of that.

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#8 h Mar 1, 2021, 3:39 PM • 1 Y

Y by Circumcircle

Beautiful.

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mihaig

#9 h Mar 3, 2021, 3:02 PM • 2 Y

Y by Circumcircle, Mango247

At equality, if we fix $B$ and $C$ then $A$ describes the union of a line and a circle.

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Wictro

#10 h Mar 10, 2021, 5:58 PM • 2 Y

Y by Circumcircle, Leartia

Consider the point $E$ , defined just like in post #7. Furthermore, let $X$ be the point where the A-Symmedian meets $(ABC)$ . It is well known that $X$ lies on the A-Apollonian circle, so $XB/XC = AB/AC = AD/AE$ . This, combined with the fact that $\angle{BXC} = \angle{DAE} = 180 - \angle{BAC}$ , gives that $\triangle{BXC} \sim \triangle{DAE}$ .
Now the simple length relations $DE/BC = AD/BX = AB/BX = AM/MC = 2AM/BC$ give that $DE = 2AM$ .
Ptolemy on $BDEC$ finishes the problem since $BE = DC, BD = \sqrt{2}AB, CE = \sqrt{2}AC$ .

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lneis1

#11 h Jul 2, 2021, 5:01 PM

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