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Source: XIII Polish Junior MO 2018 Finals - Problem 5

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#1 h Apr 24, 2018, 12:45 PM • 2 Y

Y by Adventure10, Mango247

Point $M$ is middle of side $AB$ of equilateral triangle $ABC$ . Points $D$ and $E$ lie on segments $AC$ and $BC$ , respectively and $\angle DME = 60 ^{\circ}$ . Prove that, $AD + BE = DE + \frac{1}{2}AB$ .

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#2 h Aug 4, 2018, 12:38 PM • 3 Y

Y by ike.chen, lian_the_noob12, Adventure10

Outlines of my proof
1. Let $I$ be the incenter of $ADE$ . Then show that $D$ , $M$ , $E$ , $I$ are concyclic. (Not too hard)
2. Show that $DM$ and $EM$ are angle bisectors of $\angle ADE$ and $\angle DEC$ , respectively.
3. Show that $ADEC$ has an inscribed semicircle with center $M$ and its diameter is the same line with $AB$ .
4. Since $ABC$ is equilateral, the remaining steps are not hard to proof.

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#3 h Feb 5, 2019, 6:38 PM • 2 Y

Y by Adventure10, Mango247

jrpartty,

Is I the incenter of triangle ADE or CDE.
and in your third statement
will that be ADEC or ADEB.

This post has been edited 2 times. Last edited by amar_04, Feb 5, 2019, 6:42 PM

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#4 h Jul 11, 2022, 7:38 AM

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my solution is the same as jrppartty to the 2nd steps, you can do direct proof by using the fact that AMD, MBE, and DME is a similar triangle.

Let $AM = BM =x, AD =y$ and $MD =z$ . We got $BE =\frac{x^{2}}{y}$ and $DE =\frac{z^{2}}{y}$ , substitute it to the question.

$y+\frac{x^{2}}{y}=\frac{z^{2}}{y}+x$

$x^{2} + y^{2} - xy = z^{2}$

We can prove this is indeed correct by using cosine formula on $\triangle AMD$ .
Maybe there is a non-trig solution but i just can't find it.

This post has been edited 3 times. Last edited by amogususususus, Jul 29, 2022, 3:33 AM
Reason: latex retype

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#5 h Aug 9, 2022, 9:33 PM

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$\widehat{DME}=\widehat{DCE}=60$
$\Rightarrow MDCE$ is a parallelogram

moreover oposites angles are equals and the sum of all angles in a quadrilateral is $360$
$\Rightarrow 2\widehat{MEC}=240 \Leftrightarrow \widehat{MEC}=120$

because $B,C,E$ aligned $\widehat{MEB}=180-120=60$
by the same way we get $\widehat{MDA}=60$

so easily we find : $\triangle MAD$ and $\triangle MBE$ equilateral with a comon length : $MB=MA$ so $MD=ME$ and $AD=BE$

also $\triangle MED$ isocele in $M$ but $\widehat{DME}=60$ so it is equilateral $\Rightarrow DE=MD=AM$

finally : $AD=BE=AD=AM=BM=\frac{1}{2}AB$
which concludes

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#6 h Aug 10, 2022, 5:32 AM

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Since $\widehat{DME}=90^\circ-\frac{\widehat{DCE}}2$ and $M$ lies onto the bisector of $\angle ACB$ , it is the $C-$ excenter of $\triangle CDE$ , afterwards you may continue as to the outlined proof above.

Best regards,
sunken rock

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#7 h Aug 23, 2022, 1:19 PM

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Let $(CDM)$ meet $CB$ again at $P$ and the projections of $M$ onto $CA$ and $CB$ be $X$ and $Y$ respectively. Because $CM$ bisects $\angle ACB$ , we know $MD = MP$ . Moreover, we have $\angle PME = \angle DMP - \angle DME = (180^{\circ} - \angle DAP) - 60^{\circ}$ $= (180^{\circ} - 60^{\circ}) - 60^{\circ} = 60^{\circ} = \angle DME$ so $ME$ is actually the perpendicular bisector of $DP$ .

Clearly, $\angle MXD = 90^{\circ} = \angle MYP$ . In addition, symmetry gives $MX = MY$ , implying $MDX \cong MPY$ . Now, since $D$ lies on segment $AC$ , we know $P$ lies on segment $CB$ or the extension of $CB$ past $B$ . Thus, properties of 30-60-90 triangles yield $AD + BE = (AX + XD) + (BY + YE)$ $= (AX + BY) + (YP + YE)$ $= \left( \frac{AM}{2} + \frac{BM}{2} \right) + PE = \frac{AB}{2} + DE$ as required. $\blacksquare$

Remarks:

My first idea was to use $AMD \overset{+}{\sim} BEM$ somehow. Also, $EM$ and $DM$ are tangent to $(AMD)$ and $(BME)$ respectively.
Because $\angle CMX = 60^{\circ} = \angle CMY$ we know $CD < CX$ and $CE < CY$ .
Also, if $P$ belongs to segment $CB$ , then $BP < \frac{CB}{2}$ .

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lian_the_noob12

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#8 h May 2, 2023, 8:26 PM

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BlayzyMath wrote:

$\widehat{DME}=\widehat{DCE}=60$
$\Rightarrow MDCE$ is a parallelogram

moreover oposites angles are equals and the sum of all angles in a quadrilateral is $360$
$\Rightarrow 2\widehat{MEC}=240 \Leftrightarrow \widehat{MEC}=120$

because $B,C,E$ aligned $\widehat{MEB}=180-120=60$
by the same way we get $\widehat{MDA}=60$

so easily we find : $\triangle MAD$ and $\triangle MBE$ equilateral with a comon length : $MB=MA$ so $MD=ME$ and $AD=BE$

also $\triangle MED$ isocele in $M$ but $\widehat{DME}=60$ so it is equilateral $\Rightarrow DE=MD=AM$

finally : $AD=BE=AD=AM=BM=\frac{1}{2}AB$
which concludes

Your Solution Is totally Wrong! $\widehat{DME}=\widehat{DCE}=60$ Doesn't imply that $MDCE$ is a parallelogram

This post has been edited 1 time. Last edited by lian_the_noob12, May 2, 2023, 8:26 PM

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