collinear midpoints wanted , <DAC = <DBC.

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collinear midpoints wanted , <DAC = <DBC.

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#1 h Jul 27, 2021, 9:41 PM • 1 Y

Y by HWenslawski

A point $D$ was chosen inside the acute-angled triangle $ABC$ so that $\angle DAC = \angle DBC$ . Let $K$ and $L$ be the feet of the perpendiculars drawn from the point $D$ on $AC$ and $BC$ , respectively. Prove that the midpoints of the segments $AB$ , $CD$ and $KL$ lie on the same line.

This post has been edited 1 time. Last edited by parmenides51, Jul 28, 2021, 3:11 AM
Reason: source typo

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tree_3

#2 h Jul 27, 2021, 10:50 PM • 1 Y

Y by HWenslawski

Solved with Alex Zhao, Isaac Zhu, Jeffrey Chen

Reflect D over the midpoint $M$ of $AB$ to get $E$ . By first isogonality lemma, we have that $CE, CD$ are isogonal in $\angle ACB$ . Since $CD$ is a diameter in $(CKLD)$ , then $CE$ is perpendicular to $KL$ . Taking a homothety of ratio $\frac 12$ at $D$ , we get that if the midpoint of $CD$ is $N$ then $NM$ is perpendicular to $KL.$ This implies that it is the perpendicular bisector, thus it passes through the midpoint of $KL$ as desired.

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#3 h Jul 27, 2021, 10:51 PM • 1 Y

Y by tree_3

Solved with Alex Zhao, Jeffrey Chen, Kevin Wu

Let $DL \cap AC = X, DK \cap AB = Y$ so $D$ is the orthocenter of $\triangle CXY$ . Fix $XYCKL$ and move $A$ along $XC$ at constant velocity. Note $\triangle DAX$ and $\triangle DBY$ are similar due to angle reasons, so $\frac{AX}{BY} = \frac{DX}{DY}$ is fixed. In particular, $B$ moves with constant velocity and so by gliding principle the midpoint $M$ of $AB$ moves on a line. Taking $A,B = K,L$ and $X,Y$ solves the problem (along with the well-known property that the midpoint of $XY$ is equidistant from $K$ and $L$ ).

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#4 h Jul 27, 2021, 11:14 PM • 2 Y

Y by tree_3, greenish

Solved with Alex Zhao, Isaac Zhu, Kevin Wu

Let $E = AD\cap BC, F = BD\cap AC$ . Now, since $\angle EAF = \angle FBE$ , we have $(ABEF)$ is cyclic. By Gauss line on quadrilateral $ABFE$ , we have the midpoints of $AB, FE, CD$ are collinear. Therefore, if we show midpoints of $AB, KL, FE$ are collinear, we are done.

Now, proceed with complex numbers. Let $(ABFE)$ be the unit circle. We then have (by complex foot)
$k = \frac{1}{2}(b + e + d + bc\overline{d}), l = \frac{1}{2}(a+f+d - af\overline{d})$ If we take $4$ times the complex number of each of the midpoints, the midpoint of $KL$ becomes (which we denote by the complex number $r$ )
$r = a + f + b + e + 2d - \overline{d}(be + af)$ Now, observe that by complex intersection formula,
$d = \frac{ae(b+f) - bf(a+e)}{ae - bf}, \overline{d} = \frac{b + f - a - e}{bf - ae}$ Therefore,
$r= a + f + b + e + \frac{2(ae(b+f) - bf(a+e)) + (b+f-a-e)(be+af)}{ae-bf}$ $= a + f + b + e + \frac{ae(b+f) - bf(a+e) + be(b-e) + af(f-a)}{ae - bf}$ Since we are taking four times the complex value of midpoints, we need to show that it's collinear with $2a + 2b,2e + 2f$ . This means proving that $\frac{r - (2a+2b)}{r - (2e + 2f)}$ is real. Observe that
$r - (2a + 2b) = b+e-a-f + \frac{ae(b+f) - bf(a+e) + be(b-e) + af(f-a)}{ae - bf}$ $= \frac{2afe - 2bfe + (f+e)(b^2 - a^2) + (a-b)(e^2 + f^2)}{ae-bf}$ $= \frac{(a-b)(e^2 + f^2 + 2fe - (f+e)(a+b)}{ae - bf} = \frac{(a-b)(e+f)(e+f-a-b)}{ae-bf}$ Similarly, we have $r - (2e + 2f) = \frac{(e-f)(a+b)(a+b-e-f)}{ae - bf}$ . Therefore,
$\frac{r-(2a+2b)}{r-(2e+2f)} = -\frac{(a-b)(e+f)}{(e-f)(a+b)}$ It is obvious, that by taking the conjugate, that this value is real. Therefore, we conclude that the three midpoints are collinear.

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#5 h Jul 28, 2021, 9:09 AM

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My solution at https://stanfulger.blogspot.com/2021/07/aops-collinear-midpoints.html

Best regards,
sunken rock

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#6 h Jul 28, 2021, 11:14 AM

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$[asy] /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki go to User:Azjps/geogebra */ import graph; size(10cm); real labelscalefactor = 0.5; /* changes label-to-point distance */ pen dps = linewidth(0.7) + fontsize(10); defaultpen(dps); /* default pen style */ pen dotstyle = black; /* point style */ real xmin = -69.38543559962793, xmax = 57.8516129937149, ymin = -15.494650251561545, ymax = 60.58180961234375; /* image dimensions */ pen qqwuqq = rgb(0,0.39215686274509803,0); /* draw figures */ draw((0,0)--(-20.9030609092874,-3.5090756957578133), linewidth(0.4) + blue); draw((-20.9030609092874,-3.5090756957578133)--(0,-3.5090756957578133), linewidth(0.4) + blue); draw((0,-3.5090756957578133)--(0,0), linewidth(0.4) + qqwuqq); draw((0,0)--(7.791561237284592,3.1965575201674867), linewidth(0.4) + qqwuqq); draw((-11.249873698097105,49.60977092513793)--(0,0), linewidth(0.4) + blue); draw((-11.249873698097105,49.60977092513793)--(7.791561237284592,3.1965575201674867), linewidth(0.4) + blue); draw((0,-3.5090756957578133)--(10.542607097166856,-3.5090756957578133), linewidth(0.4) + blue); draw((10.542607097166856,-3.5090756957578133)--(7.791561237284592,3.1965575201674867), linewidth(0.4) + blue); draw((-11.249873698097105,49.60977092513793)--(-20.9030609092874,-3.5090756957578133), linewidth(0.4) + blue); draw((7.791561237284592,3.1965575201674867)--(0,-3.5090756957578133), linewidth(0.4) + blue); draw((0,0)--(10.542607097166856,-3.5090756957578133), linewidth(0.4) + blue); draw((-16.076467303692255,23.050347614690057)--(5.271303548583428,-1.7545378478789067), linewidth(0.4) + blue); draw((0,22.188371723834596)--(-8.553319625023194,-3.5090756957578133), linewidth(0.4) + qqwuqq); draw((0,0)--(-8.553319625023194,-3.5090756957578133), linewidth(0.4) + qqwuqq); draw((0,22.188371723834596)--(0,0), linewidth(0.4) + qqwuqq); /* dots and labels */ dot((0,0),dotstyle); label("$D$", (0.2959506730970271,0.8667717934049873), NE * labelscalefactor); dot((0,-3.5090756957578133),dotstyle); label("$K$", (0.2959506730970271,-2.7045030692019205), NE * labelscalefactor); dot((-20.9030609092874,-3.5090756957578133),dotstyle); label("$B$", (-20.55032817607337,-2.7045030692019205), NE * labelscalefactor); dot((7.791561237284592,3.1965575201674867),dotstyle); label("$L$", (8.102923628563232,4.022782137104115), NE * labelscalefactor); dot((-11.249873698097105,49.60977092513793),linewidth(4pt) + dotstyle); label("$A$", (-10.916191337412949,50.283249543430806), NE * labelscalefactor); dot((10.542607097166856,-3.5090756957578133),linewidth(4pt) + dotstyle); label("$C$", (10.84366945335456,-2.8706088767650324), NE * labelscalefactor); dot((-16.076467303692255,23.050347614690057),linewidth(4pt) + dotstyle); label("$M$", (-15.73325975674316,23.706320333332886), NE * labelscalefactor); dot((3.895780618642296,-0.1562590877951633),linewidth(4pt) + dotstyle); label("$T$", (4.199437150830129,0.5345601782787633), NE * labelscalefactor); dot((5.271303548583428,-1.7545378478789067),linewidth(4pt) + dotstyle); label("$O$", (5.611336515116571,-1.1264978973523565), NE * labelscalefactor); dot((0,22.188371723834596),linewidth(4pt) + dotstyle); label("$A'$", (0.2959506730970271,22.875791295517327), NE * labelscalefactor); dot((-8.553319625023194,-3.5090756957578133),linewidth(4pt) + dotstyle); label("$B'$", (-8.258498416403176,-2.8706088767650324), NE * labelscalefactor); dot((-4.276659812511597,9.339648014038392),linewidth(4pt) + dotstyle); label("$M'$", (-3.939747419762297,10.002591209376147), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */ [/asy]$

Nice.

Let $A'=DK \cap AC$ , $B'=DL \cap BC$ , $M$ be the midpoint of $AB$ , $M'$ be the midpoint of $A'B'$ , $O$ the midpont of $DC$ and let $T$ be the midpoint of $KL$ .

Notice that the Newton Line of $KDLC$ is $O,T$ and $M'$ and notice that $\triangle BKD \sim \triangle ALD$ , as well as, $\triangle B'KD \sim \triangle A'LD$ .
The last two imply that:
$\frac{BK}{AL}=\frac{KD}{LD}=\frac{B'K}{A'L}$ which in turn implies that:
$\frac{BK}{B'K}=\frac{AL}{A'L}=\lambda$
We can translate this statement into homothethies, i.e. there exists a homothethy $\mathcal{H}_1\left(K,\lambda \right)$ and there exists a homothethy $\mathcal{H}_2\left( L,\lambda \right)$ .

Now let's state a lemma which will help us solve this problem.
Lemma:
Given $4$ points $O_1,O_2,A$ and $B$ and a positive real constant $\lambda$ , define $2$ homothethies $\mathcal{H}_1\left(O_1,\lambda\right)$ and $\mathcal{H}_"\left(O_2,\lambda\right)$ , such that $A \stackrel{\mathcal{H}_1}{\rightarrow} A'$ and $B \stackrel{\mathcal{H}_2}{\rightarrow} B'$ . Then the midpoints of $A'B'$ , $AB$ and $O_1O_2$ are colinear.
Proof:
$[asy] /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki go to User:Azjps/geogebra */ import graph; size(9cm); real labelscalefactor = 0.5; /* changes label-to-point distance */ pen dps = linewidth(0.7) + fontsize(10); defaultpen(dps); /* default pen style */ pen dotstyle = black; /* point style */ real xmin = -38.83077739018837, xmax = 28.993160137011987, ymin = -10.538014493526635, ymax = 27.491203741372214; /* image dimensions */ /* draw figures */ draw((-12.468312677918389,10.56667720421738)--(0,0), linewidth(0.4) + blue); draw((-9.528532796933215,-1.0734264438016645)--(1.4844601015019947,7.890637543296748), linewidth(0.4) + red); draw((-3.345023593408256,20.977423505434356)--(-21.59160176242852,0.15593090300040302), linewidth(0.4) + red); draw((-21.59160176242852,0.15593090300040302)--(-3.7842142419354476,-1.6588347041836016), linewidth(0.4) + blue); draw((3.7842142419354476,1.6588347041836016)--(-3.345023593408256,20.977423505434356), linewidth(0.4) + blue); draw((3.7842142419354476,1.6588347041836016)--(-3.7842142419354476,-1.6588347041836016), linewidth(0.4) + red); /* dots and labels */ dot((0,0),dotstyle); label("$T$", (0.17241383792096868,0.44131742993659445), NE * labelscalefactor); dot((-3.7842142419354476,-1.6588347041836016),dotstyle); label("$O_1$", (-3.280763138006986,-1.1524565589532292), NE * labelscalefactor); dot((3.7842142419354476,1.6588347041836016),linewidth(4pt) + dotstyle); label("$O_{2}$", (3.979762811379996,2.035091418826418), NE * labelscalefactor); dot((-9.528532796933215,-1.0734264438016645),dotstyle); label("$A$", (-9.3459585957266,-0.6211985626566213), NE * labelscalefactor); dot((1.4844601015019947,7.890637543296748),dotstyle); label("$B$", (1.677644827428026,8.321644375002945), NE * labelscalefactor); dot((-21.59160176242852,0.15593090300040302),linewidth(4pt) + dotstyle); label("$A'$", (-21.43207801147444,0.5298604293193624), NE * labelscalefactor); dot((-3.345023593408256,20.977423505434356),linewidth(4pt) + dotstyle); label("$B'$", (-3.147948638932834,21.33746528426984), NE * labelscalefactor); dot((-4.02203634771561,3.408605549747542),linewidth(4pt) + dotstyle); label("$M$", (-3.856292633994979,3.761679906790394), NE * labelscalefactor); dot((-12.468312677918389,10.56667720421738),linewidth(4pt) + dotstyle); label("$M'$", (-12.31214907504933,10.9336628567946), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */ [/asy]$

Let $T$ be the midpoint of $O_1O_2$ , $M$ be the midpoint of $AB$ and $M'$ be the midpoint of $A'B'$ .
Throw the whole configuration onto the complex plane, such that $t=0$ .
This implies that $o_2=-o_1$ .
Our homothethies imply that:
$a'=a\lambda + \left(1-\lambda \right) o_1 \; \; \text{and} \; \; b'=b\lambda + \left( \lambda -1 \right) o_1$ this implies that:
$m'=\frac{1}{2}\left( a'+b' \right)=m\lambda$ and this gives us that:
$\frac{m'-t}{\overline{m'-t}}=\frac{m-t}{\overline{m-t}}$ which gives us the desired condition. $\blacksquare$

Now back to the problem at hand.
By our lemma we must have that $M$ , $M'$ and $T$ are colinear, but since $M',T$ and $O$ are colinear, this implies that $M$ , $T$ and $O$ are colinear.

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#7 h Sep 20, 2021, 12:17 AM

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Denote $M$ by the midpoint of $AB$ , $N$ the midpoint of $KL$ and $P$ the midpoint of $DC$ . Since $\angle DKC= \angle DLC= 90^\circ$ , we have that $P$ is the center of $(DKCL)$ so both $N$ and $P$ lie on the perpendicular bissector $KL$ . So It suffies to prove that $MK=ML$ and for that we need some calculations ,if $x= \angle BDL = \angle ADK$ , $y=\angle MDB$ and $z=\angle MDA$ , aplying law of cosines in $\triangle MDL$ and $\triangle MDK$ :
$MK=ML \iff DL^2+MD^2-2 \cdot MD \cdot DL \cdot cos(y+x) = DM^2+KD^2-2 \cdot MD \cdot DK \cdot cos(z+x) \iff DL^2-DK^2=2\cdot MD \cdot (DL\cdot cos(x+y)- DK\cdot cos(x+z))$ Now aplying low of cosines in $\triangle MDA$ and $\triangle MBD$ :
$DL\cdot cos (x+y)= DL \cdot (\dfrac {MD^2+DB^2-MB^2}{2\cdot MD\cdot DB}) \dfrac {DL}{DB}-sen(x)\cdot sen(y)=\dfrac {MD^2+DB^2-MB^2}{2\cdot MD}\cdot (\dfrac{DL}{DB})^2-sen(x)\cdot sen(y) = \dfrac {MD^2+DB^2-MB^2}{2\cdot MD} \cdot cos^2(x)-sen(x)\cdot sen(y)(I)$ Similarly:
$DK\cdot cos(x+z)= \dfrac {MD^2+DA^2-MA^2}{2\cdot MD} \cdot cos^2(x)-sen(x)\cdot sen(z) (II)$ So: $DL\cdot cos(x+y) - DK\cdot cos(x+z)= \dfrac{cos^2(x)}{2\cdot MD}\cdot (MD^2+DB^2-MB^2-MD^2-AD^2+AM^2)-sen(x) \cdot(DL\cdot sen(y) -DK \cdot sen(z))=\dfrac{cos^2(x)}{2\cdot MD}(DB^2-AD^2)-sen(x) \cdot(DL\cdot sen(y) -DK \cdot sen(z))$ Then it's suffies to prove that:
$DL^2-DK^2=cos^2 (x) \cdot (DB^2-DA^2)-2\cdot MD \cdot sen(x) \cdot (DL\cdot sen(y)-DK \cdot sen(z))$ But: $DL^2-DK^2=BD^2cos^2 (x) -AD^2 cos^2 (x)$ Then we just need that: $DL\cdot sen(y)= DK\cdot sen(z) \iff \dfrac{DL}{DK}=\dfrac{sen(z)}{sen(y)}$ . Also, since $\triangle ADK~\triangle BDL$ , we have that: $\dfrac{DL}{DK}=\dfrac{BD}{AD}$ Then: $\dfrac{DL}{DK}=\dfrac{sen(z)}{sen(y)}\iff \dfrac{sen(z)}{sen(y)}= \dfrac{BD}{AD}$ But one can prove that this is true by just apllying two law of sines on $\triangle ADM$ and $\triangle BDM$ , so we're done!

This post has been edited 2 times. Last edited by mandemidio, Sep 20, 2021, 5:36 PM

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