MT = MP wanted, orthocenter, circumcircle, midpoint, projection

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MT = MP wanted, orthocenter, circumcircle, midpoint, projection

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Source: 2017 Maths Beyond Limits Camp - Youger Match Geo3

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#1 h Nov 25, 2020, 8:24 AM

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Let $H$ be the ortocenter of $\vartriangle ABC$ . The point $P$ lies on circumcircle of $\vartriangle ABC$ and the point $M$ is the midpoint of the side $BC$ . Let $T$ be the projection of $H$ onto line $AP$ . Prove that $MT = MP$ .

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#2 h Nov 25, 2020, 9:22 PM • 1 Y

Y by Mango247

We throw the configuration onto the complex plane and let the lowercase letters represent the complex value of the points with the same uppercase letters, and let $c=\frac{1}{b}$ .

By definition $h=2m+a$ , where $m=\frac{b+c}{2}$ , by easy computation we have that $\overline{t}=\frac{a+p-t}{ap}$ , this comes from the fact that $T\in AP$ , by also easy computation we have that $\overline{t}=h-\frac{h-t}{ap}$ , this comes from the fact that $HT \perp AP$ , thus from these two facts we have that $t=\frac{a+h+p-ahp}{2}$ .

Let $m'=\frac{p+t}{2}$ , by easy computation we have that $\frac{m-m'}{\overline{m-m'}}=-\frac{a-p}{\overline{a-p}}=ap$ , this implies that if $M'$ is the midpoint of $PT$ that $MM' \perp PT$ , this implies that $MT=MP$ .

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#3 h Nov 25, 2020, 10:09 PM

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I am going to solve this problem with moving points
$[asy] /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki go to User:Azjps/geogebra */ import graph; size(14.176985613845321cm); 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 = -13.892129150892375, xmax = 14.461842076798266, ymin = -8.26935164954028, ymax = 7.364607463493773; /* image dimensions */ pen dbwrru = rgb(0.8588235294117647,0.3803921568627451,0.0784313725490196); pen xdxdff = rgb(0.49019607843137253,0.49019607843137253,1.); pen ttffcc = rgb(0.2,1.,0.8); pen sexdts = rgb(0.1803921568627451,0.49019607843137253,0.19607843137254902); pen dtsfsf = rgb(0.8274509803921568,0.1843137254901961,0.1843137254901961); draw((-6.33,5.66)--(-7.77,-2.22)--(3.97,-1.74)--cycle, linewidth(2.) + dbwrru); /* draw figures */ draw((-6.33,5.66)--(-7.77,-2.22), linewidth(2.) + dbwrru); draw((-7.77,-2.22)--(3.97,-1.74), linewidth(2.) + dbwrru); draw((3.97,-1.74)--(-6.33,5.66), linewidth(2.) + dbwrru); draw(circle((-2.0136484426050965,0.7996514920496627), 6.500298330519953), linewidth(2.)); draw((-6.33,5.66)--(1.4465521921354356,-4.7031559689743645), linewidth(2.)); draw((-6.102703114789808,0.10069701590067366)--(-3.5792493995770283,1.9943064124532957), linewidth(2.) + xdxdff); draw((-6.33,5.66)--(-6.102703114789808,0.10069701590067366), linewidth(2.)); draw((-1.9,-1.98)--(1.4465521921354356,-4.7031559689743645), linewidth(2.) + ttffcc); draw((-0.034810120877354134,-3.1057304977105225)--(-0.41863768698720627,-3.5774254712638434), linewidth(2.) + ttffcc); draw((-1.9,-1.98)--(-3.5792493995770283,1.9943064124532957), linewidth(2.) + ttffcc); draw((-3.01971301912214,-0.11119150433611884)--(-2.459536380454886,0.12549791678941263), linewidth(2.) + ttffcc); draw(circle((-1.9,-1.98), 5.874904254538961), linewidth(1.2)); draw((3.97,-1.74)--(-7.3057386859493825,0.32054107966587525), linewidth(2.)); draw((-3.545766241840223,3.6596767174386073)--(-7.77,-2.22), linewidth(2.)); draw(circle((-6.216351557394902,2.8803485079503366), 2.7819738290933795), linewidth(1.2) + sexdts); draw((-1.9,-1.98)--(-7.3057386859493825,0.32054107966587525), linewidth(2.) + linetype("2 2") + dtsfsf); draw((-1.9,-1.98)--(-3.545766241840223,3.6596767174386073), linewidth(2.) + linetype("2 2") + dtsfsf); /* dots and labels */ dot((-6.33,5.66),dotstyle); label("$A$", (-6.239850946457547,5.920303299080578), NE * labelscalefactor); dot((-7.77,-2.22),dotstyle); label("$B$", (-7.658816441319634,-1.9600229313142195), NE * labelscalefactor); dot((3.97,-1.74),dotstyle); label("$C$", (4.072987560843696,-1.4785882098431546), NE * labelscalefactor); dot((-6.102703114789808,0.10069701590067366),linewidth(4.pt) + dotstyle); label("$H$", (-6.011802920497568,0.2951186587344525), NE * labelscalefactor); dot((-1.9,-1.98),linewidth(4.pt) + dotstyle); label("$M$", (-1.805583775013523,-1.7826522444564588), NE * labelscalefactor); dot((1.4465521921354356,-4.7031559689743645),dotstyle); label("$P$", (1.539120605732826,-4.44321254732287), NE * labelscalefactor); dot((-3.5792493995770283,1.9943064124532957),linewidth(4.pt) + dotstyle); label("$T$", (-3.4019199567333716,1.9928095186587336), NE * labelscalefactor); dot((-7.3057386859493825,0.32054107966587525),linewidth(4.pt) + dotstyle); label("$E$", (-7.202720389399677,0.5231666846944305), NE * labelscalefactor); dot((-3.545766241840223,3.6596767174386073),linewidth(4.pt) + dotstyle); label("$G$", (-3.452597295835589,3.8678710654407755), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */ [/asy]$
let $G$ and $E$ be the foot of altitudes from $B$ and $C$ .
it is easy to see that as $P$ varies the locus of $T$ is a circle with diameter $AH$ now call that circle $\Omega$ now define $T$ as the intersection of $\Omega$ and $AP$ (\neq A), now let $D$ be the projection of $M$ over $AP$ again the locus of $D$ is a circle. so $P \mapsto T \mapsto D$
.
it is enough to check three different situations to see that $PD=DT$ .( $P=C,P=B,P=AH \cap (ABC)$ )
$\boxed{\mathcal{Q.E.D}}$

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#4 h Nov 25, 2020, 10:31 PM

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iman007 wrote:

I am going to solve this problem with moving points
$[asy] /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki go to User:Azjps/geogebra */ import graph; size(14.176985613845321cm); 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 = -13.892129150892375, xmax = 14.461842076798266, ymin = -8.26935164954028, ymax = 7.364607463493773; /* image dimensions */ pen dbwrru = rgb(0.8588235294117647,0.3803921568627451,0.0784313725490196); pen xdxdff = rgb(0.49019607843137253,0.49019607843137253,1.); pen ttffcc = rgb(0.2,1.,0.8); pen sexdts = rgb(0.1803921568627451,0.49019607843137253,0.19607843137254902); pen dtsfsf = rgb(0.8274509803921568,0.1843137254901961,0.1843137254901961); draw((-6.33,5.66)--(-7.77,-2.22)--(3.97,-1.74)--cycle, linewidth(2.) + dbwrru); /* draw figures */ draw((-6.33,5.66)--(-7.77,-2.22), linewidth(2.) + dbwrru); draw((-7.77,-2.22)--(3.97,-1.74), linewidth(2.) + dbwrru); draw((3.97,-1.74)--(-6.33,5.66), linewidth(2.) + dbwrru); draw(circle((-2.0136484426050965,0.7996514920496627), 6.500298330519953), linewidth(2.)); draw((-6.33,5.66)--(1.4465521921354356,-4.7031559689743645), linewidth(2.)); draw((-6.102703114789808,0.10069701590067366)--(-3.5792493995770283,1.9943064124532957), linewidth(2.) + xdxdff); draw((-6.33,5.66)--(-6.102703114789808,0.10069701590067366), linewidth(2.)); draw((-1.9,-1.98)--(1.4465521921354356,-4.7031559689743645), linewidth(2.) + ttffcc); draw((-0.034810120877354134,-3.1057304977105225)--(-0.41863768698720627,-3.5774254712638434), linewidth(2.) + ttffcc); draw((-1.9,-1.98)--(-3.5792493995770283,1.9943064124532957), linewidth(2.) + ttffcc); draw((-3.01971301912214,-0.11119150433611884)--(-2.459536380454886,0.12549791678941263), linewidth(2.) + ttffcc); draw(circle((-1.9,-1.98), 5.874904254538961), linewidth(1.2)); draw((3.97,-1.74)--(-7.3057386859493825,0.32054107966587525), linewidth(2.)); draw((-3.545766241840223,3.6596767174386073)--(-7.77,-2.22), linewidth(2.)); draw(circle((-6.216351557394902,2.8803485079503366), 2.7819738290933795), linewidth(1.2) + sexdts); draw((-1.9,-1.98)--(-7.3057386859493825,0.32054107966587525), linewidth(2.) + linetype("2 2") + dtsfsf); draw((-1.9,-1.98)--(-3.545766241840223,3.6596767174386073), linewidth(2.) + linetype("2 2") + dtsfsf); /* dots and labels */ dot((-6.33,5.66),dotstyle); label("$A$", (-6.239850946457547,5.920303299080578), NE * labelscalefactor); dot((-7.77,-2.22),dotstyle); label("$B$", (-7.658816441319634,-1.9600229313142195), NE * labelscalefactor); dot((3.97,-1.74),dotstyle); label("$C$", (4.072987560843696,-1.4785882098431546), NE * labelscalefactor); dot((-6.102703114789808,0.10069701590067366),linewidth(4.pt) + dotstyle); label("$H$", (-6.011802920497568,0.2951186587344525), NE * labelscalefactor); dot((-1.9,-1.98),linewidth(4.pt) + dotstyle); label("$M$", (-1.805583775013523,-1.7826522444564588), NE * labelscalefactor); dot((1.4465521921354356,-4.7031559689743645),dotstyle); label("$P$", (1.539120605732826,-4.44321254732287), NE * labelscalefactor); dot((-3.5792493995770283,1.9943064124532957),linewidth(4.pt) + dotstyle); label("$T$", (-3.4019199567333716,1.9928095186587336), NE * labelscalefactor); dot((-7.3057386859493825,0.32054107966587525),linewidth(4.pt) + dotstyle); label("$E$", (-7.202720389399677,0.5231666846944305), NE * labelscalefactor); dot((-3.545766241840223,3.6596767174386073),linewidth(4.pt) + dotstyle); label("$G$", (-3.452597295835589,3.8678710654407755), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */ [/asy]$
let $G$ and $E$ be the foot of altitudes from $B$ and $C$ .
it is easy to see that as $P$ varies the locus of $T$ is a circle with diameter $AH$ now call that circle $\Omega$ now define $T$ as the intersection of $\Omega$ and $AP$ (\neq A), now let $D$ be the projection of $M$ over $AP$ again the locus of $D$ is a circle. so $P \mapsto T \mapsto D$
.
it is enough to check three different situations to see that $PD=DT$ .( $P=C,P=B,P=AH \cap (ABC)$ )
$\boxed{\mathcal{Q.E.D}}$

This is totaly wrong. You need at least 6 cases.

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#5 h Nov 25, 2020, 11:18 PM • 2 Y

Y by Merlinaeus, Mango247

or just reflect H over M and qed

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