circumcenter of an acute triangle [BT perpendicular DE]

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circumcenter of an acute triangle [BT perpendicular DE]

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Source: Russian Olympiad 2004, problem 9.8

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3647 posts

orl

#1 h May 3, 2004, 7:18 PM • 3 Y

Y by narutomath96, Adventure10, Mango247

Let $O$ be the circumcenter of an acute-angled triangle $ABC$ , let $T$ be the circumcenter of the triangle $AOC$ , and let $M$ be the midpoint of the segment $AC$ . We take a point $D$ on the side $AB$ and a point $E$ on the side $BC$ that satisfy $\angle BDM = \angle BEM = \angle ABC$ . Show that the straight lines $BT$ and $DE$ are perpendicular.

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#2 h May 5, 2004, 11:54 AM • 4 Y

Y by narutomath96, Adventure10, Mango247, and 1 other user

Assume the circumcircle of $\triangle AOC$ intersects the sides $AB$ and $BC$ again at $A'$ and $C'$ respectively. The foot of the perpendicular from $T$ to $BC$ is the midpt of $CC'$ . It's easy to show (quick angle chase) that $AC'||DM||A\"C$, where $A\"$ is the reflection of $A$ in $D$ . All of this means that $DM$ cuts $BC$ in the midpoint of $CC'$ , which is also the foot of the perpendicular from $T$ to $BC$ . In the same way we show that $EM$ cuts $AB$ in the midpt of $AA'$ , which is the foot of the perpendicular from $T$ to $AB$ .

Let the perpendicular from $T$ to $AB$ cut $BC$ in $C_1$ . Similarly we define $A_1$ . Let $X$ and $Y$ be the midpts of $AA'$ and $CC'$ respectively. The figures $BXEC_1$ and $BYDA_1$ are obviously similar, so $A_1C_1||DE$ (#).

In $\triangle BA_1C_1$ the point $T$ is the orthocenter, so $BT\perp A_1C_1$ (##).

From (#) and (##) we get the desired conclusion.

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#3 h May 14, 2004, 6:43 PM • 3 Y

Y by narutomath96, Adventure10, Mango247

Let N be the midpoint of BC and P be the midpoint of AC. Let D' be the intersection of the perpendicular bisector of MN with AB. It is easy to show that D' is the midpoint of BD. Let E' be the intersection of the perpendicular bisector of MP with BC. E' is the midpoint of BE.

Set up a coordinate system with the center in O and such that the circumcenter of ABC is the unit circle. Then, by a quick and easy computation we obtain : (we have denoted by lowercase leters the affixes of points denoted by uppercase letters)

d = 2a + b + c - ab/c
e = a + b + 2c - bc/a
t = ac/(a+c)

We must prove that (d-e)/(b-t) is an imaginary complex number. But :

d-e = (ab+bc-ac)(c-a)/ac
b-t = (ab+bc-ac)/(a+c)

Therefore z = (d-e)/(b-t) = (c-a)(c+a)/ac. It is easy to see that if z' is the conjugate of z , z+z'=0. Whence the conclusion.

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#4 h May 16, 2004, 4:22 AM • 3 Y

Y by narutomath96, Adventure10, Mango247

we have BE=RsinA+2RsinCcosB
BD=RsinC+2sinAcosB
DE <sup>2</sup> =R <sup>2</sup> [(sinC) <sup>2</sup>+6sinAsinCcosB +(sinC) <sup>2</sup> -8sinAsinC(cosB)^3]
then[ 1/(sinDEB) <sup>2</sup>] -1=
[(cos(A-B)) <sup>2</sup> /(sinC+2sinAcosB) <sup>2</sup> ]
then tg TBC=[cosBcos(A-B)]/[sin(A-B)cosB-sinA] then tg TBC=cotgDEB => we prove

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j555

#5 h Aug 29, 2006, 1:04 PM • 4 Y

Y by narutomath96, Adventure10, Mango247, and 1 other user

Consider a homothety $h$ with center in centroid of $\triangle ABC$ , s.t. $h(M)=B$ . Denote
$\angle ABC=\alpha$
$F=h(D)$ and $G=h(E)$ .
It is easy that
$\angle BFC=\angle BCF=\angle BGA=\angle BAG=\angle TCO=\angle TOC=\angle TAO=\angle TOA=\alpha$ .
Thus
$\triangle CFB \sim \triangle COT$ and $\triangle AGB \sim \triangle AOT$ , so
$\triangle CFO \sim \triangle CBT$ and $\triangle AGO \sim \triangle ABT$ . Hence
$OF=\frac{TB\cdot OC}{TC}=\frac{TB\cdot OA}{TA}=OG$ and
$\angle FOG=\angle OCT+\angle OAT=2\alpha$ .
Let $k$ be a bisector of $\angle FOG$ . We have
$\angle(OF, BT)=\angle OCT=\alpha=\angle(OF, k)$ .
Thus $BT \parallel k$ and $k\perp FG$ because $\triangle OFG$ is isocles
triangle, so $BT\perp FG$ . By homothety $DE\parallel FG$ and in the end $BT\perp DE$ .

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484 posts

#6 h Feb 13, 2008, 9:51 PM • 1 Y

Y by Adventure10

Thanhliem wrote:

we have BE=RsinA+2RsinCcosB
BD=RsinC+2sinAcosB
DE <sup>2</sup> =R <sup>2</sup> [(sinC) <sup>2</sup>+6sinAsinCcosB +(sinC) <sup>2</sup> -8sinAsinC(cosB)^3]
then[ 1/(sinDEB) <sup>2</sup>] -1=
[(cos(A-B)) <sup>2</sup> /(sinC+2sinAcosB) <sup>2</sup> ]
then tg TBC=[cosBcos(A-B)]/[sin(A-B)cosB-sinA] then tg TBC=cotgDEB => we prove

Can you involve a little bit more detail?

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7054 posts

#7 h Apr 8, 2008, 10:46 PM • 3 Y

Y by narutomath96, Adventure10, Mango247

orl wrote:

Let $C(O)$ , $C(T)$ be the circumcircles of an acute $\triangle ABC$ and $\triangle AOC$ respectively. Let $M$ be the midpoint of $[AC]$ .

Consider the points $D\in [AB]$ and $E\in [BC]$ which satisfy $\widehat {BDM}\equiv\widehat {BEM}\equiv\widehat {ABC}$ . Show that $BT\perp DE$ .

Proof (similar with the nice Grobber's proof). I"ll use the Grobber's notations. Denote the points $\left\{\begin{array}{c} A'\in BA\ ,\ A'B = A'C \\ \\ C'\in BC\ ,\ C'B = C'A\end{array}\right\|$ .

Observe that $\left \{\begin{array}{c} m(\widehat {AOC}) = 2B \\ \\ m(\widehat {AA'C}) = m(\widehat {AC'C}) = 180^{\circ} - 2B\end{array}\right\|$ , i.e. the points $A'$ , $C'$ belong to the circle $C(T)$ .

Denote the midpoints $X$ , $Y$ of the segments $[AA']$ , $[CC']$ respectively. Observe that $\left\{\begin{array}{c} MX\parallel A'C\implies E\in XM \\ \\ MY\parallel C'A\implies D\in MY\end{array}\right\|$ .

$\left \{\begin{array}{c} TX\perp BA \\ \\ TY\perp BC\end{array}\right\|\implies$ the quadrilateral $BXTY$ is ciclically (the circumcenter is the midpoint of the segement $[BT]$ ).

From the relations $m(\widehat {XDY}) = m(\widehat {XEY}) = 180^{\circ} - B$ obtain that the quadrilateral $DEYX$ is cyclically.

Thus, the perpendicular line $d$ from the point $B$ to the line $DE$ passes through the center of the circumcircle of $\triangle BXY$

(and of the quadrilateral $BXTY$ ) , i.e. the midpoint of the segment $[BT]$ . Thus, $d\equiv BT$ . In conclusion, $BT\perp DE$ .

This post has been edited 1 time. Last edited by Virgil Nicula, Dec 11, 2016, 9:08 AM

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The QuattoMaster 6000

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The QuattoMaster 6000

#8 h Jul 29, 2008, 5:27 AM • 2 Y

Y by Adventure10, Mango247

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#9 h Nov 30, 2010, 1:14 AM • 3 Y

Y by narutomath96, Adventure10, Mango247

iandrei wrote:

Let N be the midpoint of BC and P be the midpoint of AC. Let D' be the intersection of the perpendicular bisector of MN with AB. It is easy to show that D' is the midpoint of BD. Let E' be the intersection of the perpendicular bisector of MP with BC. E' is the midpoint of BE.

Set up a coordinate system with the center in O and such that the circumcenter of ABC is the unit circle. Then, by a quick and easy computation we obtain : (we have denoted by lowercase leters the affixes of points denoted by uppercase letters)

d = 2a + b + c - ab/c
e = a + b + 2c - bc/a
t = ac/(a+c)

We must prove that (d-e)/(b-t) is an imaginary complex number. But :

d-e = (ab+bc-ac)(c-a)/ac
b-t = (ab+bc-ac)/(a+c)

Therefore z = (d-e)/(b-t) = (c-a)(c+a)/ac. It is easy to see that if z' is the conjugate of z , z+z'=0. Whence the conclusion.

It's a pretty minor thing, but I think it's worth pointing out: it should be
$2d=2a+c-\frac{ab}{c}$ and
$2e=a+2c-\frac{bc}{a}.$

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NewAlbionAcademy

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NewAlbionAcademy

#10 h Jun 12, 2013, 12:21 AM • 2 Y

Y by mgrulz, Adventure10

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leader

#11 h Jun 12, 2013, 2:07 PM • 2 Y

Y by Adventure10, Mango247

Let $N,P,L$ be midpoints of $BC,BA$ and the center of the 9-point circle of $\triangle ABC$ . $S$ is the foot of perpendicular of $B$ on $AC$ . Since the composition of inversion with center $B$ and radius $\frac{BA\cdot BC}{2}$ and symmetry WRT the internal bisector of $\angle ABC$ pictures $S,N,P$ to $O,A,C$ then it pictures $\odot SNP$ to $\odot AOC$ so $\angle CBT=\angle ABL$ (1). Note that $MN||BD$ and $\angle MDB=\angle NBD$ then $MNBD$ is an isosceles trapezoid so since $LM=LN$ then $LB=LD$ Likewise $LB=LE$ so $L$ is the circumcenter of $\triangle BDE$ . So from (1) $BT\perp DE$

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#12 h Aug 25, 2014, 10:55 AM • 2 Y

Y by Adventure10, Mango247

Let the circumcircle of AOC intersects AB and CB at S and P,respectively.Now,we easy have that MD/ME=BS/BP=BC/BA(Let R and Q be the midpoints of BC and BA,then MD=MQ and MR=ME and the conclusion follows).Now,it is easy by angle chasing that <STP=2<ABC,now let point N be such that BR=BN and <RBN=<STP and <SBN=3<ABC.Now,SBN is similar to MDE,so we need to prove that(this is reduced by trivial angle chase) <PNS=<PBT,but this is obvious aplying spiral homothety with center P,so BPN is similar to STP,from which we have that SPN is similar to BPT,so we are finished.

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#13 h Dec 11, 2016, 8:37 AM • 2 Y

Y by Adventure10, Mango247

Let $F$ be the point such that $\triangle FBA \sim \triangle FCB$ . It is well-known that $BF$ is a symmedian in triangle $ABC$ and points $A,F,O,C$ lie on a circle. Let $E'$ be the intersection of lines $AB$ and $CF,$ and $D'$ be the intersection of lines $AF$ and $BC$ .

Apply inversion at $B$ of radius $\sqrt{\frac{1}{2}\cdot AB\cdot BC}$ followed by reflection in the internal bisector of angle $ABC$ . As the circle $(BDM)$ passes through the midpoint of $BC$ and $M \rightarrow F,$ we conclude that $D \rightarrow D'$ . Similarly, $E \rightarrow E'$ and we obtain $D'E' \parallel DE$ .

Finally, we note that $\angle D'BF=\angle D'CB \Longrightarrow D'B^2=D'F\cdot D'C.$ Similarly, $E'B^2=E'F\cdot E'A$ . Thus, the line $D'E'$ is the radical axis of $B$ and the circle $(AFC)$ . The conclusion $BT \perp DE$ follows.

Note: The fact that $B$ acts as a "point circle" was crucial to finish the proof.

This post has been edited 1 time. Last edited by anantmudgal09, Dec 11, 2016, 8:38 AM

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L567

#14 h Oct 9, 2021, 8:09 PM • 2 Y

Y by Aayuu, p_square

Solved with p_square, Aayuu

Let $Z$ denote the reflection of $O$ across $AC$ . Observe that $AMZD$ and $CMZE$ are cyclic. So, it suffices to show $BZ$ and $BT$ are isogonal since $BZ$ is the diameter of $(BDZE)$ .

Let $T'$ be the isogonal conjugate of $T$ in $\triangle ABC$ , so it suffices to show $Z,B,T'$ are collinear or that $BT'$ goes through the circumcenter of $\triangle AHC$ , or show that $T'$ lies on the euler liner of $\triangle AHC$ .

So we have reduced the problem to the following.

Rephrased problem wrote:

Let $ABDC$ be a quadrilateral with $\angle A = \angle B = \angle C$ . Show that $D$ lies on the Euler line of $\triangle ABC$ .

To do this, let $X = AB \cap CD$ and $Y = BD \cap AC$ . Let $P,Q$ denote midpoints of $AB,AC$ and let $G$ be the centroid. Note that $YB = YA$ and $XA = XC$ .

Black magic, by Pappus on $XBP$ and $YCQ$ , $D$ lies on $OG$ , the Euler line, done. $\blacksquare$

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#15 h Oct 17, 2021, 7:14 PM • 1 Y

Y by IAmTheHazard

Let rays $BA,BC$ meet $\odot (AOC)$ again at $S,R$ respectively; $DM\cap BC=N,DE\cap BA=K.$ Observe that $RA\parallel MD$ (analogously $SC\parallel ME$ ), since $\angle RAB=\pi-\angle ABR-\angle ARB=\angle ABR=\angle MDB.$ Hence by the Thales $N,K$ are midpoints of $CR,AS.$ Next $\angle BKT=90^{\circ}=\angle BNT,\angle KDN=\angle KEN,$ so $BNTK,DENK$ are cyclic.
Thus the diameter $BT$ of $\odot (BNK)$ is perpendicular to $DE,$ as desired.

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#16 h Aug 25, 2023, 1:54 AM • 1 Y

Y by centslordm

Let $P$ and $Q$ be the feet of the altitudes from $M$ to $\overline{BA}$ and $\overline{BC}$ respectively. Then it is clear that $D$ is the reflection of the midpoint of $\overline{BA}$ over $P$ , and a similar fact is true for $E$ . Now use complex numbers with $(ABC)$ as the unit circle, setting $A=a,B=1,C=c$ , so $M=\tfrac{a+c}{2}$ . We can compute $P=\tfrac{3}{4}a+\tfrac{1}{4}c-\tfrac{a}{4c}+\tfrac{1}{4}$ and $Q=\tfrac{3}{4}c+\tfrac{1}{4}a-\tfrac{c}{4a}+\tfrac{1}{4}$ , hence $D=a+\tfrac{1}{2}c-\tfrac{a}{2c}$ and $E=c+\tfrac{1}{2}a-\tfrac{c}{2a}$ . Furthermore, the circumcenter formula yields $t=\tfrac{ac}{a+c}$ . We wish to prove that
$\frac{2(d-e)}{t-1}=\frac{a-c-\frac{a}{c}+\frac{c}{a}}{\frac{ac-a-c}{a+c}}=\frac{(a^2-c^2)(1-\frac{a+c}{ac})}{ac-a-c}=\frac{a^2-c^2}{ac}$ is imaginary, which is clear since its conjugate is its negative.

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