Concyclic circumcentres and orthocentres

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Concyclic circumcentres and orthocentres

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Source: Indian TST Day 3 Problem 1

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#1 h Jul 11, 2014, 9:23 AM • 2 Y

Y by Adventure10, Mango247

In a triangle $ABC$ , with $AB\neq AC$ and $A\neq 60^{0},120^{0}$ , $D$ is a point on line $AC$ different from $C$ . Suppose that the circumcentres and orthocentres of triangles $ABC$ and $ABD$ lie on a circle. Prove that $\angle ABD=\angle ACB$ .

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#2 h Jul 11, 2014, 7:49 PM • 3 Y

Y by anantmudgal09, Adventure10, Mango247

Let $O,H$ denote the circumcenter and orthocenter of $\triangle ABC$ and $O_1,H_1$ denote the circumcenter and orthocenter of $\triangle ABD.$ Since $\tfrac{AH}{AO}=\tfrac{AH_1}{AO_1}=2 \cos \widehat{A}$ and $AO,AO_1$ are the isogonals of $AH,AH_1$ WRT $\widehat{BAC},$ we deduce that $\triangle AHH_1$ and $\triangle AOO_1$ are inversely similar by SAS. Hence if $M \equiv BH \cap AO$ and $N \equiv OO_1 \cap AH,$ then $\triangle AHM$ and $\triangle AON$ are inversely similar with corresponing cevians $AH_1$ and $AO_1$ or $OHNM$ is cyclic and $\overline{O_1O}:\overline{O_1N}=\overline{H_1H}:\overline{H_1M}.$ Thus if $P \equiv ON \cap HM,$ then all $\odot(PO_1H_1)$ go through the center $K$ of the spiral similarity that swaps $ON$ and $HM;$ 2nd intersection of $\odot(PMN)$ and $\odot(POH).$

If $OH_1O_1H$ is cyclic $\Longrightarrow$ $O_1H_1,$ $KP,$ $OH$ and $MN$ concur at the common radical center $Q$ of $\odot(OH_1O_1H),$ $\odot(PO_1H_1),$ $\odot(POH)$ and $\odot(OHNM),$ i.e. $O_1H_1$ must pass through $Q.$ But as $O_1 \mapsto H_1$ is an affine homography between $ON$ and $HM,$ that only happens at $O_1H_1 \equiv NM,$ $O_1H_1 \equiv OH$ and nowhere else. The former case occurs when $D \equiv C$ (contradiction) and the 1st case clearly happens when $BD$ is antiparallel to $BC$ WRT $AB,AC$ and the conclusion follows.

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#3 h Jul 16, 2014, 10:15 AM • 2 Y

Y by Adventure10, Mango247

let $O,H$ are circumcenter and orthocenter of the triangle $ABC$ , anologuosly $O',H'\in \triangle ABD$ .

Let $AO\cap BH=Z$ , $AH\cap OO'=Y$ and $BH\cap OO'=X$ . Suppose that $AB<AC$ and $H\in BZ$ . Obviously, $XH'>HH'$ .

If $Z\in HH'$ then we have that $OYHZ$ is cyclic and obviously, $O'\in OY$ . Then $2\angle ADB=\angle AO'B<\angle AYB=2 \angle B$ , $180^\circ-\angle ABD=\angle AH'D>\angle A+\angle AZH'= \angle A+\angle B$ . So $\angle C>\angle ABD$ and $\angle B>\angle ADB$ . Contradiction.

If $H'\in HZ$ then $Y\in OO'$ . Again, a contradiction.

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#4 h Sep 22, 2014, 6:46 PM • 1 Y

Y by Adventure10

Let $O,H,O',H'$ be the relevant points, $W$ its circumcircle and $l$ the angle bisector of angle A. Let $X$ be the reflection of $W$ about $l$ , $Y$ the dilation of $X$ about $A$ with factor $AO/AH$ and $Z$ the reflection of $Y$ in $l$ . Then $O,O'$ are the two points of intersection of $W,Y$ . The two points of intersection of $X,Z$ are reflections of $O,O'$ in $l$ , as well as their inverses with respect to a circle centered at $A$ (in which $X,Y$ are inverses, as are $W,Z$ ). Thus $AO,AO'$ are reflections in $l$ of each other, and the desired result follows easily.

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#5 h Sep 25, 2016, 6:23 PM • 1 Y

Y by Adventure10

Let $K,L$ be the feets of the $B$ and $C$ altitudes respectively, in triangle $ABC$ and let $M$ be the midpoint of side $AB$ . Let $D$ be a moving point on the line $AC$ . Notice that the orthocenter $H_1$ of triangle $ABD$ varies on the line $BK$ such that $DH_1 \parallel CL$ . Moreover, the circumcenter $O_1$ of triangle $ABD$ moves on the line $OM$ such that $\triangle BO_1M \sim \triangle BDK$ . The maps homothety and spiral similarity are affine, so if we denote the motion of $D$ by a parameter $z$ then the motions of $H_1$ and $O_1$ are denoted by linear functions of $z$ . In particular, the relation $\frac{PH_1}{PO_1}=\frac{PO}{PH}=\text{constant}$ is a linear equation in $z$ and so, either it has one root or it vanishes always. The latter does not hold by checking, say $D=K$ . For the former, we set $D$ such that $\angle ABD=\angle ACB$ . Let $k$ be the ratio of similarity. In this case, it is clear that points $A,O_1,H$ and $A,H_1,O$ are collinear and we get that $\frac{AO_1}{AO}=\frac{AH_1}{AH}=k$ and the concyclicity holds by applying the power of point $A$ .

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