circumcenter of OIH lies on line DE if OI// BC, incenter, orthocenter

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circumcenter of OIH lies on line DE if OI// BC, incenter, orthocenter

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Source: 2021 Ukraine NMO 10.4 11.3

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#1 h Apr 2, 2021, 6:27 PM

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Let $O, I, H$ be the circumcenter, the incenter, and the orthocenter of $\triangle ABC$ . The lines $AI$ and $AH$ intersect the circumcircle of $\triangle ABC$ for the second time at $D$ and $E$ , respectively. Prove that if $OI \parallel BC$ , then the circumcenter of $\triangle OIH$ lies on $DE$ .

(Fedir Yudin)

This post has been edited 3 times. Last edited by parmenides51, Apr 3, 2021, 9:39 PM
Reason: updated wording, circumscribed instead of described mainly and corrected proposer

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#2 h Apr 2, 2021, 8:48 PM • 1 Y

Y by Mango247

Wlog let $B > C$ . Since $AH$ and $OD$ are parallel, and $AH/AI = 2R \cos A / (r/\sin(A/2)) = 2R \cos A / (R \cos A / \sin(A/2)) = 2 \sin(A/2) = BD/DO = ID/DO$ , $AIH$ and $DIO$ are similar, so $\angle AHI = 90^\circ$ . So we have $\angle HIO = 180^\circ - (B - C)/2$ . Now consider the reflection of $O$ in $DE$ , say $O'$ . Then we have $\angle O'ED = \angle OED = 90^\circ - (B-C)/2 = 180 - \angle AED$ , so $O'$ is on $AH$ . So $DE$ is the perpendicular bisector of $OO'$ , so the circumcenter of $OO'I$ is on $DE$ . Now $\angle HO'O = \angle EO'D/2 = \angle EOD/2 = \angle EAD = (B-C)/2 = 180^\circ - \angle HIO$ , so $HIOO'$ is cyclic, and thus the conclusion of the problem follows.

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#4 h May 1, 2021, 9:23 AM • 1 Y

Y by Mango247

We use complex numbers with the assumption that $(O)$ is the unit circle. Let $A, B, C, D, I$ have complex coordinates $a^2, b^2, 1/b^2, -1, j$ respectively. Then $e=\frac{-1}{a^2}\hspace{2 mm} \wedge \hspace{2 mm} j=\frac{-(ab^2+a+b)}{b}$ Let $R$ be the circumcenter of $\triangle OIH$ . We have $\overline{j}=-j\longleftrightarrow a^{2}b^{2}+a^2+b^2+2ab+1=0\hspace{2 mm} (\diamond)$ It's enough to show that $\frac{\overline{r}+1}{r+1}=-a^2\longleftrightarrow j=\frac{-(a^2+1)(b^4+b^2+1)}{b^2(a^2-1)}$ Or $a^{4}b^{2}+a^4+2a^{3}b+2a^{2}b^2+2a^2+2ab+b^2+1\stackrel{?}{=}0\hspace{2 mm} (\star)$ But $(\star)=(a^2+1)\cdot (\diamond)=0.\hspace{3 mm} \blacksquare$

This post has been edited 4 times. Last edited by Kamran011, Aug 8, 2021, 4:42 AM

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#5 h May 7, 2021, 10:29 AM

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parmenides51 wrote:

Let $O, I, H$ be the circumcenter, the incenter, and the orthocenter of $\triangle ABC$ . The lines $AI$ and $AH$ intersect the circumcircle of $\triangle ABC$ for the second time at $D$ and $E$ , respectively. Prove that if $OI \parallel BC$ , then the circumcenter of $\triangle OIH$ lies on $DE$ .

(Fedir Yudin)

Cute!
Let $L = AI \cap (ABC), K = LO \cap (ABC), S = KI \cap (ABC), G = \ell \cap BC$ (where $\ell$ is the perpendicular line from $I$ to $AI$ ). Then $AS$ is perpendicular to $BC$ since $\angle{IOL} = \angle{IOK} = 90$ implying that $ASLK$ is an isosceles trapezoid.
Notice that $GI$ is tangent to $(BIC)$ . If we let $S' = GL \cap (ABC)$ , then
$GS'.GL = GB.GC = GI^2$ and
$\angle{LS'I} = \angle{LIG} = 90 = \angle{LS'K}$ thus $S' \equiv S$ and $G, S, L$ are collinear. Now observe that
$\angle{IGB} = \angle{IAH} = \angle{LKS} = \angle{SGB}$ hence $G,H,I$ are collinear.
Let $T$ be the reflection of $I$ across $GL$ . It suffices to show that $THIO$ is cyclic. To conclude, triangles $SIO$ and $SHI$ are similar via $\angle{SIH} = \angle{SKA} = \angle{SLI} = \angle{SOI}$ and $\angle{ISH} = \angle{OSI}$ , thus
$\frac{OS}{ST} = \frac{OS}{SI} = \frac{OI}{IH}$ Combing the previous result with $\angle{OST} = \angle{OIH}$ , we get that $\triangle OST, \triangle OIH$ are similar. Therefore $\angle{TOH} = \angle{SOI} = \angle{SIH} = \angle{TIH}$ , as required.
$\blacksquare$

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#6 h Jan 22, 2022, 7:53 AM

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I'll sketch a neat solution with Cartesian Coordinates:
Let $O$ be the origin with positive $x$ axis in the direction of $OI$ . Scale so the circumcircle is the unit circle and let $A=(a,\sqrt{1-a^2})$ . Let $B=(b,-\sqrt{1-b^2})$ and $C=(-b,-\sqrt{1-b^2})$ . WLOG $a>0$ (though we omit the case the square root is negative, that is obtuse triangles).

We get $D=(0,-1)$ and $E=(a,-\sqrt{1-a^2})$ . Reflect $E$ about $BC$ to get
$H=(a, \sqrt{1-a^2}-2\sqrt{1-b^2}).$ Now $I$ is the $x$ intercept of the line $AD$ , that is
$I=\left(\frac{a}{1+\sqrt{1-a^2}},0\right).$ Using $DI=DB$
we get a relation between $a$ and $b$ :
$(1+\sqrt{1-a^2})(1-\sqrt{1-b^2})=1.$ Now the perpendicular bisector of $OI$ has equation $x=I/2$ which intersects $DE$ at $F$ given by
$F=\left(\frac{a}{2(1+\sqrt{1-a^2})},\frac{-1-3\sqrt{1-a^2}}{2(1+\sqrt{1-a^2})}\right).$ Now it suffices to show that $OF^2=FH^2$ , or for example (let $M$ be the midpoint of $IH$ ) the slopes $FM$ and $IH$ multiply to $-1$ . Both approaches work out nicely using the relation between $a$ and $b$ .

This post has been edited 1 time. Last edited by mikestro, Jan 22, 2022, 7:58 AM
Reason: typesetting

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