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Source: Iran TST 2006

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Nima Ahmadi Pour

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Nima Ahmadi Pour

#1 h Apr 18, 2006, 4:53 PM • 2 Y

Y by Adventure10, Mango247

Let $ABC$ be a triangle such that it's circumcircle radius is equal to the radius of outer inscribed circle with respect to $A$ .
Suppose that the outer inscribed circle with respect to $A$ touches $BC,AC,AB$ at $M,N,L$ .
Prove that $O$ (Center of circumcircle) is the orthocenter of $MNL$ .

This post has been edited 1 time. Last edited by Nima Ahmadi Pour, Apr 19, 2006, 9:57 AM

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Sailor

#2 h Apr 18, 2006, 5:27 PM • 1 Y

Y by Adventure10

Nima Ahmadi Pour wrote:

Let $ABC$ be a triangle such that it's circumcircle radius is equal to the radius of outer inscribed circle with respect to $A$ .
Suppose that the outer inscribed circle with respect to $A$ touches $BC,AC,BC$ at $M,N,L$ .
Prove that $O$ (Center of circumcircle) is the orthocenter of $MNL$ .

I think you meant that it touches $AB$ at $L$ ...

Anyway, let $BI_a$ meet the circumcircle of $\triangle{ABC}$ at $K$ . Then $K$ is the midpoint of the arc $\widehat{ABC}$ , thus $OK\perp{AC}$ . Since $I_aN\perp{AC}$ and $OK=I_aN$ we conclude that $KI_a||ON$ , but $LM\perp{KI_a}$ $\Longleftrightarrow$ $LM\perp{ON}$ .

Using similar arguments we obtain the conclusion.

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Nima Ahmadi Pour

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Nima Ahmadi Pour

#3 h Apr 19, 2006, 10:06 AM • 2 Y

Y by Adventure10, Mango247

Sorry, I just made it correct.
Anyway, thanks for the notification.

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#4 h Apr 20, 2006, 5:56 PM • 2 Y

Y by Adventure10, Mango247

I thought of
$R=r_a \rightarrow R=p tg \frac{A}{2} \rightarrow |OM|^2-|ON|^2=|LM|^2-|LN|^2$ ...

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#5 h Apr 20, 2006, 6:43 PM • 2 Y

Y by Adventure10, Mango247

An interesting remark. Denote the projection $D$ of the vertex $A$ on the opposite side $[BC]$ . Then $\boxed {R=r_a\Longleftrightarrow D\in OI}\ .$

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#6 h Apr 29, 2006, 11:45 AM • 2 Y

Y by Adventure10, Mango247

this problem is way too easy for IRAN tst.
a solution very similar to sailor's but even simpler:
M on BC, N on AC, L on AB. Suffices to prove OL perpendicular to MN (because then the similar argument that OM perpendicular to LN finishes off the problem neatly). MN is perpendicular to IC with I = excentre of A-excircle. So it suffices to prove a stronger result, i.e. ICOL is an iscoceles trapezium. OC = IL (since R=r_a), so it suffices to prove <OCI=<LIC. Simple angle chasing yields if <A = 2a,etc; then <OCI=<OCB+<ICB=90-2a+90-c=2b+c; <LIC=<LIB+<BIC = b + (b+c) = 2b+c, so we are done.

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#7 h May 1, 2006, 9:20 PM • 1 Y

Y by Adventure10

hey virgil nicula, do you have a good proof of the 'interesting remark' about D,I,O collinear when R= r_a? i cant do it without side-bashing or trig-bashing.

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#8 h Mar 8, 2010, 4:58 PM • 2 Y

Y by Adventure10, Mango247

I have a solution:
Let X be the excentre opposite A and D be the point (#A) where AX intersect (ABC). Because OD // MX ( perpendicular BC),
OD=MX so ODXM is a parallelogram,AX perpendicular LN so OM perpen dicular LN.
Let the internal angle bisector of BAC intersect (ABC) at E, EBX = 90, EA = EC. Let T the point of intersection of BX and(ABC) so TA = TC and TE passes O.Let B' the foot of the perpendicular on AC from T. TB' passes O
Because OT // NX and OT= NX so TONX is a parallelogram so TX // ON hence ON perpendicular NM
Hence O is the ortheocentre of MNL

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#9 h Jun 5, 2022, 12:28 PM • 1 Y

Y by Mango247

Let $I_A$ be the $A$ -excenter ; lines $I_AA,I_AB,I_AC$ meet $\odot(ABC)$ again at points $A',B',C'$ , respectively. It isn't hard to note that:

$I_A$ is the orthocenter of $\triangle A'B'C'$ .
corresponding sides of $\triangle A'B'C'$ and $\triangle MNL$ are parallel to each other and the two triangles have opposite orientations.

Using the fact that $\triangle A'B'C'$ and $\triangle MNL$ have equal circumradius, we obtain there is a point $S$ such that reflection in $S$ (which we denote by $\mathcal R$ ) swaps $\{A',M\},\{B',N\},\{C',L\}$ . Then as $O,I_A$ are corresponding cirucmenters of $\triangle A'B'C,\triangle MNL$ , so $\mathcal R$ swaps $\{O,I_A\}$ . Since $I_A$ was the orthocenter of $\triangle A'B'C'$ , hence $O$ is the orthocenter of $\triangle MNL$ . $\blacksquare$

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