O,I,D collinear if D,X,Y collinear

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O,I,D collinear if D,X,Y collinear

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Source: Iran 3rd round 2011-geometry exam-p3

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#1 h Sep 6, 2011, 10:33 AM • 3 Y

Y by mahanmath, erfan_Ashorion, Adventure10

In triangle $ABC$ , $X$ and $Y$ are the tangency points of incircle (with center $I$ ) with sides $AB$ and $AC$ respectively. A tangent line to the circumcircle of triangle $ABC$ (with center $O$ ) at point $A$ , intersects the extension of $BC$ at $D$ . If $D,X$ and $Y$ are collinear then prove that $D,I$ and $O$ are also collinear.

proposed by Amirhossein Zabeti

This post has been edited 1 time. Last edited by goodar2006, Sep 14, 2011, 5:27 PM

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#2 h Sep 6, 2011, 4:04 PM • 16 Y

Y by bgo, mamadtakbiri, LoveMath4ever, avatarofakato, sayantanchakraborty, wiseman, bvdsf, Sx763_, Adventure10, Mango247, and 6 other users

Let the incircle $(I)$ touch $BC$ at $Z.$ $XY$ and $BC$ are the polars of $A,Z$ with respect to $(I)$ $\Longrightarrow$ $D \equiv XY \cap BC$ is the pole of $AZ$ with respect to $(I)$ $\Longrightarrow$ $ID \perp AZ.$ Since $DA$ is tangent to the circumcircle $(O)$ and the cross ratio $(B,C,D,Z)$ is harmonic, then we deduce that $AZ$ is also the polar of $D$ with respect to $(O)$ $\Longrightarrow$ $OD \perp AZ.$ Therefore, $O,I,D$ lie on a perpendicular to $AZ.$

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#3 h Sep 6, 2011, 4:22 PM • 1 Y

Y by Adventure10

really beautiful luis, I couldn't believe it the moment I was reading because I used a 3page complete brute force for solving it

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#4 h Sep 17, 2011, 12:19 AM • 6 Y

Y by Adventure10, Mango247, and 4 other users

An interesting metrical remark.

IRAN, 2011. In $\triangle ABC$ let $X$ and $Y$ be the tangent points of incircle $w=C(I,r)$ with $AB$ and $AC$ respectively. The tangent at $A$ to the circumcircle

$C(O,R)$ of $\triangle ABC$ intersects $BC$ at $D$ . Prove that $D\in XY\iff D\in OI\iff IL\parallel BC\iff\ IO\perp AZ$ $\iff$ $\ a(b+c)=b^2+c^2$

$\iff (s-a)^2=(s-b)(s-c)$ where $2s=a+b+c$ , $Z\in BC\cap w$ and $L$ is the Lemoine's point (symmedian center) .

Proof. Suppose w.l.o.g. $b\ne c$ .

$\blacktriangleright\ \boxed{D\in XY}\iff$ $\frac {DB}{DC}=\frac{ZB}{ZC}\iff$ $\frac {c^2}{b^2}=\frac {s-b}{s-c}\iff$ $\frac {b^2-c^2}{b^2+c^2}=\frac {b-c}a\iff$ $\frac {b+c}{b^2+c^2}=\frac 1a\iff$ ${\boxed{a(b+c)=b^2+c^2}}$ .

$\blacktriangleright\ \boxed{IL\parallel BC}\iff$ $[BLC]=[BIC]\iff$ $\frac {a^2}{a^2+b^2+c^2}=\frac {a}{2s}\iff$ $a^2(a+b+c)=a\left(a^2+b^2+c^2\right)\iff$ $\boxed{a(b+c)=b^2+c^2}$ .

$\blacktriangleright\ \boxed{(s-a)^2=(s-b)(s-c)}\iff$ $a^2-2a(b+c)+(b+c)^2=a^2-(b-c)^2\iff$ $\boxed{a(b+c)=b^2+c^2}$ .

$\blacktriangleright\ \boxed{IO\perp AZ}\iff$ $OA^2-OZ^2=IA^2-IZ^2\iff$ $ZB\cdot ZC=IA^2-IX^2\iff$ $\boxed{(s-b)(s-c)=(s-a)^2}$ .

$\blacktriangleright\ \boxed{D\in XY}\iff$ $Z$ is the conjugate of $D$ w.r.t. $\{B,C\}$ $\iff$ $AZ$ is polar line of $D \iff$ $DO\perp AZ$ $\iff \boxed{D\in OI}$ .

This post has been edited 7 times. Last edited by Virgil Nicula, Jan 28, 2016, 9:42 PM

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#5 h Apr 2, 2012, 9:24 PM • 2 Y

Y by Adventure10, Mango247

oh..nice problem...and nice solution by me

lemma 1:suppose that $AZ$ intersect incircle at $P$ draw tangent to incircle at $P$ proof that $D$ lie on this tangent..!proof:its not hard and im sure that see it in mathlink if U need say me i message it
lemma 2 $\frac{AB^2}{AC^2}=\frac{BZ}{ZC}$ that $Z$ is the point that incircle is tangent to $BC$
proof:
$\frac{BZ}{ZC}=\frac{BD}{DC}=\frac{DA^2}{DC^2}=\frac{{\sin{C}}^2}{{\sin{DAC}}^2}=\frac{{\sin{C}}^2}{{\sin{B}}^2}=\frac{AB^2}{AC^2}$
proof or problem:
we proof $IDC=ODC$ first we have:( $H$ is the perpendicular of $A$ on $BC$ and $M$ is midpoint of $BC$
$IDC=IPZ=IZP=ZAH$
for second we have:
$ODC=OAM$
we know that:
$MAC=ZAB$ (because that we proof by lemma 2 $AZ$ is symmedian!)and $OAC=HAB\rightarrow OAM=ZAH \rightarrow IDC=ODC$

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sayantanchakraborty

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sayantanchakraborty

#6 h Sep 1, 2014, 5:53 PM • 2 Y

Y by Adventure10, Mango247

If $D,X,Y$ are collinear,then simple angle chasing yeilds that $\angle{ADC}$ is bisected by $DXY$ .So we have $\frac{DB}{DA}=\frac{BX}{AX}=\frac{s-b}{s-a}$ and $\frac{DC}{DA}=\frac{CY}{AY}=\frac{s-c}{s-a}$ .Multiplying these two relations and noting that $DA^2=DB \times DC$ we get $(s-a)^2=(s-b)(s-c) \Rightarrow {\triangle}^2=(s-a)^3$ .Now let $IZ \perp DC$ and $OK \perp DC$ with $Z,K$ on $BC$ .Then sine rule in $\triangle{DBA}$ gives

$\frac{DB}{c}=\frac{sinC}{sin(B-C)} \Rightarrow DB=\frac{csinC}{sin(B-C)}=\frac{\frac{c^2}{2R}}{\frac{b}{2R}\frac{a^2+b^2-c^2}{2ab}-\frac{c}{2R}\frac{a^2+c^2-b^2}{2ac}}=\frac{c^2a}{b^2-c^2}$ .

$\frac{IZ}{OK}=\frac{r}{RcosA}=\frac{\frac{\triangle}{s}}{\frac{abc}{4\triangle}\frac{c^2+b^2-a^2}{2bc}}=\frac{2(b+c-a)^3}{a(c^2+b^2-a^2)(a+b+c)}$ .(Here we have used the fact that ${{\triangle}^2=(s-a)^3}$ ).

Now $\frac{DZ}{DK}=\frac{IZ}{OK} \Leftrightarrow \frac{DB+(s-b)}{DB+\frac{a}{2}}=\frac{2(b+c-a)^3}{a(c^2+b^2-a^2)(a+b+c)} \Leftrightarrow DB=\frac{c^2a}{b^2-c^2}$

which is true.So $\frac{DZ}{DK}=\frac{IZ}{OK}$ and $D,I,O$ are collinear.

I think the statement can be 'if and only if' because both the conditions yeild $(s-a)^2=(s-b)(s-c)$ .Anyways credit goes to Luis for his projective approach.

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