Perpendicular to AA1 through A1 meets B1C1 at X

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Perpendicular to AA1 through A1 meets B1C1 at X

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Source: Saint-Petersburg MO 2002

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39 posts

Andrei

#1 h Nov 1, 2004, 2:51 PM • 2 Y

Y by Adventure10, Mango247

Let $ABC$ be a triangle. The incircle of triangle $ABC$ touches the sides $BC$ , $CA$ , $AB$ at the points $A_{1}$ , $B_{1}$ , $C_{1}$ respectively. The perpendicular to the line $AA_{1}$ through the point $A_{1}$ intersects the line $B_{1}C_{1}$ at a point $X$ .
Prove that the line $BC$ bisects the segment $AX$ .

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#2 h Nov 1, 2004, 3:42 PM • 2 Y

Y by Adventure10, Mango247

Let $T$ be the second point where $AA_1$ cuts the incircle, and let $S$ be $B_1C_1\cap BC$ . Also, let $P=TS\cap A_1X, R=B_1C_1\cap AA_1$ . It's easy to see that all we need to show is that $PT\|AX$ (because $S$ is the midpt of $TP$ ).

We apply Menelaus in $PTA_1$ with transversal $RSX$ to get $\frac{XP}{XA_1}=\frac{RT}{RA_1}=\frac{AT}{AA_1}$ (the last equality holds because $A_1B_1TC_1$ is a harminic quadrilateral), and this is all we need to show that $TP\|AX$ .

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Andrei

#3 h Nov 1, 2004, 4:31 PM • 2 Y

Y by Adventure10, Mango247

Great solution Grobber

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Andrei

#4 h Nov 1, 2004, 7:58 PM • 2 Y

Y by Adventure10, Mango247

Here is my solution
Let $P,R$ be the interesction of line $A_1I$ with $B_1C_1$ and the incircle respectively. Also denote $M$ and $N$ to be the intersection of lines $AP$ and $AR$ with side $BC$ . It is easy to see that $N$ is the point where excircle of the triangle (of side $BC$ ) touches side $BC$ . It is a bit harder to prove that $M$ is the midpoint of side $BC$ ( By the way, this fact was a problem in Moldovian Olimpiad in 8-th grade a few years ago). Now let $T=B_1C_1\cap BC$ and $V=AX\cap BC$ . Let $U=AA_1\cap MI$ Obviously $AU=UA_1$ . Using Menelaos Theorem in triangle $AA_1P$ with transversal $MU$ we have that: $\frac{AU}{UA_1}\frac{A_1I}{IP}\frac{MP}{MA}=1$ and since $AU=UA_1$ we have $\frac{A_1I}{IP}\frac{MP}{MA}=1$ (1). Using the Polar is easy to prove that $IT\perp AA_1$ and thus $IT \parallel A_1X$ and so $\frac{A_1I}{IP}=\frac{XT}{TP}$ (2). Using Menelaos in triangle $AXP$ with transversal $MV$ we get $\frac{AV}{VX}\frac{XT}{TP}\frac{MP}{MA}=1$ (3). Combining the results from (1), (2)and (3) we get the desired result $AV=VX$ q.e.d.

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#5 h Nov 1, 2004, 8:42 PM • 2 Y

Y by Adventure10, Mango247

Andrei wrote:

Here is my solution
Let $P,R$ be the interesction of line $A_1I$ with $B_1C_1$ and the incircle respectively. Also denote $M$ and $N$ to be the intersection of lines $AP$ and $AR$ with side $BC$ . It is easy to see that $N$ is the point where excircle of the triangle (of side $BC$ ) touches side $BC$ . It is a bit harder to prove that $M$ is the midpoint of side $BC$ ( By the way, this fact was a problem in Moldovian Olimpiad in 8-th grade a few years ago).

Yes, I have posted a proof of this on http://www.mathlinks.ro/Forum/viewtopic.php?t=5830 .

Andrei wrote:

Now let $T=B_1C_1\cap BC$ and $V=AX\cap BC$ . Let $U=AA_1\cap MI$ Obviously $AU=UA_1$ . Using Menelaos Theorem in triangle $AA_1P$ with transversal $MU$ we have that: $\frac{AU}{UA_1}\frac{A_1I}{IP}\frac{MP}{MA}=1$ and since $AU=UA_1$ we have $\frac{A_1I}{IP}\frac{MP}{MA}=1$ (1). Using the Polar is easy to prove that $IT\perp AA_1$ and thus $IT \parallel A_1X$ and so $\frac{A_1I}{IP}=\frac{XT}{TP}$ (2). Using Menelaos in triangle $AXP$ with transversal $MV$ we get $\frac{AV}{VX}\frac{XT}{TP}\frac{MP}{MA}=1$ (3). Combining the results from (1), (2)and (3) we get the desired result $AV=VX$ q.e.d.

Both of you have very nice proofs, but the proposed solution is really beautiful. I'm attaching a PDF with a slightly rewritten form of that solution. Enjoy.

PS. Is it right that now the whole Moldavian IMO team is on MathLinks?

Darij

Attachments:

Petersburg.pdf (43kb)

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yetti

#6 h Mar 16, 2005, 4:19 AM • 2 Y

Y by Adventure10, Mango247

Let $I$ be the triangle incenter, $r$ the inradius and $P$ the intersection of the line $B_1C_1$ with the angle bisector $AI$ . From similarity of the right angle triangles $\triangle AIC_1 \sim \triangle C_1IP$ ,

$\frac{r}{IA} = \frac{IC_1}{IA} = \frac{IC_1}{IP} = \frac{IP}{r}$

Consequently, the triangles $\triangle AIA_1 \sim \triangle A_1IP$ are similar by SAS: the angle $\measuredangle AIA_1 \equiv \measuredangle A_1IP$ is common and

$\frac{r}{IA} = \frac{IA_1}{IA} = \frac{IP}{IC_1} = \frac{IP}{r}$

Thus the angles $\measuredangle IAA_1 = \measuredangle IA_1P$ are equal. The lines $AA_1 \perp A_1X$ are perpendicular by definition and the lines $AI \perp B_1C_1$ are perpendicular, because the line $B_1C_1$ is a polar of the vertex $A$ with respect to the incircle $(I)$ perpendicular to the line connecting its pole $A$ with the center of the polar circle $(I)$ . Since the point $X \equiv B_1C_1 \cap A_1X$ is on the line $B_1C_1$ , the right angle triangles $\triangle APX, \triangle AA_1X$ have the common hypotenuse $AX$ . The midpoint $M$ of this hypotenuse is the common circumcenter of the triangles $\triangle APX, \triangle AA_1X$ . The quadrilateral $APA_1X$ is cyclic with the circumcircle $(M)$ . The angles $\measuredangle PAA_1 = \measuredangle PXA_1$ are equal because they span the same arc $PA_1$ of the circle $M$ . But then the angles $\measuredangle IAA_1 \equiv \measuredangle PAA_1 = \measuredangle PXA_1$ are also equal (see above). This means that the line $IA_1$ is a tangent to the circle $(M)$ at the tangency point $A_1$ . Since $r = IA_1$ is the inradius, the lines $IA_1 \perp BC$ are perpendicular. As a result, the line $BC$ passes through the center $M$ of the circle $(M)$ , identical with the midpoint of the segment $AX$ . Q.E.D.

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