If H1H2 passes through X, then it also passes through Y

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If H1H2 passes through X, then it also passes through Y

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#1 h Nov 3, 2010, 2:36 AM • 3 Y

Y by Adventure10, Mango247, and 1 other user

Let $B$ and $C$ be arbitrary points on sides $AP$ and $PD$ respectively of an acute triangle $APD$ . The diagonals of the quadrilateral $ABCD$ meet at $Q$ , and $H_1,H_2$ are the orthocenters of triangles $APD$ and $BPC$ , respectively. Prove that if the line $H_1H_2$ passes through the intersection point $X \ (X \neq Q)$ of the circumcircles of triangles $ABQ$ and $CDQ$ , then it also passes through the intersection point $Y \ (Y \neq Q)$ of the circumcircles of triangles $BCQ$ and $ADQ.$

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#2 h Apr 6, 2011, 9:17 AM • 2 Y

Y by Adventure10, Mango247

Can anyone solve this question?I am interested in this question.Please!!!

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skytin

#3 h Apr 6, 2011, 3:28 PM • 2 Y

Y by Adventure10, Mango247

This problem is from Russian Math Olympiad 9 grade 2002-2003 Problem 8

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#4 h Apr 7, 2011, 5:56 PM • 2 Y

Y by Adventure10, Mango247

skytin wrote:

This problem is from Russian Math Olympiad 9 grade 2002-2003 Problem 8

I know but I dont have solution.Have you got solution?

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yetti

#5 h Apr 29, 2011, 5:03 AM • 2 Y

Y by Adventure10 and 1 other user

See solution at http://www.artofproblemsolving.com/Forum/viewtopic.php?f=46&t=199490.

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#6 h Apr 20, 2018, 11:42 PM • 1 Y

Y by Adventure10

Amir Hossein wrote:

Let $B$ and $C$ be arbitrary points on sides $AP$ and $PD$ respectively of an acute triangle $APD$ . The diagonals of the quadrilateral $ABCD$ meet at $Q$ , and $H_1,H_2$ are the orthocenters of triangles $APD$ and $BPC$ , respectively. Prove that if the line $H_1H_2$ passes through the intersection point $X \ (X \neq Q)$ of the circumcircles of triangles $ABQ$ and $CDQ$ , then it also passes through the intersection point $Y \ (Y \neq Q)$ of the circumcircles of triangles $BCQ$ and $ADQ.$

Let $R=\overline{AD} \cap \overline{BC}$ . Suppose $\odot(AC), \odot(BD)$ meet at points $U, V$ . Line $\overline{UV}$ is their radical axis. By Gauss-Bodenmiller configuration, we know that $\overline{H_1H_2}$ is their radical axis; hence it coincides with $\overline{UV}$ . It is also well-known that $\odot(PR), \odot(AC), \odot(BD)$ are coaxial since their centers are collinear (Newton line), so $PUVR$ is cyclic. Thus, we want $X \in \overline{UV} \implies Y \in \overline{UV}$ . Invert at point $Q$ an denote images by $'$ s appended to them. Then we have $X'$ on the cline $QU'V'$ and we want $Y'$ to be on it too.

We consider two cases:

1. $ABCD$ is non-cyclic.

Then $QB \cdot QD \ne QA \cdot QC$ hence $Q \not \in \overline{UV}$ so the cline is a circle. Note that $X'=\overline{A'B'} \cap \overline{C'D'}$ and $Y'=\overline{A'D'} \cap \overline{B'C'}$ . Thus, $U',V'$ lie on $\odot(X'Y')$ and so $X' \in \odot(QU'V') \implies X'Y'QU'V'$ is cyclic. Thus $Y' \in \odot(QU'V')$ and we're done. (Notice that $U', V'$ do not coincide with $X'$ since $\angle APD$ is acute.)

2. $ABCD$ is cyclic.

Then the cline is a line. Consequently, $X'$ has equal power in $\odot(A'C'), \odot(B'D')$ . However $X'B' \cdot X'A'=X'C' \cdot X'D'$ proving $X'A'=X'C'$ and $X'B'=X'D'$ . This is impossible and we get that the result is vacuously true in this case.

Done. $\blacksquare$

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#7 h May 24, 2019, 1:33 AM • 3 Y

Y by Infinityfun, Adventure10, Mango247

Alternatively, it's also easy to finish by simple computation once you note that $H_1H_2$ is the radical axis of $(AC), (BD).$

Upon noting this, the condition reads that $X$ has equal power to both circles, if we let $M_1, M_2$ be the midpoints of $AC, BD$ respectively then we have that
$AM_1^2 - M_1X^2 = BM_2^2 - M_2X^2.$ Since $\triangle XAC \sim \triangle XBD$ , we have that either the triangles are congruent or that $AM_1^2 - M_1X^2 = BM_2^2 - M_2X^2 = 0,$ which would imply that $\angle AXC = 90.$ However, since $APCX$ is cyclic, that would give $\angle APC = 90$ , which can't be true since $\triangle APD$ is acute. Therefore we know that the triangles and congruent and so hence $XM_1 = XM_2.$ As $H_1H_2$ is the radical axis of the two circles and $X$ is on it, we know that their radical axis is just the perpendicular bisector of $M_1M_2.$ Hence, it simply suffices to observe that $\triangle YBD \sim \triangle YCA \Rightarrow \triangle YBD \cong \triangle YCA \Rightarrow YM_1 = YM_2,$ so we're done since $Y$ lies on the radical axis of the two circles as well.

$\square$

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