Let $ABC$ be an acute triangle with orthocenter $H$. Segments $AH$ and $CH$ intersect segments $BC$ and $AB$ in points $A_1$ and $C_1$ respectively. The segments $BH$ and $A_1C_1$ meet at point $D$. Let $P$ be the midpoint of the segment $BH$. Let $D'$ be the reflection of the point $D$ in $AC$. Prove that quadrilateral $APCD'$ is cyclic. Proposed by Matko Ljulj.
Problem
Source: European Mathematical Cup 2012, Senior Division, Problem 2
Tags: geometry, geometric transformation, reflection, circumcircle, geometry proposed
pepat
27.07.2013 18:28
I suppose segments should be lines. The following solution is probably not the easiest one but anyway:
The idea is to introduce some additional points to easily make use of the very familiar configuration of triangle, orthocenter, and nine-point circle.
Observe that $P$ is the circumcenter of $\triangle A_1BC_1$ and let $M$, $N$ be the midpoints of $AC$ and $A_1C_1$, respectively. Since $M$ is the circumcenter of $AC_1A_1C$, quadrilateral $MA_1PC_1$ is a kite, hence $M$, $N$, $P$ are collinear and (as $MP$ is a diameter of nine-point circle) \[MN\cdot MP=MA_1^2 = MA\cdot MC.\] Denoting by $N'$ the reflection of $N$ about $M$ we realize $APCN'$ is cyclic (power of $M$), thus it suffices to prove $ADNC$ is cyclic.
If $\triangle ABC$ is $B$-isosceles, the conclusion is trivial. Suppose $AB<BC$ and let $AC$ and $A_1C_1$ intersect at $X$. Denoting the foot of $A$-altitude by $A_1$ we see that $DA_1MN$ is cyclic ($\angle DA_1M=90^\circ = \angle DNM$) which together with cyclic quadrilaterals $MA_1C_1B_1$ (nine-point circle) and $AC_1A_1C$ gives the result -- for instance by power of $X$ we have
\[ XD\cdot XN = XA_1\cdot XM = XC_1\cdot XA_1 = XA\cdot XC.\]
sunken rock
30.07.2013 01:09
Let $B_1=BH\cap AC$. It is known that $BB_1$ is the bisector of $\angle A_1B_1C_1$ and $A_1B_1C_1P$ is cyclic (and all those points lie onto the 9-point circle of $\Delta ABC$), hence $B_1D\cdot B_1P=A_1B_1\cdot C_1B_1\ (\ 1\ )$. This comes easily from 2 similar triangles. But $AC$ is diameter of $CAC_1A_1$ and $A_1B_1$ is symmetrical of $B_1C_1$ w.r.t. $BB_1$, hence $C_1B_1$ intersect circle $\odot (CAC_1A_1)$ at $C_2$ and $B_1C_2=B_1A_1$, consequently $(1)$ becomes $B_1D\cdot B_1P=B_1C_1\cdot B_1C_2$, but from power of point $B_1$ w.r.t. $(CAC_1A_2)$ we have $B_1C_1\cdot B_1C_2=B_1C\cdot B_1A$; under the circumstances $B_1C\cdot B_1A=B_1D\cdot B_1P=B_1D'\cdot B_1P$, done. Hope I did not mess-up the indices. Best regards, sunken rock
Panoz93
20.09.2014 01:57
It suffices to prove that $D$ is the orthocenter of triangle $APC$. For this just notice that $AD,DC$ are the polars of $C,A$ with respect to the circumcircle of $C_{1}HA_{1}B$.
bojler
18.10.2014 12:35
Let $BB_1$ be the third altitude in triangle $ABC$. Note that triangles $AB_1H$ and $BB_1C$ are similar thus $AB_1\cdot B_1C=B_1H\cdot B_1B$ . $H(B,H,D,B_1)$ so $PH^2=PD\cdot PB_1$ . $B_1P\cdot B_1D'=B_1P\cdot B_1D=B_1P\cdot (B_1P-DP)=PB_1^2-PH^2=B_1H\cdot B_1B=AB_1\cdot B_1C$.
jayme
09.09.2015 15:10
Dear Mathlinkers, for #, see http://www.artofproblemsolving.com/community/c6h1069884_orthocenter Sincerely Jean-Louis
orestis100
08.12.2016 02:27
Can someone please prove why H(B,H,D,B1)?
AlastorMoody
02.07.2019 08:30
Apply Brokard's Theorem on $\odot (BHC_1A_1) $, $D $ is the orthocenter of $\Delta PCA $, hence the conclusion
itslumi
21.03.2020 13:06
AlastorMoody wrote: Apply Brokard's Theorem on $\odot (BHC_1A_1) $, $D $ is the orthocenter of $\Delta PCA $, hence the conclusion Same solution,Can this be done whith inverzion???
jayme
27.06.2020 13:47
Dear Mathlinkers, http://jl.ayme.pagesperso-orange.fr/Docs/Orthique%20encyclopedie%207.pdf p. 73... Sincerely Jean-Louis
Olympikus
27.06.2020 20:46
orestis100 wrote: Can someone please prove why H(B,H,D,B1)? Take perspective at $C_1$ and map to $AC$. Use the Ceva-Menelaus lemma : $H(A,C;B_1,A_1C_1\cap AC)$.
Mikeglicker
27.06.2020 21:50
Why showing that D is the orthocenter of APC is suffices?
Mikeglicker
28.06.2020 13:03
Mikeglicker wrote: Why showing that D is the orthocenter of APC is suffices? Anybody?
arzhang2001
28.06.2020 17:50
do you ignore it: if we let $M$ Be the midpoint of $A_1C_1$ Then should $MDAC$ be Cyclic.
miltosk
28.06.2020 18:53
Mikeglicker wrote: Mikeglicker wrote: Why showing that D is the orthocenter of APC is suffices? Anybody? Because then $$DAC=DPC$$as acute angles with perpendicular sides and since $$DAC=D'AC$$due to the isosceles triangle $$DAD'$$, $$DPC=D'AC$$and then its obvious.
iman007
28.11.2020 22:47
$\text{Brokard kills this problem}$ $\text{basically we should prove that} \thickspace \angle{ADC}=180^{\circ} -\angle{APC}$ $\text{which means to prove that} \thickspace P\thickspace \text{is the orthocenter of the triangle} \thickspace ADC$ $\textbf{Brokard theorem} \thickspace \text{on}\thickspace (A_1BC_1H) \thickspace\text{finishes the problem}$ \(\boxed{\mathcal{Q.E.D}}\)
lelouchvigeo
12.12.2024 08:08
Nice Since $ACD$ are autopolar by Brocard's and $P$ is the center of circle $BA_1HC_1.$ We can observe that $D$ is the orthocenter of $\triangle ACP.$ Since the reflection of the orthocenter lies on the circle. We are done