ABCD=cyclic quadrilateral,$AC\cap BD=X$ AA'$\perp $BD,A'$\in$BD CC'$\perp $BD,C'$\in$BD BB'$\perp $AC,B'$\in$AC DD'$\perp $AC,D'$\in$AC Prove that: a)Prove that perpendiculars from midpoints of the sides to the opposite sides are concurrent.The point is called Mathot Point b)A',B',C',D' are concyclic c)If O'=circumcenter of (A'B'C') prove that O'=midpoint of the line that connects the orthocente of triangle XAB and XCD d)O' is the Mathot Point
Problem
Source: Third Romanian JBMO TST 2016
Tags: geometry
trungnghia215
18.06.2016 10:04
den_thewhitelion wrote: Let $ABCD$ be cyclic quadrilateral. Denote $AC\cap BD=X$ $$\begin{cases}AA'\perp BD,A'\in BD \\ CC'\perp BD,C'\in BD \\ BB'\perp AC,B'\in AC \\ DD'\perp AC,D'\in AC\end{cases}$$Prove that: a) Prove that perpendiculars from midpoints of the sides to the opposite sides are concurrent.The point is called Mathot Point b) $A',B',C',D'$ are concyclic c) If $O'$ is circumcenter of $\odot (A'B'C')$, prove that $O'$ is midpoint of the line that connects the orthocente of triangles $XAB$ and $XCD$ d) $O'$ is the Mathot Point WLOG, i will assume that $\angle BXC = \angle AXD$ is not acuted - angle. Then we can easily get that $A', B', C', D'$ lie on segment $BX, AX, DX, CX$.
Let $M, N, P, Q$ be midpoints of $AB, BC, CD, DA$, respectively. And let $l_{m}, l_{n}, l_{p}, l_{q}$ be the lines pass through $M, N, P, Q$ and be perpendicular to $CD, DA, AB, BC$, respectively. Suppose $l_{m}, l_{n}, l_{p}, l_{q}$ cut $CD, DA, AB, BC$ at $K, L, H, J$, respectively.
Note that $OM, ON, OP, OQ$ are perpendicular to $AB, BC, CD, DA$, respectively. Consider reflection through point $I$ which is the center of $MNPQ$:
Note that $OQ \parallel l_{n}$ since they are both perpendicular to $DA$. Since $N$ is reflection of $Q$ through point $I$ and $N \in l_{n}, Q \in OQ$ we conclude that $l_{n}$ is reflection of $OQ$ through point $I$. Similarly, we obtain $l_{m}, l_{p}, l_{q}$ are reflections of $OP, OM, ON$ through point $I$. Since $OM, ON, OP, OQ$ are concurrent at $O$ then $l_{m}, l_{n}, l_{p}, l_{q}$ are also concurrent at $J$ where $I$ is midpoint of $IJ$.
Since $\angle BB'C = BC'C = 90^{\circ}$ then $BB'C'B$ is cyclic quadrilateral. Hence $\angle B'C'A' = \angle B'C'B = \angle B'CB = \angle ACB$ since $A', C' \in BD$ and $B', D' \in AC$. Similarly, we also obtain $AA'D'D$ is also a cyclic quadrilateral, then $\angle B'D'A' = \angle ADB = \angle ACB$ since $ABCD$ is cyclic. Thus, $A'B'C'D'$ is also cyclic. We're done.
Given $\triangle ABC$. $BE, CF$ are altitudes of $\triangle ABC$ ($E \in CA, F \in AB$). $M, N$ are midpoints of $BC, AH$ ($H$ is orthencenter of $\triangle ABC$, respectively. Then $MN$ is perpendicular bisector of $EF$.
Proof$M, N$ are circumcenter of quadrilaterial $BCEF, AEHF$, respectively. The result follows as desired.
Denote $H, K$ are orthocenters of $\triangle XAB, XCD$, respectively. And $U, V$ are midpoints of $XH, XK$.
I will show that midpoint of $HK$ - $O'$ is also circumcenter of $A'B'C'D'$.
We can obtain easily that $B'C'\parallel DA; A'D'\parallel BC; A'B'\parallel CD; C'D'\parallel AB$ since part (b) and our assumptions.
From the lemma, we get that $PV$ is perpendicular bisector of $C'D'$ (this also means $PV \perp C'D'$). On other hand, $O', V$ are midpoints of $KH, KX$ then $O'V\parallel HX$. Note that $(HX, C'D') \equiv (HX, AB) + (AB, C'D')\equiv (HX, AB) \equiv \frac{\pi}{2}\pmod{\pi}$. Hence $O'V \perp C'D'$, we conclude that $O'V \parallel PV$. In other words, we can say that $O', V, P$ are collinear, or $O'$ lies on the perpendicular bisector of $C'D'$.
Simiarly, we also get $O'$ also lies on perpendicular bisector of $A'B'$. Since $A', B', C', D'$ lie on the circle then $O'$ must be the circumcenter of $A'B'C'D'$. We're done.
We have $O'P$ passes through $P$ and be perpendicular to $C'D' \parallel AB$, similarly, we also get ... . According to part (a), we conclude that $O'$ is the Mathot point of cyclic quadrilaterial $ABCD$.
Attachments:
jayme
18.06.2016 12:09
Dear Mathlinkers, for a little variation, http://jl.ayme.pagesperso-orange.fr/Docs/Du%20cercle%20des%20huit%20points%20au%20cercle%20des%20neuf%20points.pdf p. 4... Sincerely Jean-Louis
electrovector
07.04.2020 12:28
trungnghia215 wrote:
Note that $(HX, C'D') \equiv (HX, AB) + (AB, C'D')\equiv (HX, AB) \equiv \frac{\pi}{2}\pmod{\pi}$.
I couldn't understand this part. Can you explain it a little more? How did you use mod $\pi$ and edges?
pyramid_fan90
23.04.2022 19:15
den_thewhitelion wrote: ABCD=cyclic quadrilateral,$AC\cap BD=X$ AA'$\perp $BD,A'$\in$BD CC'$\perp $BD,C'$\in$BD BB'$\perp $AC,B'$\in$AC DD'$\perp $AC,D'$\in$AC Prove that: a)Prove that perpendiculars from midpoints of the sides to the opposite sides are concurrent.The point is called Mathot Point b)A',B',C',D' are concyclic c)If O'=circumcenter of (A'B'C') prove that O'=midpoint of the line that connects the orthocente of triangle XAB and XCD d)O' is the Mathot Point
Let $E$ and $F$ be the intersections of the perpendiculars from $M$ and $P$ on $(DC)$ and $(AB)$ and the perpendiculars from $N$ and $Q$ on $(AD)$ and $(BC).$ We can easily see that $OMEP$ and $ONFQ$ are parallelograms so $\overrightarrow{OE}=\overrightarrow{OM}+\overrightarrow{OP}=\frac{1}{2}\left(\overrightarrow{OA}+ \overrightarrow{OB}+ \overrightarrow{OC}+ \overrightarrow{OD}\right)$ and analogously $\overrightarrow{OF}=\frac{1}{2}\left(\overrightarrow{OA}+ \overrightarrow{OB}+ \overrightarrow{OC}+ \overrightarrow{OD}\right)$ which implies that $E\equiv F$ and the conclusion follows.