Let $ABCD$ be an inscribed quadrilateral. Let the circles with diameters $AB$ and $CD$ intersect at two points $X_1$ and $Y_1$, the circles with diameters $BC$ and $AD$ intersect at two points $X_2$ and $Y_2$, the circles with diameters $AC$ and $BD$ intersect at two points $X_3$ and $Y_3$. Prove that the lines $X_1Y_1, X_2Y_2$ and $X_3Y_3$ are concurrent. Maxim Didin
Problem
Source: Tournament of Towns, Senior A-Level Paper, Spring 2020 , p5
Tags: geometry, concurrency, concurrent, cyclic quadrilateral, circles
62861
03.06.2020 22:32
Using vectors with the circumcenter of $ABCD$ at the origin, we claim that $p = \tfrac12(a+b+c+d)$ lies on all three radical axes. Indeed,
\begin{align*}
\mathrm{Pow}(P, (AB)) & = |p - \tfrac12(a+b)|^2 - |a - \tfrac12(a+b)|^2\\
& = |\tfrac12(c+d)|^2 - |\tfrac12(a-b)|^2\\
& = \tfrac14[|c+d|^2 - |a-b|^2]\\
& = \tfrac12[a\cdot b + c\cdot d]
\end{align*}which will be equal to $\mathrm{Pow}(P, (CD))$.
Of course, this argument can easily be made "synthetic." To do that, let $O$ be the circumcenter of $ABCD$ and let $M_{AB}$ be the midpoint of $\overline{AB}$ and define other expressions similarly. Then $\overline{M_{AB}M_{CD}}$, $\overline{M_{AC}M_{BD}}$, $\overline{M_{AD}M_{BC}}$ share a common midpoint $M$.
Let $P$ be the reflection of $O$ over $M$, so $OM_{AB}PM_{CD}$ is a parallelogram. We claim $P$ lies on the radical axis of $(AB)$ and $(CD)$. Indeed,
\begin{align*}
AM_{AB}^2 + OM_{AB}^2 & = DM_{CD}^2 + OM_{CD}^2\\
\implies AM_{AB}^2 + PM_{CD}^2 & = DM_{CD}^2 + PM_{AB}^2\\
\implies PM_{AB}^2 - AM_{AB}^2 & = PM_{CD}^2 - DM_{CD}^2
\end{align*}as desired.
UltraInstinct
27.08.2020 22:50
CantonMathGuy wrote:
Using vectors with the circumcenter of $ABCD$ at the origin, we claim that $p = \tfrac12(a+b+c+d)$ lies on all three radical axes. Indeed,
\begin{align*}
\mathrm{Pow}(P, (AB)) & = |p - \tfrac12(a+b)|^2 - |a - \tfrac12(a+b)|^2\\
& = |\tfrac12(c+d)|^2 - |\tfrac12(a-b)|^2\\
& = \tfrac14[|c+d|^2 - |a-b|^2]\\
& = \tfrac12[a\cdot b + c\cdot d]
\end{align*}which will be equal to $\mathrm{Pow}(P, (CD))$.
Of course, this argument can easily be made "synthetic." To do that, let $O$ be the circumcenter of $ABCD$ and let $M_{AB}$ be the midpoint of $\overline{AB}$ and define other expressions similarly. Then $\overline{M_{AB}M_{CD}}$, $\overline{M_{AC}M_{BD}}$, $\overline{M_{AD}M_{BC}}$ share a common midpoint $M$.
Let $P$ be the reflection of $O$ over $M$, so $OM_{AB}PM_{CD}$ is a parallelogram. We claim $P$ lies on the radical axis of $(AB)$ and $(CD)$. Indeed,
\begin{align*}
AM_{AB}^2 + OM_{AB}^2 & = DM_{CD}^2 + OM_{CD}^2\\
\implies AM_{AB}^2 + PM_{CD}^2 & = DM_{CD}^2 + PM_{AB}^2\\
\implies PM_{AB}^2 - AM_{AB}^2 & = PM_{CD}^2 - DM_{CD}^2
\end{align*}as desired.
What is the motivation for choosing $(a+b+c+d)/2$?
i3435
27.08.2020 22:59
UltraInstinct wrote: CantonMathGuy wrote:
Using vectors with the circumcenter of $ABCD$ at the origin, we claim that $p = \tfrac12(a+b+c+d)$ lies on all three radical axes. Indeed,
\begin{align*}
\mathrm{Pow}(P, (AB)) & = |p - \tfrac12(a+b)|^2 - |a - \tfrac12(a+b)|^2\\
& = |\tfrac12(c+d)|^2 - |\tfrac12(a-b)|^2\\
& = \tfrac14[|c+d|^2 - |a-b|^2]\\
& = \tfrac12[a\cdot b + c\cdot d]
\end{align*}which will be equal to $\mathrm{Pow}(P, (CD))$.
Of course, this argument can easily be made "synthetic." To do that, let $O$ be the circumcenter of $ABCD$ and let $M_{AB}$ be the midpoint of $\overline{AB}$ and define other expressions similarly. Then $\overline{M_{AB}M_{CD}}$, $\overline{M_{AC}M_{BD}}$, $\overline{M_{AD}M_{BC}}$ share a common midpoint $M$.
Let $P$ be the reflection of $O$ over $M$, so $OM_{AB}PM_{CD}$ is a parallelogram. We claim $P$ lies on the radical axis of $(AB)$ and $(CD)$. Indeed,
\begin{align*}
AM_{AB}^2 + OM_{AB}^2 & = DM_{CD}^2 + OM_{CD}^2\\
\implies AM_{AB}^2 + PM_{CD}^2 & = DM_{CD}^2 + PM_{AB}^2\\
\implies PM_{AB}^2 - AM_{AB}^2 & = PM_{CD}^2 - DM_{CD}^2
\end{align*}as desired.
What is the motivation for choosing $(a+b+c+d)/2$? I think it's because that's the anticenter.
UltraInstinct
27.08.2020 23:23
i3435 wrote: UltraInstinct wrote: CantonMathGuy wrote:
Using vectors with the circumcenter of $ABCD$ at the origin, we claim that $p = \tfrac12(a+b+c+d)$ lies on all three radical axes. Indeed,
\begin{align*}
\mathrm{Pow}(P, (AB)) & = |p - \tfrac12(a+b)|^2 - |a - \tfrac12(a+b)|^2\\
& = |\tfrac12(c+d)|^2 - |\tfrac12(a-b)|^2\\
& = \tfrac14[|c+d|^2 - |a-b|^2]\\
& = \tfrac12[a\cdot b + c\cdot d]
\end{align*}which will be equal to $\mathrm{Pow}(P, (CD))$.
Of course, this argument can easily be made "synthetic." To do that, let $O$ be the circumcenter of $ABCD$ and let $M_{AB}$ be the midpoint of $\overline{AB}$ and define other expressions similarly. Then $\overline{M_{AB}M_{CD}}$, $\overline{M_{AC}M_{BD}}$, $\overline{M_{AD}M_{BC}}$ share a common midpoint $M$.
Let $P$ be the reflection of $O$ over $M$, so $OM_{AB}PM_{CD}$ is a parallelogram. We claim $P$ lies on the radical axis of $(AB)$ and $(CD)$. Indeed,
\begin{align*}
AM_{AB}^2 + OM_{AB}^2 & = DM_{CD}^2 + OM_{CD}^2\\
\implies AM_{AB}^2 + PM_{CD}^2 & = DM_{CD}^2 + PM_{AB}^2\\
\implies PM_{AB}^2 - AM_{AB}^2 & = PM_{CD}^2 - DM_{CD}^2
\end{align*}as desired.
What is the motivation for choosing $(a+b+c+d)/2$? I think it's because that's the anticenter. what is the anticenter? is there any handout covering this? is this important? never heard of it.
UltraInstinct
28.08.2020 13:48
what is the anticenter? is there any handout covering this? is this important? never heard of it.[/quote] anyone, please?
UltraInstinct
29.08.2020 14:59
i3435 wrote: UltraInstinct wrote: CantonMathGuy wrote:
Using vectors with the circumcenter of $ABCD$ at the origin, we claim that $p = \tfrac12(a+b+c+d)$ lies on all three radical axes. Indeed,
\begin{align*}
\mathrm{Pow}(P, (AB)) & = |p - \tfrac12(a+b)|^2 - |a - \tfrac12(a+b)|^2\\
& = |\tfrac12(c+d)|^2 - |\tfrac12(a-b)|^2\\
& = \tfrac14[|c+d|^2 - |a-b|^2]\\
& = \tfrac12[a\cdot b + c\cdot d]
\end{align*}which will be equal to $\mathrm{Pow}(P, (CD))$.
Of course, this argument can easily be made "synthetic." To do that, let $O$ be the circumcenter of $ABCD$ and let $M_{AB}$ be the midpoint of $\overline{AB}$ and define other expressions similarly. Then $\overline{M_{AB}M_{CD}}$, $\overline{M_{AC}M_{BD}}$, $\overline{M_{AD}M_{BC}}$ share a common midpoint $M$.
Let $P$ be the reflection of $O$ over $M$, so $OM_{AB}PM_{CD}$ is a parallelogram. We claim $P$ lies on the radical axis of $(AB)$ and $(CD)$. Indeed,
\begin{align*}
AM_{AB}^2 + OM_{AB}^2 & = DM_{CD}^2 + OM_{CD}^2\\
\implies AM_{AB}^2 + PM_{CD}^2 & = DM_{CD}^2 + PM_{AB}^2\\
\implies PM_{AB}^2 - AM_{AB}^2 & = PM_{CD}^2 - DM_{CD}^2
\end{align*}as desired.
What is the motivation for choosing $(a+b+c+d)/2$? I think it's because that's the anticenter. Okay I understand that the anticenter is just another name of the euler's point. However, why would it motivate you to try to show that those three lines concur at this point? I don't see how it explains the motivation.
UltraInstinct
29.08.2020 23:53
i3435 wrote: UltraInstinct wrote: CantonMathGuy wrote:
Using vectors with the circumcenter of $ABCD$ at the origin, we claim that $p = \tfrac12(a+b+c+d)$ lies on all three radical axes. Indeed,
\begin{align*}
\mathrm{Pow}(P, (AB)) & = |p - \tfrac12(a+b)|^2 - |a - \tfrac12(a+b)|^2\\
& = |\tfrac12(c+d)|^2 - |\tfrac12(a-b)|^2\\
& = \tfrac14[|c+d|^2 - |a-b|^2]\\
& = \tfrac12[a\cdot b + c\cdot d]
\end{align*}which will be equal to $\mathrm{Pow}(P, (CD))$.
Of course, this argument can easily be made "synthetic." To do that, let $O$ be the circumcenter of $ABCD$ and let $M_{AB}$ be the midpoint of $\overline{AB}$ and define other expressions similarly. Then $\overline{M_{AB}M_{CD}}$, $\overline{M_{AC}M_{BD}}$, $\overline{M_{AD}M_{BC}}$ share a common midpoint $M$.
Let $P$ be the reflection of $O$ over $M$, so $OM_{AB}PM_{CD}$ is a parallelogram. We claim $P$ lies on the radical axis of $(AB)$ and $(CD)$. Indeed,
\begin{align*}
AM_{AB}^2 + OM_{AB}^2 & = DM_{CD}^2 + OM_{CD}^2\\
\implies AM_{AB}^2 + PM_{CD}^2 & = DM_{CD}^2 + PM_{AB}^2\\
\implies PM_{AB}^2 - AM_{AB}^2 & = PM_{CD}^2 - DM_{CD}^2
\end{align*}as desired.
What is the motivation for choosing $(a+b+c+d)/2$? I think it's because that's the anticenter. See above comment please
dungnguyentien
19.09.2020 15:49
Let $M,N,P,Q,R,S$ be the midpoints of $AB,BC,CD,DA,AC,BD$. $MP,NQ,RS$ have the same midpoint $I$. Let $(O)$ be the circumcircle of the quadrilateral $ABCD$. Since $AB,CD,X_1Y_1$ are the radical axes of three circles $(O), (M;\dfrac{AB}{2}), (P;\dfrac{CD}{2})$, these lines are concurrent at $E$. Denote by $T$ the reflection of $O$ through $I$. $MOPT$ is a parallelogram. Therefore, $MT\parallel OP\perp EP$ and $PT\parallel OM \perp EM$. Then, $T$ is the orthocenter of $\Delta EMP$ and $ET\perp MP$. From $ET\perp MP$ and $X_1Y_1\perp MP$, we can see that $X_1Y_1$ passes through $T$. Similarly, $X_1Y_1,X_2Y_2,X_3Y_3$ are concurrent at $T$.
Attachments:
Modesti
14.12.2020 05:44
UltraInstinct wrote: what is the anticenter? is there any handout covering this? is this important? never heard of it. Probably the motivation is that it seemed to be that point in an accurate diagram. Experience and good diagrams are the secret to most of the amazing solutions, in my opinion
Lukaluce
28.03.2021 22:03
Let $WXYZ$ denote the complete quadrilateral $WXYZST$ where $\{S\} = WX \cap YZ$ and $\{T\} = XY \cap WZ$. Let $\{P\} = AB \cap CD$ and $\{Q\} = BC \cap AD$. It is well known that the circles with diameters $AC$, $BD$, and $PQ$ are coaxial with their common axis being the Aubert line of $ABCD$, therefore $X_3Y_3$, $X_2Y_2$, and $X_1Y_1$ are the Aubert lines of $ABCD$, $ABDC$, and $ACBD$, respectively. Consequently, showing the concurrence of $X_3Y_3$, $X_2Y_2$, and $X_1Y_1$ is equivalent to showing the Aubert lines of $ABCD$, $ABDC$, and $ACBD$ concur. Let $H_1, H_2, H_5,$ and $H_6$ be the orthocenters of $\triangle CDQ$, $\triangle ABQ$, $\triangle ACQ$, and $\triangle BDQ$, respectively. Since $H_1H_6 \perp BQ \perp H_2H_5$, therefore $H_1H_6 \parallel H_2H_5$, and similarly $H_1H_5 \perp H_2H_6$, therefore $H_1H_2H_5H_6$ is a parallelogram, ergo the Aubert line of $ACBD$ bisects $H_1H_2$. Similarly, the Aubert line of $ABDC$ bisects $H_3H_4$, where $H_3$ and $H_4$ are the orthocenters of $ADP$ and $BCP$, respectively. Therefore, we have to prove that the midpoint of $H_1H_2$ is also the midpoint of $H_3H_4$ (as $H_1-H_2-H_3-H_4$ is the Aubert line of $ABCD$). Note that the midpoint of $H_1H_2$ is the intersection of the maltitude from the midpoint of $BC$ in $ABCD$ with the maltitude from the midpoint of $AD$ in $ABCD$ (the maltitude in a quadrilateral is the perpendicular line from a midpoint of a side in the quadrilateral to the opposite side). FInally, this means that the 4 maltitudes of $ABCD$ must be concurrent, which holds as $ABCD$ is cyclic (to show this well known lemma holds just note the maltitudes from $M$ and $P$ concur at the reflection of the circumcenter of $(ABCD)$ w.r.t. the intersection of the diagonals in the parallelogram $MNPQ$, as do the maltitudes from $N$ and $Q$, where $M$, $N$, $P$, and $Q$ are the midpoints of $AB, BC, CD,$ and $DA$, respectively).
erkhes_mng
19.12.2022 08:38
dungnguyentien wrote: Let $M,N,P,Q,R,S$ be the midpoints of $AB,BC,CD,DA,AC,BD$. $MP,NQ,RS$ have the same midpoint $I$. Let $(O)$ be the circumcircle of the quadrilateral $ABCD$. Since $AB,CD,X_1Y_1$ are the radical axes of three circles $(O), (M;\dfrac{AB}{2}), (P;\dfrac{CD}{2})$, these lines are concurrent at $E$. Denote by $T$ the reflection of $O$ through $I$. $MOPT$ is a parallelogram. Therefore, $MT\parallel OP\perp EP$ and $PT\parallel OM \perp EM$. Then, $T$ is the orthocenter of $\Delta EMP$ and $ET\perp MP$. From $ET\perp MP$ and $X_1Y_1\perp MP$, we can see that $X_1Y_1$ passes through $T$. Similarly, $X_1Y_1,X_2Y_2,X_3Y_3$ are concurrent at $T$. I think this solution is the best solution I have ever seen.
v4913
20.12.2022 02:14
Let M1, M2, N1, N2 = midpoints of AB, CD, AD, BC respectively, and P, Q, R = AB \cap CD, AD \cap BC, AC \cap BD; then the first two radical axes are the lines through P, Q perp to M1M2, N1N2 respectively, and they intersect at the point W = (A+B+C+D)/2 - O if O is the circumcenter, because this point is the orthocenter of both PM1M2 and QN1N2 (since it’s the reflection of O over midpoint of M1M2, N1N2). The third radical axis is the line through R perp to KL if K, L are midpoints of AC, BD so it suffices to show that RW \perp KL but R is O-antipode wrt (OKL) and OKWL is a parallelogram so W is the orthocenter of RKL, and we are done.