Let $ABCD$ be a convex quadrilateral, and $P$ be a point in its interior. The projections of $P$ on the sides of the quadrilateral lie on a circle with center $O$. Show that $O$ lies on the line through the midpoints of $AC$ and $BD$.
Problem
Source: 2009 Balkan Shortlist BMO G3 - difficult
Tags: diagonals, projections, geometry, midpoints, convex quadrilateral, Circumcenter, Isogonal conjugate
18.05.2020 21:45
Someone told me This problem can be solved using $\angle APB + \angle CPD$ = 180 But well Bump
19.05.2020 00:58
19.05.2020 04:16
ppanther wrote:
19.05.2020 05:57
Since $\angle APB + \angle CPD =180 $, $P$ has an isogonal conjugate like $Q$ where the conic with foci $P$ and $Q$ touches the sides of $ABCD$. Let this conic be $\omega$ with centre $O$. We restate the problem as below. Quote: Let $\omega$ be a conic with centre $O$ inscribed in triangle $ABC$ which touches the sides $BC,AC,AB$ in $D,E,F$ respectively. Let $G$ be a point on $\omega$ such that $D,F,G,E$ lie in that order. If the tangent at $G$ to $\omega$ cut $AB,AC$ in $K,L$, prove that $O$ lies on the line passing through midpoints of $KC$ and $BL$. Animate $G$. Since $K,L$ move projectively $deg(K)=deg(L)=1$ so the midpoints of $KC$ and $BL$ are of degree $1$ too. Note that $deg(O)=0$ so the collinearity statement is of degree $2$. We will show that the statement in $E,F$ and $G$ such that the tangent at $G$ is parallel to $BC$. The later case is obvious and $E,F$ are symmetric so checking the statement at $E$ will be enough. In that case we should prove that $O$ lies on the line passing through midpoints of $AC$ and $BE$. But one can project from the infinity point of the line $AC$ to make $\omega$ a circle. In that case the problem becomes the well-known incircle lemma.
19.05.2020 20:51
Bump.....
19.05.2020 20:59
Seeing the isogonal conjugate of P is kinda obvious but the midpoint condition seems to be eluding me (perhaps gauss line would be of any use here?)
20.05.2020 23:41
I didn't read the solution of @matinyousefi so sorry if it's the same call the foot of perpendiculars from $P$ to sides $ AB , BC, CD, DA$ , $X,Y,Z,T$ respectively. Claim:$X,Y,Z,T$ are concyclic is equivalent to $P$ has a isogonal conjugate wrt $ABCD$.
Call the isogonal conjugate of $P$ $\rightarrow$ $Q$ . Call the foot of perpendiculars from $Q$ to $AB,BC,CD,DA$ $\rightarrow$ $ L,K,J,H$ respectively. Claim: $ L,K,J,H,X,Y,Z,T$ are concyclic.(call the circle $\omega$)
Call center of $\omega$ $\rightarrow$ $O$. Claim: O is the midpoint of $ PQ $
Claim:$P,Q$ are focis of an ellipse thangent to sides of $ABCD$.
now We restate the problem as Quote: In convex quadrilateral $ABCD$ .$ M$ is midpoint of $AC$ and $N$ is midpoint of $BD$. An ellipse with focis $P,Q$ and center $O$ is tangent to $AB,BC,CD,DA$. Prove $O,M,N$ are collinear.
Do an appropriate projective map (sth like ratio of axises of the ellipse ) so the ellipse become circle, well the midpoints(center) and tangent lines are still the same thing so the now the problem becomes Quote: Circle $\omega$ is tangent to sides $AB,BC,CD,DA$ of convex quadrilateral $ABCD$. Prove midpoint of $AC$ and $BD$ and the center of $\omega$ are collinear. Finally. this is The Newton Theorem . $ \mathcal{QED} $
21.05.2020 00:45
An elementary solution but a long-winded one. I would be very happy if someone would re-write my proof using directed angles. First, we present a well-know lemma about pedal circles and isogonal conjugates. Lemma $P'$, the isogonal conjugate of P in $ABCD$'s projection on the sides of $ABCD$ are the second intersections of the circle $XYWZ$ with the sides of $ABCD$
We now have O re-defined as the mid-point of PP', since it is precisely the intersection of the perpendicular bisectors of YY', XX', WW', and ZZ'. Then the midpoint conditions come into play. We reflect P' across M, N, the mid-points of the diagonals, and denote its reflections $P_2, P_3$, respectively. We discover that parallelograms $CP_2AP'$, $DP_3BP'$ have formed. We now prove that $P_2, P_3, P$ are collinear. Namely, we prove $\angle CPD + \angle CPP_2 +\angle DPP_3=\pi$, denote $K$ as the intersection of $CP_2, DP_3$ (this can actually be avoided by using directed angles). $\angle DKC = \angle AP'B$ $\angle CPD=\pi - \angle PCD - \angle PDC $ $\angle CPP_2= \angle KP_2P - \angle PCP_2$ $\angle DPP_3= \angle KP_3P - \angle PDP_3$ Now we sum all three equations to obtain: $Sum= 2\pi - (\angle PCD+\angle PDC)-(\angle PCA-\angle P'AC+\angle PDB+\angle P'BD)- \angle AP'B$ After chasing some angles using the isogonal conjugates and parallelograms, we arrive at $Sum=\angle P'BA + \angle P'AB+ \angle AP'B= \pi$ Therefore, $P_2,P_3,P$ are collinear. So we have arrived at the desired conclusion. $\hfill\square$
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21.05.2020 00:53
The proof for the properties of isogonal conjugates is cleaner herehttps://usamo.wordpress.com/2014/11/30/three-properties-of-isogonal-conjugates/