A convex hexagon $ABCDEF$ with an interior point $P$ is given. Assume that $BCEF$ is a square and both $ABP$ and $PCD$ are right isosceles triangles with right angles at $B$ and $C$, respectively. Lines $AF$ and $DE$ intersect at $G$. Prove that $GP$ is perpendicular to $BC$. Proposed by Patrik Bak - Slovakia
Problem
Source: IGO 2023 Intermidiate P2
Tags: geometry
18.01.2024 14:09
Let $X=CP \cap AF$ and $Y=BP \cap DE$. We have that $\angle CBP = 90 - \angle PBF = \angle ABF$. Also, $CB=BF$ and $BP=AB$ ,because of the square and the isosceles triangle. So, triangles $BPC$, $BAF$ are equal, hence $\angle CPB = \angle BAF = \angle BAX$ from where we conclude that $BAXP$ is cyclic. So, $\angle CXF=\angle PXF = \angle ABP =90= \angle CBF$ and $X$ lies on $(BCEF)$. Analogously, $CDYP$ is cyclic and $Y$ lies on $(BCEF)$. Also, $\angle PXG = 90 = \angle PYG$, so $PYGX$ is cyclic. From the cyclic quadrilaterals and the equal triangles we proved before we get that: $\angle PGF = \angle PGX = \angle PYX = \angle BYX = \angle BCX= \angle BCP = \angle BFA => GP \parallel BF => GP \perp BC$, as needed.
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18.01.2024 16:08
My 100th post! $\angle FBC =90$, $\angle ABP = 90 \implies \angle ABF = \angle PBC$ Since $AB=BP$ and $BF=BC \implies \triangle AFB \equiv \triangle PCB$. $\implies \angle PCB = \angle AFB$, $\angle FAB = \angle BPC$, $AF=PC$ $\angle BCE =90$, $\angle PCD = 90 \implies \angle ECD = \angle BCP$ Since $EC = BC$ and $CD=CP \implies \triangle ECD \equiv \triangle BCP$. $\implies \angle CED = \angle PBC$, $\angle EDC = \angle BPC$, $ED=BP$ $\implies \angle FAB = EDC \implies \angle GAP = \angle GDP = \angle BPC - 45$ Let $\angle PBC = a$, $\angle PCB = b$ $\implies \angle BPC = 180 -a -b \implies \angle GAP = \angle GDP = 135-a-b$ $\angle FGE = 180 - \angle GFE - \angle GEF = 180 - (180 -90-b)-(180-90-a)=a+b$ Let $Y$ be the intersection of $PD$ with the circle with centre $B$ and radius $AB$. Let $X$ be the intersection of $PA$ with the circle with centre $C$ and radius $CD$. So, $\angle AYP = \angle PXD = 45$. Let $XD$ intersect $AF$ at $G'$. $\angle AG'X = 180 - \angle AXD - \angle FAX = 180 - 45 - (135 - a - b) =a+b$ $\implies AGX = \angle AG'X$ Since A, G, G' are on the same line, we get $G'=G$. So, $G, D, X$ are on the same line. Similar, $G, F, Y$ are on the same line. $\angle AYD = DXA = 45 \implies A, Y, X, D$ are on the same circle $\implies GA \cdot GY = GD \cdot GX$ So, the power of $G$ to the circle centered at $B$ (radius $AB$) and the circle centered at $C$ (radius $CD$) is equal, so $G$ is on the radical axis. But $P$ is on the radical axis of the 2 circles, so $GP$ is the radical axis. So, $GP$ is perpendicular to $BC$. $\blacksquare$
18.01.2024 23:08
19.01.2024 05:58
Here is my solution: Let $H$ the foot of the altitud from $P$ in the triangle $BPC$; if $H$,$P$,$G$ are collinear then $PG$ is perpendicular to $BC$. Since we have $AB=BP$, $BF=BC$ $\angle ABF= \angle PBC$ triangles $ABF$ and $PBC$ are congruent, also since $DC=PC$, $CE=CB$, $\angle DCE= \angle PCB$ then triangles $ABF$, $PBC$ and $DCE$ are congruent. With angle chasing isn't hard to get $\angle FGE + \angle BAF =\angle FGE + \angle EDC=180°$ then $AB//GD$ and $AG//CD$ Let $AG\cap CF = X$ and $GD \cap BP = Y$ and $\angle PXG=\angle PYG=90°$ and is important to notice that $PXGY$ is cyclic. Now look at cuadrilaterals $ABPX$ and $DCPY$, both are defined similarly ($\angle BAP=\angle CDP$, $\angle PAX=\angle PDY$ and continue with the other angles) , they are similarly, then $\frac{BP}{CP}=\frac{XP}{PY}$ an inmediatly $XBCY$ is a cyclic cuadrilateral. Finally by the cyclics $XBCY$ and $PXGY$: $\angle HBP =\angle CBY =\angle CXY= \angle PXY= \angle PGY$, then $\angle BPH=\angle GPY$, the desired collinearity is immediate.
26.01.2024 14:03
Easily coord bashable
30.01.2024 17:20
Coordinate Bashing... Some congruency and Equations gives us the desired result.
15.02.2024 20:10
Very nice problem : Simple angle chasing gives that triangles $BPC$, $BAF$, and $EDC$ are congruent with vertices in that order. Now overly triangles $FEG$ and $BCP$ aligning segments $BC$ and $EF$. We need to show that $GP$ is perpendicular to $BC$ in the overlayed diagram but this follows from simple angle chasing that shows that $G$ is the orthocenter of $BPC$.
14.05.2024 09:04
Take $(BCEF)$ unit circle. \[b=1,c=i,e=-1,f=-i\]Let $p=m+ni$ \[a=i(m-1)-n+1\]\[m+ni=i(d-i)+i\iff d=i(1-m)+n-1\]\[g=\frac{i(2n-2)(i-im+n)-(im-n+1)(2i-2im)}{(-n+1-im)(i-im+n)-(im-n-1)(n-i+im)}\]\[g=\frac{ni-nim+n^2-i+im^2-mn+m-1}{1-m-n}\]\[g-p=\frac{n^2-i+im^2+m^2-1+in^2}{1-m-n}=\frac{(i+1)(m^2+n^2-1)}{1-m-n}\]\[\overline{g-p}=\frac{(1-i)(m^2+n^2-1)}{1-m-n}\]\[\frac{g-p}{\overline{g}-\overline{p}}=\frac{i+1}{1-i}=i\]Thus $\frac{1}{2}(g+e+f-ef\overline{g})=\frac{1}{2}(p+e+f-ef\overline{p})\iff GP\perp EF$ as desired.$\blacksquare$
14.05.2024 17:57
Lmao, funny problem took me a while to realize what was going on. We denote by $H$ the intersection of lines $\overline{AB}$ and $\overline{CD}$. Let $\Gamma_B$ and $\Gamma_C$ be the circle with center $B$ and $C$ and radius $BP$ and $CP$ respectively. We prove the following key claims. Claim : Quadrilateral $HBPC$ is cyclic. Proof : It is clear that since \[\measuredangle PCH+ \measuredangle HBP = \frac{\pi}{2}+ \frac{\pi}{2} = \pi\]the quadrilateral $HBPC$ is cyclic as desired. Claim : Quadrilateral $HAGD$ is a parallelogram. Proof : Note that $CD = CF$ and $CE=BC$ the latter since a square has all four sides of equal length. Further, \[\measuredangle DCE = \frac{\pi}{2} + \measuredangle PCE = \measuredangle PCB\]Thus, $\triangle DCE \cong \triangle PCB$. Similarly we can show that $\triangle ABF \cong \triangle BPC$. Thus, \[\measuredangle GDH = \measuredangle EDC = \measuredangle BPC = \measuredangle BHC = \measuredangle AHD\]Thus, $\overline{AH} \parallel \overline{GD}$. Similarly, we can show that $\overline{AG}\parallel \overline{HD}$ from which it follows that $HAGD$ is indeed a parallelogram as claimed. Now we are almost there. Note that then, if we let $X$ and $Y$ be the intersections of $DP$ with $AG$ and $AP$ with $DG$ respectively, we have \[\measuredangle AXP = \measuredangle GXP = \measuredangle GXD = \measuredangle HDX = \measuredangle CDP = \frac{3\pi}{2}\]Thus, $X$ must lie on $\Gamma_B$. Similarly, we can show that $Y$ lies on $\Gamma_C$. Further it trivially follows that since \[\measuredangle AYD = \measuredangle PYD = \frac{3\pi}{2}=\measuredangle AXP = \measuredangle AXD\]the quadrilateral $AXYD$ must be cyclic. We denote by $\Gamma$ the circumcircle of this quadrilateral. Now, by Radical Center Theorem on circles $\Gamma_B$ , $\Gamma_C$ and $\Gamma$, we can note that $AX$ , $DY$ and the radical axis of $\Gamma_B$ and $\Gamma_C$ concur. But clearly, $\overline{AX} \cap \overline{DY}=G$. Thus, $G$ lies on the radical axis of $\Gamma_B$ and $\Gamma_C$. Further, $P$ also trivially lies on this radical axis. Thus, $\overline{GP}$ is the radical axis of $\Gamma_B$ and $\Gamma_C$. But, note that since the radical axis is known to be perpendicular to the line joining the centers, and $B$ and $C$ are respectively the centers of $\Gamma_B$ and $\Gamma_C$, we can conclude that $\overline{GP} \perp \overline{BC}$ and we are done.