we have a triangle $ ABC $ and make rectangles $ ABA_1B_2 $ , $ BCB_1C_2 $ and $ CAC_1A_2 $ out of it. then pass a line through $ A_2 $ perpendicular to $ C_1A_2 $ and pass another line through $ A_1 $ perpendicular to $ A_1B_2 $. let $ A' $ the common point of this two lines. like this we make $ B' $ and $ C' $. prove $ AA' $ , $ BB' $ and $ CC' $ intersect each other in a same point.
Problem
Source:
Tags: geometry, IGO
TelvCohl
05.09.2015 17:38
acupofmath wrote: we have a triangle $ ABC $ and make rectangles $ ABA_1B_2 $ , $ BCB_1C_2 $ and $ CAC_1A_2 $ out of it. then pass a line through $ \color{red}{ C_2 }\normalcolor $ perpendicular to $ \color{red}{ A_1C_2 }\normalcolor $ and pass another line through $ \color{red}{ B_1 }\normalcolor $ perpendicular to $ \color{red}{ A_2B_1 }\normalcolor $. let $ A' $ the common point of this two lines. like this we make $ B' $ and $ C' $. prove $ AA' $ , $ BB' $ and $ CC' $ intersect each other in a same point. Let $ G $ be the centroid of $ \triangle ABC $ and $ X $ be the intersection of $ AA' $ with the perepndicular bisector of $ BC. $ Notice $ \triangle GBC $ and $ \triangle A'C_2B_1 $ are homothetic and hence congruent $ \color{blue} ^{[1]} $ $ , $ so $ XB_1 \parallel CC_2, $ $ XC_2 \parallel BB_1. $ Let $ H_b ,H_c $ be the projection of $ B,C $ on $ B_1B_2, $ $ C_1C_2, $ resp. and let $ \triangle DEF $ be the triangle with the sides $ A_1A_2, $ $ B_1B_2, $ $ C_1C_2. $ Clearly, $ H_b, H_c $ lie on the circumcircle $ \Omega_A $ of $ BCB_1C_2. $ Since $ \odot (ACH_b) $ passes through the midpoint $ M_b $ of $ B_1B_2 $ $ \color{blue} ^{[2]} $ and $ \triangle XB_1C_2 $ $ \stackrel{+}{\sim} $ $ \triangle M_bCA $ $ \color{blue} ^{[3]} $ $ , $ so by simple angle chasing $ \Longrightarrow $ the intersection $ U $ of $ AH_b $ and $ C_2X $ lies on $ \Omega_A. $ Analogously, the intersection $ V $ of $ AH_c, B_1X $ lies on $ \Omega_A, $ so by Pascal's theorem for $ B_1H_bUC_2H_cV $ we get $ D \in AX $ $ \Longrightarrow $ $ AA' $ passes through the centroid of $ \triangle DEF $ $ \color{blue} ^{[4]} $ $.$ ____________________________________________________________ $ \bullet $ $ \qquad $ I state the properties that I used in my proof here, they are well-known and easy to prove so I'll not prove them here. [1] Given two triangles $ \triangle ABF, \triangle ACE $ such that $ AB \perp AF, $ $ AC \perp AE, $ $ \triangle ABF \stackrel{-}{\sim} \triangle ACE. $ Then $ EF $ is perpendicular to the A-median of $ \triangle ABC. $ [2] Given two triangles $ \triangle ABE, \triangle ACF $ such that $ AB \perp BE, $ $ AC \perp CF, $ $ \triangle ABE \stackrel{-}{\sim} \triangle ACF. $ Let $ H $ be the projection of $ A $ on $ EF $ and let $ M $ be the midpoint of $ EF. $ Then $ B,C,H,M $ are concyclic. [3] Given three triangles $ \triangle BDC, $ $ \triangle CEA, $ $ \triangle AFB $ such that $ \triangle BDC \stackrel{+}{\sim} \triangle CEA \stackrel{+}{\sim} \triangle AFB, $ $ DB=DC, $ $ EC=EA, $ $ FA=FB. $ Let $ X $ be the reflection of $ D $ in $ BC. $ Then $ AEXF $ is a parallelogram. [4] Given three triangles $ \triangle BDC, $ $ \triangle CEA, $ $ \triangle AFB $ such that $ \triangle BDC \stackrel{+}{\sim} \triangle CEA \stackrel{+}{\sim} \triangle AFB, $ $ DB=DC, $ $ EC=EA, $ $ FA=FB. $ Let $ M $ be the midpoint of $ BC $ and let $ X $ be the image of $ D $ under the translation $ \mathbf{T}( 2\overrightarrow{MD} ). $ Then $ AX $ is parallel to the D-median of $ \triangle DEF. $
MRF2017
20.09.2015 17:34
Proposed by Alexi Zaslaveski , Russia
anantmudgal09
27.09.2016 13:12
Excellent Problem!
Lemma 1Lemma 1. Lines $B'C'$ and $B_2C_1$ are parallel.
Proof:- Fix the points $A,B,C,A_1,A_2,B_2,C_1$ and vary the points $C_2,B_1$ such that $BCB_1C_2$ is a rectangle. The map $C_2 \rightarrow B'$ obtained by rotating at $90^{\circ}$ around $A_1$ and projecting on the line perpendicular to $B_2C_1$ at $B_2$ is a projective map (cross-ratio is preserved) and similarly, is the map sending $B_1$ to $C'$. Thus, points $B',C'$ are projectively related. It suffices to showing that $B'C' \parallel B_2C_1$ for three distinct positions of the pair $C_2,B_1$. Indeed, when $C_2=B$ and $B_1=C$ the lines $B'C'$ and $B_2C_1$ coincide and there is nothing to prove. Otherwise, we see that it suffices to proving the following two statements:
1. $A_1C_2 \parallel B_2C_1$ if and only if $A_2B_1 \parallel B_2C_1$. In this case the line $B'C'$ is the line at infinity and $B'C' \parallel B_2C_1$ for clear reasons.
2. $A_1B' \parallel BC$ if and only if $A_2C' \parallel BC$. In this case, line $C_2B_1$ is taken as the line at infinity.
Indeed, let $\alpha=\angle AB_2C_1$ and $\beta=\angle AC_1B_2$ and let us denote angles $ABC$ and $ACB$ by $B$ and $C$ respectively. Notice that $(B+\alpha)+(C+\beta)=(B+C)+(\alpha+\beta)=(180^{\circ}-\angle BAC)+\angle BAC=180^{\circ}$ and so, $\sin (B+\alpha)=\sin (C+\beta)$.
1. $A_1C_2 \parallel B_2C_1$ iff $\angle C_2A_1B=\alpha$ and $\angle C_2BA_1=B$; $BC_2=BA_1\cdot \frac{\sin \alpha}{\sin (C+\beta)}$. Similarly, $A_2B_1 \parallel B_2C_1$ if and only if $CB_1=CA_2\cdot \frac{\sin \beta}{\sin (B+\alpha)}$. By sine rule, $CA_2\sin \beta=AC_1\sin \beta=AB_2\sin \alpha=BA_1\sin \alpha$ and we see that both the iff conditions are equivalent.
2. $A_1B' \parallel BC$ iff $\angle B'A_1B_2=\alpha$ and $\sin \angle B_2B'A_1=\sin (B+\alpha)$; $B_2B'=A_1B_2\cdot \frac{\sin B}{\sin (B+\alpha)}=AB\cdot \frac{\sin B}{\sin (B+\alpha)}$. Similarly, $A_2C' \parallel BC$ iff $C_1C'=AC\cdot \frac{\sin C}{\sin (C+\beta)}$. As $AB\sin B=AC\sin C$, we see that both the conditions are equivalent.
The Lemma is proven.
Lemma 2Lemma 2. Let $A_3=A_1C_2 \cap A_2B_1$ and define the points $B_3,C_3$ analogously. The lines $A'A_3,B'B_3,C'C_3$ are concurrent at a point (say $T$).
Proof: Applying the Ceva's Theorem in triangle $A_3B_3C_3$ in the distance form, we see that proving the product of ratios like $\frac{A'B_1}{A'C_2}=\frac{\sin \angle A'C_2B_1}{\sin \angle A'B_1C_2}=\frac{\sin \angle A_1C_2B}{\sin \angle A_2B_1C}$ is equal to $1$. This product on rearranging the terms is, in fact, equal to the product of ratios $\frac{BC_2}{BA_1}$ cycling through the vertices, which is equal to $1$ as the quadrilaterals here are rectangles and the opposite sides have equal lengths. The Lemma is proven.
Lemma 3Lemma 3. Triangles $ABC$ and $A'B'C'$ are orthological and the centers of orthology coincide.
Proof: Let $T'$ be the isogonal conjugate of $T$ is triangle $A'B'C'$. Notice that by Lemma 1., isogonality in angle $B'A'C'$ for a pair of lines passing through $A'$ is equivalent to their isogonality in the angle $B_2A'C_1$. As the lines $A'T'$ and $A'A_3$ are isogonal in angle $B'A'C'$ we see that $A'T'$ and $A'A_3$ are isogonal in angle $B_2A'C_1$ and thus, $A'T' \perp B_2C_1$. We conclude that $A'T' \perp BC$ and similar relations hold. Point $T'$ is thus, one of the orthology centers of the triangles $ABC$ and $A'B'C'$. Notice that by Lemma 1., lines $B'C'$ and $B_2C_1$ are parallel. By construction, $AB_2 \parallel B'T'$ and $AC_1 \parallel C'T'$. Thus, line $AT'$ is perpendicular to the line $B'C'$. The Lemma is proven.
By degenerate Sondat's Theorem, the triangles $ABC$ and $A'B'C'$ are perspective, i.e., lines $AA',BB',CC'$ are concurrent.
Durjoy1729
23.08.2018 18:24
https://drive.google.com/file/d/1mLs9yBbA6yajTImWlYSFAYwBms33yW0b/view?usp=drivesdk
mzy
06.11.2019 16:23
[asy][asy] /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki go to User:Azjps/geogebra */ import graph; size(8cm); real labelscalefactor = 0.3; /* changes label-to-point distance */ pen dps = linewidth(0.2) + fontsize(3); pen dotstyle = black; /* point style */ real xmin = -42.301043410924784, xmax = 32.73072609798678, ymin = -18.6882664304738, ymax = 25.638074581274033; /* image dimensions */ pen qqwuqq = rgb(0,0.39215686274509803,0); draw(arc((-11.473944448124983,-1.4422005494156393),1.1675067869643915,-55.32222283110016,-14.909333625747541)--(-11.473944448124983,-1.4422005494156393)--cycle, linewidth(0.5) + qqwuqq); draw(arc((-8.463364626929115,-5.7936350842490825),1.1675067869643915,89.622231103771,124.67777716889983)--(-8.463364626929115,-5.7936350842490825)--cycle, linewidth(0.5) + qqwuqq); draw(arc((0.6366353730708854,-5.853635084249082),1.1675067869643915,64.28598374037537,89.622231103771)--(0.6366353730708854,-5.853635084249082)--cycle, linewidth(0.5) + qqwuqq); 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label("$\gamma$", (0.7967077067577038,-4.4836005224071105), NE * labelscalefactor,qqwuqq); label("$\delta$", (3.2317212806864872,0.06967594675399238), NE * labelscalefactor,qqwuqq); label("$\epsilon$", (-4.473823513278497,7.308218025933182), NE * labelscalefactor,qqwuqq); label("$\zeta$", (-8.443346588957428,4.973204452004411), NE * labelscalefactor,qqwuqq); label("$\alpha_1$", (-7.898510088374046,-1.1367477331092057), NE * labelscalefactor,qqwuqq); label("$\beta_1$", (-7.1590891232965985,-2.0318362697819014), NE * labelscalefactor,qqwuqq); label("$\gamma_1$", (-1.3604720813734532,-2.2318362697819014), NE * labelscalefactor,qqwuqq); label("$\delta_1$", (-1.204804509778201,-1.6091659834008958), NE * labelscalefactor,qqwuqq); label("$\epsilon_1$", (-5.952665443433394,2.7327753425697044), NE * labelscalefactor,qqwuqq); label("$\zeta_1$", (-6.886670873004907,2.404689520682763), NE * labelscalefactor,qqwuqq); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */ [/asy][/asy] Construct point $P$ as in the diagram, where $\angle PBA=\angle BA_1C_2, \angle PBC=\angle BC_2A_1, \cdots$. It is easy to show that point $P$ exists, using the converse of ceva's theorem. By some simple angle chasing, we note that $\angle PBC=\angle BC_2A_1=\angle A'C_2B_1$. Likewise, $\angle PCB=\angle A'B_1C_2$, so $\triangle PBC \cong \triangle A'C_2B_1$, so $PBC_2A'$ is a parallelogram. Hence, $PA'\parallel BC_2\perp BC$, and similarly, $PB'\perp AC$, $PC'\perp AB$. Using a similar argument, $PBA_1C'$ is a parallelogram, so $A_1C'=C_2A'$ implies $A_1C'A'C_2$ is a rectangle. Hence, $PB\parallel A'C_2\perp A'C'$. Similarly, $PC\perp A'B'$ and $PA\perp B'C'$. Now, fix points $A,B,C,P,$ and vary $A',B',C'$ such that the perpendicular conditions ($PA'\perp BC$, $PA\perp B'C'$, etc.) still hold. We notice a few special cases: When $A'=B'=C'=P$, the concurrency point of $AA',BB',CC'$ is $P$. When $A',B',C' \rightarrow \infty$ , they concur at $H$, the orthocenter of $\triangle ABC$. When $A'$ lies on $AC$, since $PB'\perp AC$ and $A'B'\perp PC$, $B'$ is the orthocenter of $\triangle PA'C$, which means $B'C\perp A'P\perp BC$, so $B'$ lies on $BC$. The concurrency point is $C$. [asy][asy] import graph; size(8cm); real labelscalefactor = 0.5; /* changes label-to-point distance */ pen dps = linewidth(0.7) + fontsize(10); defaultpen(dps); /* default pen style */ pen dotstyle = black; /* point style */ real xmin = -10.30503212408729, xmax = 2, ymin = -4, ymax = 5.147858768119547; /* image dimensions */ /* draw figures */ draw((-6.72,4.21)--(-8.44,-2.25), linewidth(0.5)); draw((-8.44,-2.25)--(0.66,-2.31), linewidth(0.5)); draw((-6.72,4.21)--(0.66,-2.31), linewidth(0.5)); draw((-6.72,4.21)--(-4.756538908246226,0.2984362791679838), linewidth(0.5)); draw((-4.756538908246226,0.2984362791679838)--(-8.44,-2.25), linewidth(0.5)); draw((-4.756538908246226,0.2984362791679838)--(0.66,-2.31), linewidth(0.5)); draw((-2.906559902478888,-0.19412779047904705)--(-4.742292085327908,2.4592044217794466), linewidth(0.5) + blue); draw((-2.906559902478888,-0.19412779047904705)--(-7.012608645620668,-2.2552132088417353), linewidth(0.5) + blue); draw((-7.012608645620668,-2.2552132088417353)--(-4.742292085327908,2.4592044217794466), linewidth(0.5) + blue); draw((xmin, -0.885264990462881*xmin-1.7389807359105604)--(xmax, -0.885264990462881*xmax-1.7389807359105604), linewidth(0.5) + red); /* line */ draw((xmin, -0.0036522631482535624*xmin-2.280825100971261)--(xmax, -0.0036522631482535624*xmax-2.280825100971261), linewidth(0.5) + red); /* line */ draw((xmin, -0.5932529572965661*xmin-1.9184530481842665)--(xmax, -0.5932529572965661*xmax-1.9184530481842665), linewidth(0.5) + red); /* line */ /* dots and labels */ dot((-6.72,4.21),linewidth(1pt) + dotstyle); label("$A$", (-6.857779215830774,4.381815071943619), NE * labelscalefactor); dot((-8.44,-2.25),linewidth(1pt) + dotstyle); label("$B$", (-8.680379234736962,-2.5274959088189144), S* labelscalefactor); dot((0.66,-2.31),linewidth(1pt) + dotstyle); label("$C$", (0.7971408635752191,-2.4612195444950533), NE * labelscalefactor); dot((-4.756538908246226,0.2984362791679838),linewidth(1pt) + dotstyle); label("$P$", (-4.687228284224313,0.3389568481880838), NE * labelscalefactor); dot((-4.742292085327908,2.4592044217794466),linewidth(1pt) + dotstyle); label("$A'$", (-4.5546755555765905,2.2609714135800596), NE * labelscalefactor); dot((-7.012608645620668,-2.2552132088417353),linewidth(1pt) + dotstyle); label("$B'$", (-6.940624671235601,-2.229252269361539), S * labelscalefactor); dot((-2.906559902478888,-0.19412779047904705),linewidth(1pt) + dotstyle); label("$C'$", (-3.2125791780183963,-0.10840861099797953), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); [/asy][/asy] Similarly, we see that we have two other cases where the concurrency point is $B$ and $A$ respectively. So we guess the concurrency point lies on the conic passing through $A,B,C,P,H$. Let $AA'$ and $BB'$ intersect the conic at $A''$ and $B''$. It suffices to show $A''=B''$. Note that $A'\rightarrow AA'\rightarrow AA''\rightarrow A''$ is a projective transformation, and since $A' \rightarrow B'$ is a linear (also projective) transform, $A'\rightarrow B'\rightarrow B''$ is also a projective transform. Since $A''=B''$ matches at 5 distinct cases (we actually only need 3), the transformation $A\rightarrow A''$ and $B\rightarrow B''$ are the same, so we always have $A''=B''$. [asy][asy] /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki go to User:Azjps/geogebra */ import graph; size(15cm); real labelscalefactor = 0.5; /* changes label-to-point distance */ pen dps = linewidth(0.7) + fontsize(10); defaultpen(dps); /* default pen style */ pen dotstyle = black; /* point style */ real xmin = -10, xmax = 4, ymin = -6, ymax = 7; /* image dimensions */ pen qqzzqq = rgb(0,0.6,0); /* draw figures */ draw((-6.72,4.21)--(-8.44,-2.25), linewidth(0.5)); draw((-8.44,-2.25)--(0.66,-2.31), linewidth(0.5)); draw((-6.72,4.21)--(0.66,-2.31), linewidth(0.5)); draw((-6.72,4.21)--(-4.756538908246226,0.2984362791679838), linewidth(0.5)); draw((-4.756538908246226,0.2984362791679838)--(-8.44,-2.25), linewidth(0.5)); draw((-4.756538908246226,0.2984362791679838)--(0.66,-2.31), linewidth(0.5)); draw((-8.777394487185102,1.369004637647065)--(-4.787503809596287,-4.397907092258011), linewidth(0.5) + blue); draw((-8.777394487185102,1.369004637647065)--(0.14693867810322261,5.848691583962299), linewidth(0.5) + blue); draw((0.14693867810322261,5.848691583962299)--(-4.787503809596287,-4.397907092258011), linewidth(0.5) + blue); draw((xmin, -4.454294468989227*xmin-25.722858831607606)--(xmax, -4.454294468989227*xmax-25.722858831607606), linewidth(0.5) + red); /* line */ draw((xmin, 0.9431407265797813*xmin + 5.710107732333353)--(xmax, 0.9431407265797813*xmax + 5.710107732333353), linewidth(0.5) + red); /* line */ draw((xmin, -0.38983266436967645*xmin-2.0527104415160133)--(xmax, -0.38983266436967645*xmax-2.0527104415160133), linewidth(0.5) + red); /* line */ pair hyperbolaLeft1 (real t) {return (2.0329451352789323*(1+t^2)/(1-t^2),2.0329451352789323*2*t/(1-t^2));} pair hyperbolaRight1 (real t) {return (2.0329451352789323*(-1-t^2)/(1-t^2),2.0329451352789323*(-2)*t/(1-t^2));} draw(shift((-6.047314939971949,2.2630200321111817))*rotate(-77.48821331360655)*graph(hyperbolaLeft1,-0.99,0.99), linewidth(0.5) + qqzzqq); draw(shift((-6.047314939971949,2.2630200321111817))*rotate(-77.48821331360655)*graph(hyperbolaRight1,-0.99,0.99), linewidth(0.5) + qqzzqq); /* hyperbola construction */ /* dots and labels */ dot((-6.72,4.21),linewidth(1pt) + dotstyle); label("$A$", (-6.865975323306674,4.376433075293477), NE * labelscalefactor); dot((-8.44,-2.25),linewidth(1pt) + dotstyle); label("$B$", (-8.705933437631015,-2.5494469965500164), NE * labelscalefactor); dot((0.66,-2.31),linewidth(1pt) + dotstyle); label("$C$", (0.806302851517472,-2.462656519459246), NE * labelscalefactor); dot((-4.756538908246226,0.2984362791679838),linewidth(1pt) + dotstyle); label("$P$", (-4.678855300619247,0.3319968428635673), NE * labelscalefactor); dot((-4.787503809596287,-4.397907092258011),linewidth(1pt) + dotstyle); label("$A'$", (-4.592064823528476,-4.5977022558922025), NE * labelscalefactor); dot((0.14693867810322261,5.848691583962299),linewidth(1pt) + dotstyle); label("$B'$", (0.2161276073002301,5.886587376672885), NE * labelscalefactor); dot((-8.777394487185102,1.369004637647065),linewidth(1pt) + dotstyle); label("$C'$", (-9.087811536830408,1.460273045043585), NE * labelscalefactor); dot((-6.749980641611705,-0.3370639777752116),linewidth(2pt) + dotstyle); label("$H$", (-7.022198182070061,-0.32761078302628926), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */ [/asy][/asy]
jred
25.12.2020 09:45
TelvCohl wrote: acupofmath wrote: we have a triangle $ ABC $ and make rectangles $ ABA_1B_2 $ , $ BCB_1C_2 $ and $ CAC_1A_2 $ out of it. then pass a line through $ \color{red}{ C_2 }\normalcolor $ perpendicular to $ \color{red}{ A_1C_2 }\normalcolor $ and pass another line through $ \color{red}{ B_1 }\normalcolor $ perpendicular to $ \color{red}{ A_2B_1 }\normalcolor $. let $ A' $ the common point of this two lines. like this we make $ B' $ and $ C' $. prove $ AA' $ , $ BB' $ and $ CC' $ intersect each other in a same point. Let $ G $ be the centroid of $ \triangle ABC $ and $ X $ be the intersection of $ AA' $ with the perepndicular bisector of $ BC. $ Notice $ \triangle GBC $ and $ \triangle A'C_2B_1 $ are homothetic and hence congruent $ \color{blue} ^{[1]} $ , ...... If anyone would like to explain why $ \triangle GBC $ and $ \triangle A'C_2B_1 $ are homothetic？