73. A nice problems with a square and a perpendicularity.

by Virgil Nicula, Aug 2, 2010, 12:16 AM

http://www.artofproblemsolving.com/Forum/viewtopic.php?f=46&t=359124

PP1. Let $ABCD$ be a square. For an arbitrary $M\in [BC]$ consider $E\in AM\cap CD$ , $F\in DM\cap BE$ . Prove that $CF\perp AE$ .

Proof 1. Let $N\in AE\cap BD$ , $P\in CN\cap AB$ , $Q\in AB\cap CF$ . Apply Blanchet's theorem in $\triangle BDE$ , where $BC\perp DE$ , $M\in BC$ and obtain $\widehat{BCP}\equiv\widehat{BCQ}$ ,

i.e. $BP=BQ$ . From symmetry w.r.t. $BD$ obtain and $BP=BM$ . Hence $BQ=BM$ . A rotation around $B$ realizes $CQ\rightarrow AM$ . In conclusion $CQ\perp AM$ , i.e. $CF\perp AE$ .

Remark. I can prove $\widehat{BCP}\equiv\widehat{BCQ}$ without Blanchet's theorem and rotation. Let $G\in CF\cap AE$ and $H\in BG\cap DC$ . Thus, division $(D,C,H,E)$ - harmonically $\implies$

$(N,M,G,E)=B(D,C,H,E)\cap AE$ - harm. $\implies$ $(N,M,G,E)$ - harm. Since $CM\perp CE$ obtain $\widehat{MCG}\equiv\widehat{MCN}$ $\implies$ $\widehat{BCP}\equiv\widehat{BCQ}$ . Thus $\triangle CPQ$ is isosceles $\implies$

$BP=BQ=BM$ (last equality from symmetry w.r.t. $BD$). Therefore, $\triangle BAM\equiv\triangle BCQ$ $\implies$ $\widehat{ BAM}\equiv\widehat{BCQ}$ $\implies$ $\widehat{ BAG}\equiv\widehat{BCG}$ $\implies$ $ABGC$ is cyclic $\implies$ $CF\perp AE$ .

Proof 2 (metric). Suppose w.l.o.g. $AB=1$ and let $BM=m\in [0,1]$ . Denote $L\in DM\cap AB$ and $G\in AE\cap CF$ . Prove easily that $DE=\frac 1m$ , $CE=\frac {1-m}{m}$ , $BL=\frac {m}{1-m}$ .

Denote $m\left(\widehat{BAG}\right)=x$ and $m\left(\widehat{BCG}\right)=y$ . Thus, $\underline{\tan x=m}$ and $\frac {m^2}{1-m}=$ $\frac {\frac {m}{1-m}}{\frac 1m}=$ $\frac {BL}{CE}=$ $\frac {FB}{FE}=$ $\frac {CB}{CE}\cdot\frac {\sin y}{\cos y}=$ $\frac {1}{\frac {1-m}{m}}\cdot\tan y=$ $\frac {m}{1-m}\cdot\tan y$ , i.e. $\tan y=m$ . Hence

$\underline{\tan y=\tan x}$ $\Longleftrightarrow$ $x=y$ $\Longleftrightarrow$ $\widehat {BAG}\equiv \widehat{BCG}$ , i.e. the quadrilateral $ABGC$ is cyclic, i.e. $CF\perp AE$ .

Remark. If $R$ is the center of given square and $S\in RM\cap BE$ , then $CS\perp BE$ .

Proof. Let $T\in RM\cap AB$ , Apply Menelaus's theorem to $:\ \left\{\begin{array}{cccc} \overline{TRM}/ABC\ : & \frac {TA}{TB}\cdot\frac {MB}{MC}\cdot\frac {RC}{RA}=1 & \implies & \frac {TA}{TB}=\frac {1-m}{m}\ .\\\ \overline{TMS}/ABE\ : & \frac {TA}{TB}\cdot\frac {MB}{MC}\cdot\frac {RC}{RA}=1 & \implies & \frac {SB}{SE}=\left(\frac {m}{1-m}\right)^2\ .\end{array}\right\|$ Denote $m\left(\widehat{CBE}\right)=x$ , $m\left(\widehat {SCE}\right)=z$ .

Observe that $\tan x=\frac {1-m}{m}$ and $\left(\frac {m}{1-m}\right)^2=\frac {SB}{SE}=\frac {CB}{CE}\cdot \frac {\sin\widehat{SCB}}{\sin\widehat{SCE}}=$ $\frac {m}{1-m}\cdot\cot z$ $\implies$ $\tan z=\frac {1-m}{m}$ $\implies$ $\tan z=\tan x$ $\implies$ $x=z$ $\implies$ $CS\perp AE$ .

Another proof. Denote $L\in AB\cap DM$ and $S'\in BE\cap CL$ . Apply Pappus' theorem to $AB$ , $DC$ and to $\{A,B,L\}\subset AB$ , $\{D,C,E\}\subset DC$ . Since $R\in BD\cap AC$ ,

$M\in LD\cap AE$ , $S'\in LC\cap BE$ obtain $S'\in RM$ . Thus, $S'\in RM\cap BE$ , i.e. $S'\equiv S$ . From $\frac {CE}{AB}=\frac {MC}{MB}$ and $\frac {LB}{CD}=\frac {MB}{MC}$ obtain $CE\cdot BL=BC^2$ . Prove easily that in any

right trapezoid $XYZT$ $(XT\parallel YZ$ , $XY\perp XT)$ exists $XT\cdot YZ=XY^2$ $\Longleftrightarrow$ $XZ\perp YT$ . Apply it to the right trapezoid $BCEL$ and get $CL\perp BE$ , i.e. $CS\perp BE$ .

Proof 3. Suppose w.l.o.g. $AB=1$ and denote $BM=m\in [0,1]$ . Define $G\in AE\cap CF$ , $H\in DC\cap BG$ . From $MC\parallel AD$ in $\triangle AED$ obtain easily $\frac {EC}{1-m}=\boxed{ED=\frac 1m}$ .

Since division $(D,C,H,E)$ is harmonic, i.e. $\frac {HC}{HE}=\frac {DC}{DE}$ (is an well-known property or prove easily) obtain $\frac {HC}{HE}=m$ , i.e. $\frac {HC}{m}=$ $\frac {HE}{1}=$ $\frac {CE}{1+m}=$ $\frac {1-m}{m(1+m)}$ $\implies$

$\boxed{HC=\frac {1-m}{1+m}}$ . Denote $x=m\left(\widehat{CBH}\right)$ , $y=m(\angle AED)$ . Thus $\tan x=\frac {HC}{BC}=\frac {1-m}{1+m}$ and $\tan y=\frac {AD}{DE}=m$ . Therefore, $\tan x=\frac {1-\tan y}{1+\tan y}=\tan\left(45^{\circ}-y\right)$ $\implies$

$\boxed{x+y=45^{\circ}}$ . Thus, $m\left(\widehat{GAC}\right)=m\left(\widehat{EAC}\right)=45^{\circ}-y=x=m\left(\widehat{GBC}\right)$ . In conclusion, $G$ belongs to circumcircle of $\triangle ABC$ $\implies$ $CG\perp GA$ $\implies$ $CF\perp AE$ .

Proof 4. Suppose w.l.o.g. $AB=1$ and denote $BM=m\in [0,1]$ . Define $G\in AE\cap CF$ , $H\in DC\cap BG$ . From $MC\parallel AD$ in $\triangle AED$ obtain easily $\frac {EC}{1-m}=\boxed{ED=\frac 1m}$ .

Since division $(D,C,H,E)$ is harmonic, i.e. $\frac {HC}{HE}=\frac {DC}{DE}$ ( prove easily) obtain $\frac {HC}{HE}=m$ . Apply Menelaos' to $\overline{BGH}/CEM\ :\ \frac {BM}{BC}\cdot\frac {HC}{HE}\cdot\frac {GE}{GM}=1$ $\Longleftrightarrow$

$m\cdot m\cdot \frac {GE}{GM}=1$ $\Longleftrightarrow$ $\frac {GM}{GE}=m^2$ $\Longleftrightarrow$ $\frac {GM}{GE}=\left(\frac{CM}{CE}\right)^2$ $\implies$ $G$ is the foot of $C$-symmedian in $C$-right triangle $CEM$ $\implies$ $CG\perp EM$ $\implies$ $CF\perp AE$ .

Otherwise, denote $N\in AE\cap BD$ and $(D,C,H,E)$ is harmonically $\implies$ $(N,M,G,E)$ is harmonically $\implies$ $\frac {GM}{GE}=$ $\frac {NM}{NE}=$ $\frac {NM}{NA}\cdot\frac {NA}{NE}=$ $\frac {BM}{AD}\cdot\frac {AB}{DE}=$

$m\cdot\frac {1}{\frac 1m}=$ $m^2=$ $\left(\frac {CM}{CE}\right)^2$ $\implies$ $\frac {GM}{GE}=\left(\frac{CM}{CE}\right)^2$ $\implies$ $G$ is the foot of $C$-symmedian in $C$-right triangle $CEM$ $\implies$ $CG\perp EM$ $\implies$ $CF\perp AE$ .

Otherwise, observe that $(D,C,H,E)$ is harmonically $\implies$ $(N,M,G,E)$ is harmonically. Observe that $NM\cdot NE=$

$\frac {NM}{NA}\cdot\frac {NE}{NA}\cdot NA^2=$ $\frac {BM}{AD}\cdot\frac {DE}{AB}\cdot NA^2=$ $m\cdot\frac 1m\cdot NA^2=NA^2$ $\implies$ $NA=NG$ $\implies$ $CF\perp AE$ .

Remarks. If denote $L\in DF\cap AB$ and $ABCD$ is a rectangle, then $BL\cdot CE=AB^2$ and $\tan \gamma\cdot\sin 2\beta=\tan 2\phi$ ,

where $\gamma=m\left(\widehat{AGC}\right)$ , $\phi=m\left(\widehat {BAC}\right)$ . In the particular case when $\phi =45^{\circ}$ obtain $\gamma=90^{\circ}$, i.e. the proposed problem.

Final. Let $ABCD$ be a square. For an arbitrary $M\in [BC]$ consider the points $\left\{\begin{array}{c} E\in AM\cap CD\\\ L\in DM\cap AB\\\ S\in BE\cap CL\\\ F_1\in DM\cap BE\\\ F_2\in AM\cap CL\\\ R\in AC\cap BD\end{array}\right\|$ . Prove that $\boxed{\begin{array}{c} S\in RM\\\ F_1F_2\parallel BC\\\ BE\perp CL\\\ CF_1\perp AM\\\ BF_2\perp DM\end{array}}$ .

Remark. If $G_1\in CF_1\cap ME$ , $G_2\in BF_2\cap ML$ , then polygons $RBSC$ (quadrilateral, in circle with diameter $[BC]$), $SF_1G_2G_1F_2$ (pentagon, in circle with diameter $[F_1F_2]$) and

$ABG_2G_1CD$ (hexagon, in circle with diameter $[AC]$) are cyclically. From here can obtain another interesting properties of this figure. For example, the circumcircles of $RBSC$ and

$SF_1G_2G_1F_2$ are tangent in $S$ . Indeed, since $F_1F_2\parallel BC$ obtain that the circumcircles of $\triangle SF_1F_2$ , $SBC$ are tangent in $S$ . Thank very much to Mateescu Constantin for his nice draw.


PP2. Let $ABC$ be an $A$-isosceles and right triangle. Let $\{M,N\}\subset (BC)$ be two points such that $N\in (MC)$ and $MB^2 + NC^2 = MN^2$. Ascertain the measure of the angle $\widehat{MAN}$ .

Proof 1 (metric). Suppose w.l.o.g. $AB=AC=1$ , i.e. $BC=\sqrt 2$ . Denote $\left\{\begin{array}{c} m\left(\widehat{MAB}\right)=u\\\\ m\left(\widehat{NAC}\right)=v\end{array}\right\|$ and

$\left\{\begin{array}{ccccccc} X\in (AB) & , & MX\perp AB & ; & BX=MX=x & ; & MB=x\sqrt 2\\\\ Y\in (AC) & , & NY\perp AC & ; & CY=NY=y & ; & NC=y\sqrt 2\end{array}\right\|\implies$

$\left\{\begin{array}{ccc} AX=1-x & ; & \tan u=\frac {x}{1-x}\\\\ AY=1-y & ; & \tan v=\frac {y}{1-y}\end{array}\right\|$ , $MN=(1-x-y)\sqrt 2$ , where $\left\{\begin{array}{c} \{x,y\}\subset (0,1)\\\ x+y<1\end{array}\right\|$ . Remark that $\boxed{1+2xy=2(x+y)\ \iff\ (1-x)(1-y)=\frac 12}\ (*)$ .

$\blacktriangleright\ MB^2+NC^2=MN^2\iff$ $x^2+y^2=(1-x-y)^2\iff$ $0=1+2xy-2(x+y)\iff\ \boxed{1+2xy=2(x+y)}$ .

$\blacktriangleright\ m\left(\widehat{MAN}\right)=45^{\circ}\iff$ $u+v=45^{\circ}\iff$ $\tan (u+v)=1\iff$ $\tan u+\tan v=1-\tan u\tan v\iff$

$\frac {x}{1-x}+\frac {y}{1-y}=1-\frac {xy}{(1-x)(1-y)}\iff$ $x(1-y)+y(1-x)+xy=(1-x)(1-y)\iff$ $\boxed{1+2xy=2(x+y)}$ .

Remark. The area $[XAY]=\frac 12\cdot [ABC]$ (constant) $\iff$ $2\cdot AX\cdot AY=AB^2\iff$ $\frac {(1-x)(1-y)}{2}=\frac 14\iff$

$(1-x)(1-y)=\frac 12$ $\stackrel{(*)}{\iff}$ $1+2xy=2(x+y)\iff$ $MB^2+NC^2=MN^2\iff $ $m\left(\widehat{MAN}\right)=45^{\circ}$ .

Proof 2 (synthetic). Rotate around about the point $A$ for $90^\circ$ to get $\triangle ACC'\implies$ $CM'=BM$ , $NM'=\sqrt{NC^2+CM'^2}=$

$\sqrt{NC^2+BM^2}=MN\implies$ $\triangle AMN\cong\triangle AM'N\implies$ $m\left(\angle MAM'\right)=90^\circ\iff$ $m\left(\widehat{MAN}\right)=45^\circ$ .
This post has been edited 64 times. Last edited by Virgil Nicula, Dec 2, 2015, 11:12 AM

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153. Geometrical minimum/maximum.
152. Concurrence, cyclically, metrics, symmedian a.s.o.
151. Miscellaneous problems for the admission at math.
150. Some (5) problems with fixed points.
149. Some (8) problems with collinearity or concurrence.
148. A special class of systems.
147. A problem with two polynominals.
146. Synthetic, metric, trigonometric and analitycal proofs.
145. Spain MO P5 - Point on median.
144. A conditioned concurrency in a triangle.
143. Assess of a ratio (nice and difficult problem).
142. Inscribed/circumscribed quadrilateral w.r.t. ABC.
141. N is interior <==> 1/3<a/b<3 (CRUX Mathematicorum).
140. A synthetical property <==> linear/angled relation.
139. A caracterization of a right angle in a triangle.
138. Generalized "slicing" problems (own).
137. Stewart's relation.
136. A geometrical interpretation of a system (own).
135. Three strong geometrical inequalities.
+ September 2010
134. Some problems with projections/perpendiculars (2).
133. Some problems with projections/perpendiculars (1).
132. Asupra unei inegalitati intr-un triunghi.
131. A stronger Form of the Steiner-Lehmus Theorem (own).
130. A triangle and a fixed point.
129. BX=CY and three properties.
128. Value of IG and two inequalities in neighbourhood of 0.
127. Pagina educativa.
126. Routh's theorem.
125. I like very much this Darij Grinberg's message.
124. A nice and easy problem with a right-angled triangle.
123. My extension and its proof.
122. An identity for an equilateral triangle. Extension.
121. Viete's relations with cos A , cos B , cos C.
120. Two equal neighbouring angles/sides in a quadrilateral.
119. Inspired by a Miculita's problem.
118. An usual remarkable geometrical locus.
117. Some applications of harmonic division/quadrilateral.
116. The length of AE (nine methods !).
115. An difficult equation with one radical.
114. A property of tangential triangle for ABC.
113. Range for a function.
112. The equivalence relation ".s.s."
111. Simple, but interesting ...
110. German TST 2004 - Exam V , Problem 1
109. Two equal areas ==> find A.
108. Old, short and nice proof of Euler's relation.
107. XY parallel BC and X,Y are isogonal/izotomic points.
106. OLM - Bucuresti (geometrie).
105. A problem with three circles.
104. Concursul Revistei de Matematica din Galati (RMG).
103. Metrical usual relations in triangle. Applications.
102. Characterization of a metrical relation in a triangle.
101. O problema cu numere complexe.
100. Bobiller's theorem.
99. Some simple and nice geometry problems.
98. Symmedian & median (metrical relations).
97. Problem of butterfly.
+ August 2010
96. Integrals.
95. Three circles in an angle.
94. Equations of angle bisectors (interior and exterior).
93. The Appolonius' construct problems.
92. CA , CB tangent to parabola - OIMU 2008 Problem 4.
91. JBMO 2010, Problem 3.
90. Nice and difficult inequality in ABC (own).
89. Similarity (parallel/antiparallel).
88. A nice geometric locus conditioned metrically.
87. An interesting vectorial identity.
86. A very nice maxmin.
85. Problem "slicing" : 60-24-12.
84. IRAN national math olympiad - 2010
83. A metrical characterization of concurrence.
82. AA' , BN , CM concur in Gergonne point.
81. Sawayama and Thebault’s theorem.
80. Three concurrent perpendiculars (the Carnot's lemma).
79. A quadrilateral with many perpendicularities.
78. Jakobi's theorem.
77. Two equivalent perpendicularities.
76. A parallelogram and a circle (nice !).
75. Extended Steinbart Theorem.
74. Inegalitatea Erdos-Mordell.
73. A nice problems with a square and a perpendicularity.
+ July 2010
72. Colinearity in a triangle - H , P , M .
70. Colinearity with pole/polar - H\in PQ .
71. An angle questions (synthetic/trigonometric proofs).
69*1. Slicing problem 3.
69. Position of x1, x2 w.r.t. a given real number k.
68. Some nice applications of "pole/polar".
67. Remarkable algebraic/geometrical inequalities (1).
66. Remarkable perpendicularities in a triangle.
65. Parallel to BC through I .
64. Problems of the type "instant" (at most one minute).
63. A nice problem with radical center (livetolove212).
62. Nice and easy problem with a right triangle.
61. IMO Shortlist 2009 (very nice problem and its proof).
60. Mediteranean M.C. 2010 Problem 3.
59. IMO 2010, problem 4.
58. A cevian and two incircles.
57. Opinii si comentarii relativ la inv. rom. II
56. A extremum problem with areas.
55. Two important circles - mixtilinear incircle/excircle.
54. Cinci grade de libertate.
53. An remarkable concurrency.
52. Harmonical division.
51*1. JBTST-3/2010 Aplicatii la proprietatea fluturelui.
51. A nice problem with perpendicularity from Jaime.
50. IMAC 2010 seniors, the first day.
49. The midpoint of a cevian in an acute triangle.
+ June 2010
48. An inequality in a triangle : h_a+max{b,c} ...
47. Interesting and difficult identities in a triangle.
46. Outside of a quadrilateral.
45. \cos B+\cos C=1 <==> nice property.
44. Relatia IA=IO sau IA=IM intr-un triunghi ABC etc.
43. A "slicing" problem with a parameter.
42. IMAC 2010 Problema 2.
41. ONM 2010 Calarasi Problema 3.
+ May 2010
40. Lema lui Haruki.
39. A nice and remarkable problem with Brocard's point.
38. Minimum area of triangle touching with semicircle.
37. Own extension of the Steiner-Lehmus' problem.
36. ONM (juniori) - 2010 Iasi, problema 4.
35. Nice problem from Arqady (Eduard_Tur).
34. IMO Shortlist 2002 geometry problem 7.
+ April 2010
32. Linkuri diverse.
31.Some nice metrical problems.
29. Inegalitate integrala de la Kunny.
28. A fost odată ...
27.Problem of Darij Grinberg (I - centroid of triangle DEF).
26. Outside of a triangle.
25. Geometric equations.
24. A fine and cool inequalities.
23. A very nice exponential inequality (own - from CRUX).
22. Some interesting limits (sequence/function).
21 bis. Some problems with complex numbers.
21. Probleme de geometrie (Yetty & I).
20. O ineg.cu suma de AG/AI .
19. Own proposed problems in CRUX Mathematicorum - Canada.
18. O problema a lui Toshio Seimiya.
17. O inegalitate "tare" intr-un triunghi.
16. Length of the (interior) angle bisector (10 proofs).
15. Inequalities stronger than Panaitopol's.
14. Opinii si comentarii relativ la inv. rom.
13. Inegalitatea de la Sorin Borodi.
12. Inegalitatea I. Maftei & S. Radulescu (2).
11. Inegalitatea I. Maftei & S. Radulescu (1).
10. Walker's inequality and its extension.
9. Inegalitatea HO+HA+HB+HC >= 3R .
8. The first problem Shortlist ONM 2010 Romania.
7. Some interesting "slicing" problems.
6. Interesting problems from JBMO.
5. Theoretical preliminary for inequalities.
4. Remarkable geometrical inequalities (2).
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