221. Some nice and simple "slicing" problems.

by Virgil Nicula, Feb 15, 2011, 2:04 PM

PP1. Let $ABC$ be an $A$-isosceles triangle with $A=108^{\circ}$ . Denote $\left\|\begin{array}{ccc} D\in (BC) & ; & DB=DC\\\ E\in (CA) & ; & \widehat{EBA}\equiv\widehat{EBC}\end{array}\right\|$ . Prove that $BE=2\cdot AD$ .

Proof 1 (trig.). $\frac {BE}{AD}=$ $\frac {BE}{AB}\cdot\frac {AB}{AD}=$ $\frac {\sin\widehat{BAE}}{\sin \widehat{BEA}}\cdot\frac {1}{\cos\widehat{BAD}}=$ $\frac {\sin 2x}{\sin\left(45^{\circ}+\frac {3x}{2}\right)}\cdot\frac {1}{\cos x}=$ $2\cdot\frac {\sin x}{\sin\left(45^{\circ}+\frac {3x}{2}\right)}$ . For $x=54^{\circ}$ obtain that $\frac {BE}{AD}=2$ .

Proof 2 (syn.). Denote $F\in AD$ so that $\left\|\begin{array}{c} D\in (AF)\\\ DF=DA\end{array}\right\|$ . Prove easily that $ABFE$ is a isosceles trapezoid for which $BF\parallel AE$ . In conclusion, $BE=2\cdot AD$ .


PP2. Let $ABC$ be a triangle. Denote $D\in (BC)$ so that $\widehat{DAB}\equiv\widehat{DAC}$ . Suppose that $DB\cdot DC=AD^2$ and $m\left(\widehat{ADB}\right)=45^{\circ}$ . Ascertain $A$ , $B$ , $C$ .

Proof 1 (trigonometric). Denote $m\left(\widehat{DAB}\right)=m\left(\widehat{DAC}\right)=x$ . Thus, $B=135^{\circ}-x$ and $C=45^{\circ}-x$ . Apply the Sinus' theorem

in the triangles : $\left\|\begin{array}{ccc} \triangle DAB & ; & \frac {DA}{\sin\left(135^{\circ}-x\right)}=\frac {DB}{\sin x}\\\\ \triangle DAC & ; & \frac{DA}{\sin \left(45^{\circ}-x\right)}=\frac{DC}{sinx}\end{array}\right\|\ \bigodot\implies$ $\sin^{2}x=\sin\left(45^{\circ}-x\right) \cdot \sin\left(45^{\circ}+x\right)$ . I used the given relation

$DA^2=DB \cdot DC$ . Therefore, $1-\cos 2x=\cos 2x \iff$ $\cos 2x=\frac 12$ $\iff$ $x=30^{\circ}$ . In conclusion, $A=60^{\circ}$ , $B=105^{\circ}$ and $C=15^{\circ}$ .

Remark. Can use the well-known relation $\boxed{\ AD^2=bc-DB\cdot DC\ }$ .

Proof 2 (metrico-synthetic). Denote the midpoint $M$ of the side $[BC]$ and the diameter $[EF]$ in the circumcircle $w=C(O,R)$ of $\triangle ABC$

so that $M\in EF$ and the sideline $BC$ separates $E$ , $A$ . Using the power of $D$ w.r.t. $w$ obtain that $AD^2=DB\cdot DC=DA\cdot DE$ , i.e.

$DA=DE$ $\iff$ $OD\parallel AF$ , i.e. $M$ is the midpoint of $[OE]$ $\iff$ the triangle $BFC$ is equilateral $\implies$ $A=60^{\circ}$ , $B=105^{\circ}$ and $C=15^{\circ}$ .


PP3. Let $ABC$ be an equilateral triangle. Consider the points $\left\|\begin{array}{ccc} D\in (BC) & ; & DC=2\cdot DB\\\ E\in (CA) & ; & EC=EA\\\ F\in (AB) & ; & FA=3\cdot FB\end{array}\right\|$ . Ascertain $m(\angle DEF)$ .

Proof 1 (metric). Suppose w.l.o.g. that $AB=12$ . Thus, $\left\|\begin{array}{ccc} DB=8 & ; & DC=4\\\ EC=6 & ; & EA=6\\\ FA=9 & ; & FB=3\end{array}\right\|$ and using the generalized Pytagoras' theorem obtain that $ED=2\sqrt 7$ , $EF=3\sqrt 7$ and $DF=7$ . Can suppose w.l.o.g. again that $ED=2$ , $EF=3$ and $DF=\sqrt 7$ . In conclusion, using again the generalized Pytagoras' theorem

in $\triangle DEF$ for the side $[DF]$ obtain that $\cos\widehat{DEF}=\frac {ED^2+EF^2-DF^2}{2\cdot ED\cdot EF}=$ $\frac {4+9-7}{2\cdot 2\cdot 3}=\frac 12$ , i.e. $m\left(\widehat{DEF}\right)=60^{\circ}$ .

Proof 2 (synthetic - Sunken rock). Denote the point $M$ for which $MCAB$ is a rhombus. Observe that $M\in ED$ because $\frac {DB}{DC}=\frac {MB}{CE}=2$ .

Denote the intersection $N\in EF\cap BC$ and the midpoint $P$ of $[AB]$ . Show easily that $BEPN$ is a parallelogram $\implies$ $BN=CE\implies$

$\triangle MBN\equiv \triangle MCE$ and a $60^\circ$ rotation around $M$ will map $C$ to $B$ and $E$ to $N$, i.e. $\triangle MEN$ is equilateral $\implies$ $m\left(\widehat{DEF}\right)=60^{\circ}$ .


PP4. Let $ABC$ be an $A$-isosceles triangle with $A=100^\circ$ . Denote $\left\|\begin{array}{ccc} D\in (BC) & ; & m\left(\widehat{DAC}\right)=40^\circ\\\ E\in (AB) & ; & m(\widehat{BCE})=10^\circ\end{array}\right\|$ . Find $m\left(\widehat{DEC}\right)$ .

Proof. Note that $m(\angle BAD)=60^{\circ}$ $\implies$ $E'\in (AB)$ such that $\triangle ADE'$ is equilateral. Note that $m(\angle ACD)=m(\angle DAC)=$ $40^{\circ}\implies DA=DC$ .

Hence $CD=AD=DE'$ . Since we have $m(\angle CDE')=160^{\circ}$ , then $m(\angle DCE')=m(\angle DE'C)=10^{\circ}$ $\implies$ $E'=E$. So $m(\angle DEC)=10^{\circ}$ .


PP5. Let $ ABCD$ be a quadrilateral for which $ BA=BD$ , $ AC=AD$ , $ m(\angle BAC)<30^{\circ}$ and $ m(\angle BAC) +m(\angle BDA) = 60^{\circ}$ . Find $ m(\angle BCA)$ .

Proof. Denote $ x=m(\angle BAC)$ . Observe that $ \underline {BA=BD}$ and $ \underline {m(\angle BAC) +m(\angle BDA) = 60^{\circ}}$ $ \implies$ $ m(\angle BAD)=m(\angle BDA)=60^{\circ}-x$ ,

$ m(\angle CAD)=60-2x$ and $ m(\angle DBA)=60+2x$ . Construct the equilateral $ \triangle ABE$ so that $ AD$ separates $ B$ , $ E$ . Observe that $ m(\angle DBE)=2x$ ,

$ m(\angle DAE)=x$ , $\triangle DBE$ is $ B$-isosceles, i.e. $ m(\angle BED)=m(\angle BDE)=90^{\circ}-x$ and $ \boxed {\ m(\angle ADE)=30^{\circ}\ }$ . Observe that $ ABC\equiv AED$ because

$ AB=AE$ , $ \underline{AC=AD}$ and $ m(\angle BAC)=m(\angle EAD)=x$ . Hence $ BC=ED$ and $ m(\angle BCA)=m(\angle EDA)=30^{\circ}$ , i.e. $ m(\angle BCA)=30^{\circ}$ .

Particular case. For $ m(\angle BAC)=10^{\circ}$ we"ll obtain an well-known problem.


PP6. Se considera doua triunghiuri isoscele $ ABC$ , $ MNP$ astfel incat $ AB=AC$ , $ MN=MP$ si $ N\in (AC)$ , $ B\in (MP)$ .

Notam $ R\in AB\cap MN$ , $ S\in BC\cap NP$ . Sa se arate ca $ BP=CN\ \Longleftrightarrow$ semidreapta $ [RS$ este bisectoarea unghiului $ \widehat {BRN}$ .

Dem. Notam $ X\in BC\cap MN$ , $ Y\in AB\cap NP$ , $ Z\in NB\cap RS$ . Aplicam teorema Menelaus transversalelor $\overline{XBC}/\triangle ANR$ , $ \overline{YPN}/\triangle BMR$ respectiv:

$ \left|\begin{array}{ccc} \frac {XR}{XN}\cdot\frac {CN}{CA}\cdot\frac {BA}{BR}=1 & \implies & \frac {XN}{XR}=\frac {CN}{BR}\\\\ \frac {YR}{YB}\cdot\frac {PB}{PM}\cdot\frac {NM}{NR}=1 & \implies & \frac {YR}{YB}=\frac {NR}{PB}\end{array}\right|$ . Aplicam teorema lui Ceva punctului $ S$ si triunghiului $ BRN\ : \ \ \frac {TB}{TN}\cdot\frac {XN}{XR}\cdot\frac {YR}{YB}=1$ $ \implies$

$ \frac {TN}{TB}=\frac {XN}{XR}\cdot\frac {YR}{YB}$ $ \implies$ $ \frac {TN}{TB}=\frac {NC}{PB}\cdot\frac {RN}{RB}$ . In concluzie, $ NC=PB$ $ \Longleftrightarrow$ $ \frac {TN}{TB}=\frac {RN}{RB}$ $ \Longleftrightarrow$ semidreapta $ [RS$ este bisectoarea unghiului $ \widehat {BRN}$ .


PP7. Let $ ABC$ be an $ A$-isosceles triangle. Let $ M\in BC$ , $ N\in (AC)$ , $ P\in (AB)$

so that $ AN=BP$ and $ MN=MP$ . Prove that $ \triangle\ MNP\ \sim\ \triangle\ ABC$ .

Proof. $ \left\|\begin{array}{c} R\in (AB)\ ,\ AR=BP\\\\ X\in (NR)\ ,\ XN=XR\\\\ T\in (NP)\ ,\ TN=TP\\\\ S\in (BC)\ ,\ PS\perp BC\end{array}\right\|\implies\ NR\parallel BC$ , $ AX=PS$ , $ AX\parallel PS$ , $ XT\parallel AB$ ;

$ APSX$ is a parallelogram $ \implies XS\parallel AB$ ; $ XT\parallel AB\ \implies T\in XS$ ; $ TM\perp PN\ \implies$

$ MSPT$ is cyclically$ \implies\widehat {MNP}\equiv\widehat {MPT}\equiv\widehat {MST}\equiv\widehat{MBA}$ . In conclusion, $ m(\widehat {MNP})=B$ .


PP8. Let $ABC$ be a triangle with an acute $\angle A$ . Denote angle bisector $AE$ and the altitude

$BH$ , where $E\in BC$ , $H\in AC$ . Suppose $m(\angle AEB)=45^{\circ}$ . Ascertain $m(\angle EHC)$ .

Proof. Denote $D\in AB$ for which $CD\perp AB$ and $x=m(\angle BAE)$ . Observe that $m(\angle BCH)=45^\circ-x$ , $m(\angle CBD)=45^\circ+x$

and $m(\angle BCD)=45^\circ-x$ . Thus, $\widehat {BCH}\equiv\widehat {BCD}$ , i.e. the lines $AE$ and $CE$ are angle bisectors. So the line $DE$ is also an angle bisector

of $\widehat{ADC}$ . From $\triangle BCH\equiv\triangle BCD$ obtain that $m(\angle EHC)=m(\angle EDC)=45^\circ$ .


An easy extension. Consider in the triangle $ABC$ two points $E\in (BC)$ and $H\in (AC)$

so that $\widehat{EAB}\equiv\widehat{EAC}$ and $m(\angle AHB)=2\cdot m(\angle AEB)=2\phi$ . Ascertain $m(\angle EHC)$ .

Proof 1. Denote $x=m(\angle BAE)=m(\angle CAE)$ and $m(\angle AEB)=\phi$ . Observe that $m(\angle AHB)=2\phi$ , $m(\angle BCA)=\phi -x$ and

$m(\angle CBH)=\phi +x$ , $m(\angle ADC)=180^{\circ}-2\phi $. Denote $D\in AB$ so that $B\in (AD)$ and $\widehat{BCA}\equiv\widehat{BCD}$ . Thus $E$ is the incenter of

$\triangle ACD$ . So the line $DE$ is also an angle bisector of $\widehat{ADC}$ . From $\triangle BCH\equiv\triangle BCD$ obtain that $m(\angle EHC)=m(\angle EDC)=90^\circ -\phi$ .

Remark. Since $\phi =\frac A2+C$ obtain that $m(\angle EHC)=m(\angle EDC)=90^\circ -\phi =$ $90^\circ -\left(\frac A2+C\right)\implies$ $\boxed{m(\angle EHC)=\frac {B-C}{2}}$ .

Proof 2. Let $I$ be the incenter of $\triangle AHB$ . Observe that $\angle EAB={A\over 2}\implies \phi=\pi-B-{A\over 2}={A\over 2}+C$ and $\angle AHI=\angle BHI=$ ${A\over 2}+C\implies$

$\angle BHC=$ $\pi-A-2C=B-C$ . Thus, $\angle HBC=\pi-\angle BHC-C=$ $\pi-B\implies $ $\angle ABI=$ $\angle HBI=$ $B-{\pi\over 2}$ . Hence $\angle IBE={\pi\over 2}$

and $\angle IHB=\angle IEB$ $\implies$ the quadrilateral $IHEB$ is cyclic. Hence $\angle IHE={\pi\over 2}$ $\implies $ $\angle EHC=$ $\pi-{\pi\over 2}-\phi=$ ${\pi\over 2}-\phi$ $\left(={B-C\over 2}\right)$ .
This post has been edited 70 times. Last edited by Virgil Nicula, Nov 22, 2015, 3:10 PM

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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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