298. Some nice problems from Crux Mathematicorum.

by Virgil Nicula, Jul 17, 2011, 1:43 AM

Proposed problem 2011. Let $\triangle ABC$ with the incircle $w=C(I,r)$ . Let $\begin{array}{c} D\in BI\cap AC\\\\ E\in CI\cap AB\end{array}$ and $\begin{array}{c} P\in DE\ ;\ IP\perp DE\\\\ Q\in IP\cap BC\end{array}$ . Prove that $IQ=2\cdot IP\implies A=60^{\circ}$ .

Proof 1 (trigonometric). Denote $\left\{\begin{array}{c} x=m\left(\widehat{IDE}\right)\\\\ y=m\left(\widehat{IED}\right)\\\\ z=x+y=\frac {B+C}{2}\end{array}\right\|$ . Thus, $\left\{\begin{array}{c} m\left(\widehat{AEI}\right)=z+\frac B2\\\\ m\left(\widehat{BQI}\right)=90^{\circ}+x-\frac B2\end{array}\right\|$ . Therefore, $IQ=2\cdot IP\iff$ $\frac {r}{IE}=2\cdot\frac {IP}{IE}\cdot\frac {r}{IQ}\iff$

$\sin\left(z+\frac B2\right)=2\sin y\cos\left(\frac B2-x\right)\iff$ $\sin\left(z+\frac B2\right)=\sin \left(\frac B2+y-x\right)+\sin\left(x+y-\frac B2\right)=$

$\sin \left(\frac B2+y-x\right)+\sin\left(z-\frac B2\right)\iff$ $\sin \left(\frac B2+y-x\right)=\sin\left(z+\frac B2\right)-\sin\left(z-\frac B2\right)=$ $2\cos z\sin\frac B2\iff$

$\boxed{\sin\frac B2\cos(y-x)+\cos\frac B2\sin(y-x)=2\cos z\sin\frac B2}\ (3)$ . Similarly obtain that $\boxed{\sin\frac C2\cos(y-x)-\cos\frac C2\sin(y-x)=2\cos z\sin\frac C2}\ (4)$ .

Multiply the relation $(3)$ by $\sin\frac C2$ and the relation $(4)$ by $\sin\frac B2$ we get $\left(\sin \frac C2\cos\frac B2+\cos\frac C2\sin\frac B2\right)\sin(y-x)=0\iff$

$\sin\frac {B+C}{2}\sin(y-x)=0\iff$ $\sin z\sin(y-x)=0\iff$ $y=x\stackrel{(3)}{\iff}$ $\sin\frac B2=2\cos z\sin\frac B2\iff$ $\cos z=\frac 12\iff$ $z=60^{\circ}\iff$ $A=60^{\circ}$ .

Remark. The relations $(3)$ , $(4)$ can write thus $:\left\{\begin{array}{c} \tan\frac B2\cdot (1+\tan x\tan y)+ (\tan y-\tan x)=2(1-\tan x\tan y)\cdot \tan\frac B2\\\\ \tan\frac C2\cdot (1+\tan x\tan y)+\cdot (\tan x-\tan y)=2(1-\tan x\tan y)\cdot\tan\frac C2\end{array}\right\|$ $\iff$

$\tan\frac B2=-\tan C2=\frac {\tan y-\tan x}{1-3\tan x\tan y}\iff$ $\left\{\begin{array}{c} \tan x=\tan y\\\\ \tan x\tan y=\frac 13\end{array}\right\|$ $\iff$ $x=y=\frac {\pi}{6}\iff$ $B+C=120^{\circ}\iff$ $A=60^{\circ}$ .

Proof 2 (metric). Observe that $IP=ID\cdot\sin x$ . Apply the Sinus' theorem in $\triangle IBQ\ :\ \frac {IQ}{\sin\frac B2}=\frac {IB}{\sin\left(\frac B2+90^{\circ}-x\right)}\iff$ $IQ=\frac {IB\cdot\sin\frac B2}{\cos\left(x-\frac B2\right)}$ .

Therefore, $IQ=2\cdot IP\iff$ $2\sin x\cos\left(x-\frac B2\right)=\frac {IB}{ID}\cdot \sin\frac B2\iff$ $\sin\left(2x-\frac B2\right)+\sin \frac B2=\frac {a+c}{b}\cdot\sin\frac B2$ . Since $\frac {a+c}{b}\cdot\sin\frac B2=$

$\frac {(\sin A+\sin C)\cdot\sin\frac B2}{\sin B}=$ $\frac {2\sin\frac {A+C}{2}\cos\frac {A-C}{2}\sin\frac B2}{2\sin \frac B2\cos\frac B2}=\cos\frac {A-C}{2}$ , obtain that $\sin\left(2x-\frac B2\right)+\sin \frac B2=\cos\frac {A-C}{2}\iff$ $\sin\left(2x-\frac B2\right)=$

$\cos\frac {A-C}{2}-\cos \frac {A+C}{2}\implies$ $\boxed{\sin\left(2x-\frac B2\right)=2\sin \frac A2\sin \frac C2}$ . Analogously obtain that $\boxed{\sin\left(2y-\frac C2\right)=2\sin \frac A2\sin \frac B2}$ .

In conclusion, $\sin\left(2x-\frac B2\right)\sin\frac B2=\sin\left(2x-\frac C2\right)\sin\frac C2\iff$ $\cos (2x-C)-\cos 2x=\cos (2y-B)-\cos 2y\iff$

$\cos (2x-C)-\cos (2y-B)=\cos 2x-\cos 2y\stackrel{(2x-C)+(2x-B)=0}{\iff}$ $\cos 2x=\cos 2y\iff$ $x=y=\frac {\pi}{6}\iff$ $B+C=120^{\circ}\iff$ $A=60^{\circ}$ .


Proposed problem 2153. Suppose that $\{a, b, c\}\subset \mathbb R$ . If for all $x\in [-1,1]$ we have $\left|ax^2+bx+c\right|\le 1$ , then prove that $\left|cx^2 + bx + a\right|\le 2$ .

Proof. Let $f(x) = ax^2+bx+c$ and $g(x) = cx^2+bx+a$ . Then by assumption $|a + b + c|= |f(1)|\le 1$ , $|a-b+c|= |f(-1)|\le 1$

and $|c|=|f(0)|\le 1$ . Hence $|g(x)| =$ $\left|c\cdot\left(x^2-1\right)+(a+b+c)\cdot\frac {1+x}{2}+(a-b+c)\cdot \frac{1-x}{2}\right|\le $ $\left|x^2-1\right|+$ $\frac {|1+x|}{2}+$ $\frac {|1-x|}{2}=$

$1-x^2+$ $\frac {1+x}{2}+$ $\frac {1-x}{2}=$ $2-x^2\le 2$ for all $x\in [-1,1]$ .


Proposed problem 2151. $\triangle ABC$ is a triangle with $B=2C$ . Let $H$ be the foot of the perpendicular from $A$

to $BC$ and let $D$ be the point on the side $[BC]$ where the excircle touches $BC$ . Prove that $AC=2\cdot HD$ .

Proof 1 (synthetic). Since $b+CD=c+BD=s$ obtain that $CD=\frac {a+c-b}{2}$ . Construct the point $E\in BC$ so that $B\in (EC)$ and $BE=BA$ . Observe that

$B=2C\implies$ the triangle $AEC$ is $A$-isosceles and $HC=HE=\frac {a+c}{2}$ . Thus, $HD=HC-DC=\frac {a+c}{2}-\frac {a+c-b}{2}=\frac b2$ , i.e. $AC=2\cdot HD$ .

Proof 2 (metric). From an well-known (or prove easily) property $\boxed{B=2C\iff b^2=c(c+a)}\ (*)$ , $HC=b\cdot\cos C\ (C<60^{\circ})$

and $CD=s-b$ obtain that $HD=HC-CD=$ $b\cdot\cos C-(s-b)=$ $\frac {a^2+b^2-c^2-a(a+c-b)}{2a}=$ $\frac {s(b-c)}{a}$ .

Thus, $AC=2\cdot HD\iff$ $ab=(b-c)(a+b+c)\iff$ $ab=a(b-c)+b^2-c^2=$ $b^2=c(c+a)\iff$ $B=2C$ .

Here is a proof of the well-known property $(*)$ . Denote $F\in AC$ for[/b] which $\widehat{FBA}\equiv\widehat{FBC}$ , i.e. $\frac {FA}{c}=\frac {FC}{a}=\frac {b}{a+c}$ . Therefore, $FA=\frac {bc}{a+c}$

and $\triangle AFB\sim\triangle ABC\iff$ $\frac {AF}{AB}=\frac {AB}{AC}\iff$ $c^2=\frac {b^2c}{a+c}\iff$ $b^2=c(c+a)$ . Otherwise. Construct the point $E\in BC$ so that $B\in (EC)$

and $BE=BA$ . Observe that $BE=c$ , $AE=b$ , $EC=c+a$ and $\triangle EAB\sim\triangle ECA\iff$ $\frac {EA}{EC}=\frac {AB}{CA}\iff$ $b^2=c(c+a)$ .


Proposed problem 3009. Let $ABC$ be a triangle with the incircle $I$ . Denote $\left\{\begin{array}{c} E\in BI\cap AC\\\ F\in CI\cap AB\end{array}\right\|$ . Prove that

$AE\cdot AF$ is greater than, equal to or less than $AI^2$ according is $A$ greater than, equal to or less than $90^{\circ}$ .

Proof. $AE\cdot AF-AI^2=\frac {bc}{a+c}\cdot\frac {bc}{a+b}-$ $\frac {bc(s-a)}{s}\ \boxed{.s.s.}\ \ \frac {bc}{(a+b)(a+c)}-1$ $+\frac as\ \boxed{.s.s.}\ \ \frac as-\frac {a^2+a(b+c)}{(a+b)(a+c)}\ \boxed{.s.s.}$

$\frac 1s-$ $\frac {a+b+c}{(a+b)(a+c)}\ \boxed{.s.s.}\ \ 2(a+b)(a+c)$ $-(a+b+c)^2\ \boxed{.s.s.}\ \ a^2$ $-\left(b^2+c^2\right)$ , where $X\ \boxed{.s.s.}\ \ Y\ \iff\ X=Y=0\ \vee\ XY>0$ .

For example, $AE\cdot AF\ >\ AI^2\ \iff\ a^2\ >\ b^2\ +\ c^2\ \iff\ A\ >\ 90^{\circ}$ .
This post has been edited 78 times. Last edited by Virgil Nicula, Nov 21, 2015, 7:51 AM

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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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    by MrMustache, Nov 11, 2020, 4:52 PM

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    by piphi, Jun 10, 2020, 11:44 PM

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  • Total entries: 456
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