225. An interesting problems from Kunny.

by Virgil Nicula, Feb 18, 2011, 4:30 PM

Proposed problem 1. Let $P$ , $Q$ be two points on the circle $x^2+y^2=1$ . Two points $P$ , $Q$ move in the first quadrant, the second quadrant respectively

so that $OP\perp OQ$ . Consider the triangle enclosed the tangent lines of $C$ at the points $P$ , $Q$ and the $x$-axis. Find the minimum area of this triangle.

Proof. Denote the intersections $R\in PP\cap QQ$ , $M\in PP\cap Ox$ , $N\in QQ\cap Ox$ and $PM=x$ ,

$QN=y$ . Observe that $\frac {PO}{RN}=$ $\frac {MO}{MN}$ $=\frac {MP}{MR}$ $\implies$ $\frac {1}{1+y}=\frac {x}{x+1}\implies$ $\boxed {\ xy=1\ }\ (1)$ . Therefore,

$[MRN]$ is minimum $\Longleftrightarrow$ $[OPM]+[OQN]$ is minimum $\Longleftrightarrow$ $x+y$ is minimum $\stackrel{(1)}{\Longleftrightarrow}$ $x=y$ .

Extension. Let $P$ , $Q$ be two points on the circle $x^2+y^2=1$ . Two points $P$ , $Q$ move in the first quadrant,

the second quadrant respectively so that $m\left(\widehat{PRN}\right)=2\phi$ is constant and $\phi\le 45^{\circ}$ . Consider the triangle

enclosed the tangent lines of $C$ at the points $P$ , $Q$ and the $x$-axis. Find the minimum area of this triangle.

Method 1. Denote $R\in PP\cap QQ$ , $M\in PP\cap Ox$ , $N\in QQ\cap Ox$ and $u=m(\widehat {POM})$ , $v=\widehat{QON}$ . Thus $u+v=2\phi$ ,

$PM=\tan u$ and $QN=\tan v$ . Therefore, $[MRN]$ - min. $\Longleftrightarrow$ $[OPM]+[OQN]$ - min. $\Longleftrightarrow$ $\tan u+\tan v$ - minimum.

Since $\left\{u\ ,\ v\right\}\subset \left[0\ ,\ 90^{\circ}\right)$ obtain $\tan u+\tan v\ge $ $2\cdot\tan\frac {u+v}{2}=$ $2\cdot\tan\phi$ with equality iff $u=v=\phi$ . In conclusion

the area of the triangle $MRN$ is minimum $\Longleftrightarrow$ $u=v=\phi$ $\Longleftrightarrow$ $OR\perp MN$ $\Longleftrightarrow$ the triangle $MRN$ is isosceles.

Remark. I used the well-known inequality $\boxed {\ \{\ u\ ,\ v\ \}\subset \left[0\ ,\ 90^{\circ}\right)\ \implies\ \tan u+\tan v\ge 2\cdot\tan\frac {u+v}{2}\ }\ (*)$ .

Indeed, denote $\tan \frac u2=p<1$ , $\tan \frac v2=q<1$ and the inequality $(*)$ becomes $\frac {2p}{1-p^2}+\frac {2q}{1-q^2}\ge\frac {p+q}{1-pq}$ $\Longleftrightarrow$

$(1-pq)^2(p+q)\ge (p+q)\left(1-p^2\right)\left(1-q^2\right)$ $\Longleftrightarrow$ $(1-pq)^2\ge (1-p^2)(1-q^2)$ $\Longleftrightarrow$ $p^2+q^2\ge 2pq$ .

Method 2. Denote the intersections $R\in PP\cap QQ$ , $M\in PP\cap Ox$ , $N\in QQ\cap Ox$ and $RM=x$ , $RN=y$ ,

$RP=RQ=k\ge 1$ . Observe that $xy\sin 2\phi=2\cdot [MRN]=x+y$ $\Longleftrightarrow$ $\boxed {\ \frac 1x+\frac 1y=\sin 2\phi\ }\ (2)$ . Since

$(x+y)\left(\frac 1x+\frac 1y\right)\ge 4$ obtain $x+y\ge \frac {4}{\sin 2\phi}$ . Prove easily that if $\triangle MRN$ is isosceles, then $x_0=y_0=\frac {1}{\sin\phi\cos\phi}$

and in this case $x_0+y_0=\frac {4}{\sin 2\phi}$ . In conclusion $\min\ [MRN]=\frac {2}{\sin 2\phi}$ when the triangle $MRN$ is isosceles.


Proposed problem 2. Prove that the equation of the parabola which touches the $x$-axis and $y$-axis at

$A(a,0)$ and $B(0,b)$ , where $a>0$ , $b>0$ is $\boxed{(bx-ay)^2-2ab(bx+ay)+a^2b^2=0}\ (*)$ .

Proof. Consider that the equation of the required parabola $\mathbb P$ is $p(x,y)\equiv Ax^2+2Bxy+Cy^2+2Dx+2Ey+F=0$ , where $\underline{B^2=AC}$ . Therefore,

$2Ax+2B(y+xy')+2Cyy'+2D+2Ey'=0\implies$ the slope of the tangent line to $\mathbb P$ in the point $T(u,v)\in \mathbb P$ is $y'(u,v)=-\frac {Au+Bv+D}{Bu+Cv+E}$ .

$\blacktriangleright\ p(a,0)=0$ and $y'(a,0)=0\ \implies\ \underline{Aa^2+2Da+F=0}$ and $\underline{Aa+D=0}\ \implies\ D=-Aa$ and $F=Aa^2$ .

$\blacktriangleright\ p(0,b)=0$ and $y'(0,b)=\infty\ \implies\ \underline{Cb^2+2Eb+F=0}$ and $\underline{Cb+E=0}\ \implies\ E=-Cb$ and $F=Cb^2$ .

Choose w.l.o.g. $\boxed{A=b^2}$ . Obtain that $\boxed{C=a^2}$ , $\boxed{F=a^2b^2}$ , $\boxed{D=-b^2a}$ , $\boxed{E=-ba^2}$ and $B^2=AC\implies B=\epsilon\cdot ab$ , where

$\epsilon\in\{\pm 1\}$ . If $\epsilon =1$ , then the equation of $\mathbb P$ becomes $(bx+ay)^2-2ab(bx+ay)+a^2b^2=0$ , i.e. $(bx+ay-ab)^2=0$ , what is the

line $\frac xa+\frac yb=1$ , absurd. In conclusion, $\boxed{B=-ab}$ and the equation of the required parabola is $(bx-ay)^2-2ab(bx+ay)+a^2b^2=0$ .

Remark. The equation $(*)$ of the parabola $\mathbb P$ is equivalently with the equation $(bx+ay-ab)^2=4abxy$ .


Proposed problem 3. For positive real numbers $a\ ,\ b\ ,\ c$ prove that $\frac{b^2+c^2}{a}+\frac{c^2+a^2}{b}+\frac{a^2+b^2}{c}\ge 2(a+b+c)$ .

Proof. $\left\|\begin{array}{c} \frac {b^2}{a}\ge 2b-a\\\\ \frac {c^2}{a}\ge 2c-a\end{array}\right\|\ \bigoplus\ \implies \frac {b^2+c^2}{a}\ge 2(b+c-a)$ . In conclusion, $\sum\frac {b^2+c^2}{a}\ge 2(a+b+c)$ .


Proposed problem 4. For positive real numbers $a\ ,\ b\ ,\ c$ prove that $\frac{b^3+c^3}{a}+\frac{c^3+a^3}{b}+\frac{a^3+b^3}{c}\ge 6(ab+bc+ca)-4\left(a^2+b^2+c^2\right)$ .

Proof. $t>0\implies t^3+2=t^3+1+1\ge 3\sqrt [3]{t^3\cdot 1\cdot 1}=3t\implies t^3\ge 3t-2$ . So $\left\|\begin{array}{ccc} t:=\frac ba & \implies & \frac {b^3}{a}\ge 3ab-2a^2\\\\ t:=\frac ca & \implies & \frac {c^3}{a}\ge 3ac-2a^2\end{array}\right\|\ \bigoplus\ \implies$

$ \frac {b^3+c^3}{a}\ge 3a(b+c)-4a^2$ a.s.o. In conclusion, $\frac{b^3+c^3}{a}+\frac{c^3+a^3}{b}+\frac{a^3+b^3}{c}\ge 6(ab+bc+ca)-4\left(a^2+b^2+c^2\right)$ .
This post has been edited 37 times. Last edited by Virgil Nicula, Nov 22, 2015, 2:53 PM

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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 tigerzhang, Aug 1, 2021, 12:02 AM

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    by samrocksnature, Apr 14, 2021, 5:16 AM

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    by the_mathmagician, Jan 17, 2021, 7:43 PM

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    by OlympusHero, Dec 30, 2020, 6:08 PM

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

  • 372554 views!

    by mrmath0720, Sep 28, 2020, 1:11 AM

  • wow... i am lost.

    369302 views!

    -piphi

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

  • That was a lot! But, really good solutions and format! Nice blog!!!! :)

    by CSPAL, May 27, 2020, 4:17 PM

  • impressive :D
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    by OlympusHero, May 14, 2020, 8:43 PM

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  • Joined: Jun 22, 2005
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  • Total entries: 456
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