109. Two equal areas ==> find A.

by Virgil Nicula, Sep 11, 2010, 5:08 PM

Quote:
Proposed problem. Let $ABC$ be a triangle with the incenter $I$ . Denote the midpoints $M , N$ of the sides $[AB]$ , $[AC]$ respectively
and the intersections $X\in MI\cap AC$ , $Y\in NI\cap AB$ . Prove that $k\cdot [XAY]=[ABC]$ $\iff$ $\cos A=1-\frac k2$ .
I will solve this interesting problem with the following very usual lemma (you can prove easily).

Lemma. Given are $\triangle ABC$ and $\left|\begin{array}{cc}
D\in BC & U\in AB\\\
V\in AC & W\in AD\end{array}\right|$. Then $\boxed {\ W\in UV\Longleftrightarrow \frac{\overline {UB}}{\overline {UA}}\cdot \overline {DC}+\frac{\overline {VC}}{\overline {VA}}\cdot \overline {BD}=\frac{\overline {WD}}{\overline {WA}}\cdot \overline {BC}\ }\ (*)$ .

Remark. For $W:=I$ the remarkable relation $(*)$ becomes $\boxed {\ I\in UV\Longleftrightarrow a+b\cdot \frac{\overline {UB}}{\overline {UA}}+c\cdot \frac{\overline {VC}}{\overline {VA}}=0\ }$ .


Proof of the proposed problem. Observe that $\frac {\overline{BD}}{c}=\frac {\overline{DC}}{b}=\frac {\overline{BC}}{b+c}$ . Using this lemma for $W: =I$, we will ascertain the ratios $\frac{\overline {XA}}{\overline {CA}}$ and $\frac{\overline {YA}}{\overline {BA}}\ :$

$\overline {MIX}\ \blacktriangleright\ a-b+c\cdot \frac{\overline {XC}}{\overline {XA}}=0\iff$ $\frac {\overline{XC}}{b-a}=\frac {\overline{XA}}{c}=\frac {\overline{AC}}{b-a-c}$ $\implies\frac {\overline{AX}}{\overline{AC}}=\frac {c}{a+c-b}$ .

$\overline {NIY}\ \blacktriangleright\ a-c+b\cdot \frac{\overline {YB}}{\overline {YA}}=0\iff$ $\frac {\overline{YB}}{c-a}=\frac {\overline{YA}}b=\frac {\overline{AB}}{c-a-b}\implies\frac {\overline{AY}}{\overline{AB}}=\frac {b}{a+b-c}$ .

Denote $k\cdot [AXY]=[ABC],\ k\in \overline {1,3}$ . Thus, $k\cdot \frac {\overline {AX}}{\overline {AC}}\cdot \frac {\overline{AY}}{\overline{AB}}=1$ $\iff$ $kbc=(a+c-b)(a+b-c)$ $\iff$

$a^2=(b-c)^2+kbc$ $\iff$ $kbc=(b^2+c^2-2bc\cdot\cos A)-(b-c)^2$ $\iff$ $\cos A=1-\frac k2$ . In conclusion,

$\boxed {\ k\cdot [AXY]=[ABC]\ \iff\ \cos A=1-\frac k2\ }$ . In particular, $\left\|\begin{array}{ccc}
k=1 & \iff &  A=60^{\circ}\\\
k=2  & \iff &  A=90^{\circ}\\\
k=3 & \iff & A=120^{\circ}\end{array}\right\|$ .

Quote:
An easy extension. Let $P$ be a point with the barycentrical coordinates $(x,y,z)$ w.r.t. $\triangle ABC$ . Denote the midpoints $M , N$

of $[AB]$ , $[AC]$ respectively and $X\in MP\cap AC$ , $Y\in NP\cap AB$ . Prove that $[XAY]=\frac {yz}{x^2-(y-z)^2}\cdot [ABC]$ .

In particular, $[XAY]=[ABC]\ \iff\ x^2=y^2+z^2-yz$ .
Quote:
Proposed problem. Solve in $\mathbb N$ the equation $m^2=n^2+p^2-np$ , i.e. find a triangle $ABC$ with $A=60^{\circ}$ and $\{a,b,c\}\subset\mathbb N$ .
This post has been edited 29 times. Last edited by Virgil Nicula, Nov 23, 2015, 8:12 AM

Comment

0 Comments

Own problems or extensions/generalizations of some problems which was posted here.

avatar

Virgil Nicula
Archives
+ October 2017
+ September 2017
+ December 2016
+ October 2016
+ February 2016
+ September 2013
+ October 2010
+ September 2010
Shouts
Submit
  • orzzzzzzzzz

    by mathMagicOPS, Jan 9, 2025, 3:40 AM

  • this css is sus

    by ihatemath123, Aug 14, 2024, 1:53 AM

  • 391345 views moment

    by ryanbear, May 9, 2023, 6:10 AM

  • We need virgil nicula to return to aops, this blog is top 10 all time.

    by OlympusHero, Sep 14, 2022, 4:44 AM

  • :omighty: blog

    by tigerzhang, Aug 1, 2021, 12:02 AM

  • Amazing blog.

    by OlympusHero, May 13, 2021, 10:23 PM

  • the visits tho

    by GoogleNebula, Apr 14, 2021, 5:25 AM

  • Bro this blog is ripped

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

  • Holy- Darn this is good. shame it's inactive now

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

  • godly blog. opopop

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

  • long blog

    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
    awesome. 358,000 visits?????

    by OlympusHero, May 14, 2020, 8:43 PM

72 shouts
Tags
About Owner
  • Posts: 7054
  • Joined: Jun 22, 2005
Blog Stats
  • Blog created: Apr 20, 2010
  • Total entries: 456
  • Total visits: 404395
  • Total comments: 37
Search Blog
a