21. Probleme de geometrie (Yetty & I).

by Virgil Nicula, Apr 22, 2010, 2:30 AM

PP1 (Yetty). Fie cercurile $w_k\ ,\ k\in\overline {1,3}$ care au un singur punct comun $D$, adica $\{D\}=w_1\cap w_2\cap w_3$ .

Sa se arate ca $\left\|\begin{array}{ccc} \{D,A\}=w_2\cap w_3 & , & A_1\in (AD\cap w_1\\\\ \{D,B\}=w_3\cap w_1 & , & B_1\in (BD\cap w_2\\\\ \{D,C\}=w_1\cap w_2 & , & C_1\in (CD\cap w_3\end{array}\ \right\|\ \ \Longrightarrow$ $\frac {AD}{AA_1}+\frac {BD}{BB_1}+\frac {CD}{CC_1}=1$.

Demonstratie. Notam $\left\{\begin{array}{c} \{B,C_2\}=A_1B\cap w_3\\\\ \{C,B_2\}=A_1C\cap w_2\end{array}\right\|$ . Se arata usor ca $A\in B_2C_2$ si $\left\{\begin{array}{c} B_1B_2\parallel A_1C_2\\\\ C_1C_2\parallel A_1B_2\end{array}\right\|$ . Notam $\left\{\begin{array}{c} B_0\in DB_2\cap A_1C_2\\\\ C_0\in DC_2\cap A_1B_2\end{array}\right\|$ . Deci

$\left\{\begin{array}{c} B_1B_2\parallel A_1C_2\ \Longrightarrow\ \frac {BD}{BB_1}=\frac {B_0D} {B_0B_2}\\\\ C_1C_2\parallel A_1B_2\ \Longrightarrow\ \frac {CD}{CC_1}=\frac {C_0D}{C_0C_2}\end{array}\right\|$ $\Longrightarrow$ $\sum \frac {AD}{AA_1}=\frac {AD}{AA_1}+\frac {B_0D}{B_0B_2}+\frac {C_0D}{C_0C_2}=1$ deoarece $D\in AA_1\cap B_0B_2\cap C_0C_2$ in $\triangle A_1B_2C_2$ .

Observatie.. Desi aceasta frumoasa problema are un caracter general, nu are un grad mare de dificultate.

Situatii particulare ale ei le-am intalnit ca probleme propuse in diferite reviste sau culegeri de probleme.

PP2 (own). Let $ABCD$ be a convex quadrilateral with $AD\not\parallel BC$. Define the points $E=AD \cap BC$ and $I = AC\cap BD\ .$

Prove that the triangles $EDC$ and $IAB$ have the same centroid if and only if $AB \parallel CD$ and $IC^{2}= IA \cdot AC$.

Proof. Let $I$ be the origin of the vectorial system, i.e. for any $X$ we have the vectors $X=\overrightarrow{IX}$ , $I=\overrightarrow 0$. Thus, $\ (\exists)$ $m,n\in R^{+}- \{ 0,1\}$ , $ IC=m\cdot IA$ , $ID=n\cdot IB$ $\Longleftrightarrow$

$ C=-mA$ , $D=-nB$ . Therefore, $E\in AD\cap BC$ $\Longleftrightarrow$ $(\exists )x,y\in R$ , $x,y\not\in \{ 0,1\}$ such that $E=(1-x)B+xC=(1-y)A+yD$ $\Longleftrightarrow E=-xmA+(1-x)B=$

$(1-y)A-ynB\Longrightarrow$ $x=\frac{1+n}{1-mn}$ , $y=\frac{1+m}{1-mn}$ , $E=\frac{m(1+n)}{mn-1}\cdot A+\frac{n(1+m)}{mn-1}\cdot B$ . Thus, $\triangle ABI$ and $\triangle CDE$ have the same centroid $\Longleftrightarrow$ $E+C+D=A+B$ $\Longleftrightarrow $

$\frac{m(1+n)}{mn-1}\cdot A+\frac{n(1+m)}{mn-1}\cdot B-mA-nB=A+B$ $\Longleftrightarrow$ $mn^2=2n+1\ \wedge \ nm^2=2m+1$ $\Longleftrightarrow$ $m=n$ and $n^3=2n+1$ $\Longleftrightarrow$ $m=n$ and $n^2=n+1$ $\Longleftrightarrow$

$AB\parallel CD$ and $\left( \frac{IC}{IA}\right) ^2=$ $\frac{IC}{IA} +1$ $\Longleftrightarrow$ $AB\parallel CD$ and $IC^2=IA\cdot AC$ .


PP3. Let $w_1=C(r_1)$ and $w_2=C(r_2)$ be two circles tangent externally at $M$ and tangent internally to a circle $w$ at $A$ , $B$ respectively.

The common tangent of $w_1$ and $w_2$ at $M$ meet $w$ at $C$ and $D$. Show that $CB\cdot AD=AC\cdot BD\ (*)$ and $\frac {MC\cdot MD}{CD}=\sqrt {r_1r_2}$ .

Proof. Denote the intersection $T$ of the tangents in $A$ , $B$ to $w$ . Observe that $TA=TB$ and $T\in CD$ , i.e. $T$ is the radical center of $w$ , $w_1$ , $w_2$ . Thus,

$\left\{\begin{array}{c} \widehat {TAC}\equiv\widehat{TDA}\implies TAC\sim TDA\implies\frac {TA}{TD}=\frac {AC}{DA}\\\\ \widehat {TBC}\equiv\widehat{TDB}\implies TBC\sim TDB\implies\frac {TB}{TD}=\frac {BC}{DB}\end{array}\right\|\implies (*)\ .$ Apply the Casey's theorem to $w_1$ , $w_2$ , $C$ , $D\implies$ $MC\cdot MD=CD\cdot\sqrt {r_1r_2}$ .
This post has been edited 9 times. Last edited by Virgil Nicula, Apr 9, 2016, 2:50 AM

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    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
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About Owner
  • Posts: 7054
  • Joined: Jun 22, 2005
Blog Stats
  • Blog created: Apr 20, 2010
  • Total entries: 456
  • Total visits: 404427
  • Total comments: 37
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