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Source: Estonia IMO TST 1997 Day 1 P1

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IstekOlympiadTeam

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IstekOlympiadTeam

#1 h Nov 27, 2015, 8:28 PM • 1 Y

Y by Adventure10

In a triangle $ABC$ points $A_1,B_1,C_1$ are the midpoints of $BC,CA,AB$ respectively,and $A_2,B_2,C_2$ are the midpoints of the altitudes from $A,B,C$ respectively. Show that the lines $A_1A_2,B_1B_2,C_1,C_2$ are concurrent.

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9775 posts

jayme

#2 h Nov 28, 2015, 8:29 AM • 2 Y

Y by Adventure10, Mango247

Dear Mathlinkers,

you can see

http://jl.ayme.pagesperso-orange.fr/Docs/The%20cross-cevian%20point.pdf p. 20...

Sincerely
Jean-Louis

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4071 posts

#3 h Nov 28, 2015, 5:23 PM • 2 Y

Y by Adventure10, Mango247

IstekOlympiadTeam wrote:

In a triangle $ABC$ points $A_1,B_1,C_1$ are the midpoints of $BC,CA,AB$ respectively,and $A_2,B_2,C_2$ are the midpoints of the altitudes from $A,B,C$ respectively. Show that the lines $A_1A_2,B_1B_2,C_1,C_2$ are concurrent.

All of them pass through Symmedian point $K$ of triangle $ABC$ .
For the proof, see Theorem 11 on page 10 of this article.

This post has been edited 3 times. Last edited by ThE-dArK-lOrD, May 22, 2018, 10:24 AM

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eshan

#4 h Nov 28, 2015, 6:05 PM • 8 Y

Y by purumehta, Euler149, IstekOlympiadTeam, mssmath, rkm0959, aops1221math, Adventure10, Mango247

My Solution.

Let $AD,BE,CF$ be the altitudes of $\triangle ABC.$

As $A_2$ is the mid-point of $AD,~C_1$ is the mid-point of $AB$ and $B_1$ is the mid-point of $AC,$ we have $\triangle AA_2C_1\cong ADB,~\triangle AB_1C_2\cong ADC.$

So $C_1,~A_2,~B_1$ are sollinear.

Which means, $A_2\in B_1C_1.$

Similarly, $B_2\in C_1A_1,~C_2\in A_1B_1.$

So we can say that $A_1A_2,~B_1B_2,~C_1C_2$ are $3$ cevians of $\triangle A_1B_1C_1.$

Now, $\frac{C_1A_2}{A_2B_1}=\frac{BD}{DC},$ which we get by mid-point theorem, as $BD=2C_1A_2,~DC=2A_2B_1.$

Similarly, $\frac{B_1C_2}{C_2A_1}=\frac{AF}{FB},~\frac{A_1B_2}{B_2C_1}=\frac{CE}{EA}.$

So $\frac{C_1A_2}{A_2B_1}\cdot \frac{B_1C_2}{C_2A_1}\cdot \frac{A_1B_2}{B_2C_1}=\frac{BD}{DC}\cdot \frac{AF}{FB}\cdot \frac{CE}{EA}=1,$ as altitudes of a triangle are concurrent.

Thus by converse of Ceva's theorem, $A_1A_2,~B_1B_2,~C_1C_2$ are concurrent. $\Square$

Attachments:

This post has been edited 6 times. Last edited by eshan, Feb 16, 2016, 5:52 PM

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suli

#5 h Nov 28, 2015, 6:54 PM • 3 Y

Y by mssmath, Adventure10, Mango247

Notice that by homothety $A_2, B_2, C_2$ are points on the medial triangle, i.e. $A_1 A_2, B_1 B_2, C_1 C_2$ are cevians of $A_1 B_1 C_1$ which are concurrent by Ceva's theorem and the homothety which states $\frac{B_1 A_2}{C_1 A_2} = \frac{BD}{CD}$ where $AD$ is altitude in $ABC$ . (And yes, another application of Ceva's theorem on the orthic triangle is needed.)

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#6 h May 22, 2018, 12:55 AM • 2 Y

Y by Adventure10, Mango247

This problem is just a particular case of Cevian Nest Theorem.

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#7 h Aug 12, 2020, 4:20 PM

Y by

My bary solution.
Let $K = AA_2 \cap BC$ , $\alpha = \angle BAC$ , $\beta = \angle CBA$ , $\gamma = \angle ACB$ and $d$ is the diameter of the circumcircle of $\triangle ABC$ .
$BK = AB\cdot cos \beta = d\cdot sin \gamma cos \beta$ , $CK = AC\cdot cos \gamma = d\cdot sin \beta cos \gamma$ .
$\implies K = [0: sin \beta cos \gamma: sin \gamma cos \beta] \implies A_2 = [sin \beta cos \gamma + sin \gamma cos \beta:sin \beta cos \gamma: sin \gamma cos \beta] = [sin \alpha: sin \beta cos \gamma: sin \gamma cos \beta]$ .
$A_1 = [0:1:1]$ .
$L = [sin^2 \alpha : sin^2 \beta : sin^2 \gamma]$ .
$sin^2\alpha = sin\alpha (sin\alpha) + 0(sin^2 \beta \cdot sin^2 \gamma - sin \beta cos \gamma \cdot sin \gamma cos \beta)$ .
$sin^2 \beta = sin^2\beta \cdot sin^2\gamma + sin^2\beta \cdot cos^2\gamma = (sin\beta cos\gamma + sin \gamma cos \beta)(sin\beta cos\gamma) + sin^2\beta \cdot sin^2\gamma - sin \beta cos \gamma \cdot sin \gamma cos \beta$
$= sin\alpha (sin\beta cos\gamma)+ 1(sin^2\beta \cdot sin^2\gamma - sin \beta cos \gamma \cdot sin \gamma cos \beta)$ .
Similarly, $sin^2\gamma= sin\alpha (sin\gamma cos\beta)+ 1(sin^2\beta \cdot sin^2\gamma - sin \beta cos \gamma \cdot sin \gamma cos \beta)$ .
$\implies L \in A_1A_2$
Similarly, $L \in B_1B_2$ and $L \in C_1C_2$ .

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