Proving collinearity by constructing homotheties

by XmL, Jul 26, 2014, 4:55 AM

I got the following two problems from a really good friend of mine(a good geometer too), and I happen to solve them using "similar" methods(pun intended).

First Problem:

Problem Statement: Let $B',C'$ denote the reflection of $B,C$ over $AC,AB$, let the line tangent to $(AB'C')$ at $A$ intersect $BC,B'C'$ at $Y,X$, prove that $AX=AY$.

//cdn.artofproblemsolving.com/images/2d6a9abc1b5bcbd627ee223e11143b39f1b1e1f3.jpg

Solution


Second Problem:

Problem Statement: $(O)$ is the circumcircle of $\triangle ABC$, $P$ moves on segment $BC$, the circle with $BP$ as diameter meets $(O)$ at S, $AC\cap BS=M$, H is the orthocenter of $ABP$, $MH$ meets the line perpendicular to $BC$ through $B$ at $N$. Prove that $N$ is a fixed point.

//cdn.artofproblemsolving.com/images/acc51d4bf0186e2ccff92ec69eebf39657ae64b5.jpg

Solution
This post has been edited 3 times. Last edited by XmL, Aug 12, 2014, 11:13 PM

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3 Comments

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Absolutely nice solution to both parts!

Could you tell how you got the idea of reflecting stuff over $A$ in the first question? Also, what motivated you to construct $B_2$?

Sorry if my questions are silly. It's just that I'm not very good at synthetic geometry, especially problems that require considering new points.
This post has been edited 2 times. Last edited by AnonymousBunny, Jul 26, 2014, 8:36 PM

by AnonymousBunny, Jul 26, 2014, 8:35 PM

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Hey man there's no such thing as silly question, plus yours is really good.

$B_1,C_1$ are reflected so that we can convert it into a concurrence problem, also there reflections are more "well known" than $Y$ if you know what I mean, as simple as that.

There are two things going on before I decide to construct $B_2,A_1$. First of all there are a couple of parallels. Secondly since we want to know more about the line $AY'$(prove that it's tangent to the circle), what's a better way to do that than to construct a triangle homothety to $ABB_1$ knowing that $A_1$ would lie on $AY'$ hence now we just have to analyze the line $AA_1$.

My friend also showed me that the construction of $B_2,A_1$ can be avoided by extending $CC_1,BB_1$ so that they meet and get a parallelogram and then use menelaus to prove the tangent. However, you know I'm a synthetic guy so~~

Anyway hope this helps :lol: btw from what I've seen of you on your blog and Brilliant you seem to be really good at combo, can you give us some tips on that subject? Thank you! Feel free to ask more questions if you have any.

by XmL, Jul 27, 2014, 4:12 AM

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Thanks for the really great insightful answer! That explains a lot about how you got the idea of constructing the additional points. Thanks! It's not easy (at least for me) to figure out these constructions; but your explanation makes it look so easy. :) You're amazingly awesome at synthetic geo!

Regarding the second point, I don't consider myself to be good at combinatorics at all. I can only solve some of the combinatorics problems which are easy to visualize. Basically, my algorithm for solving a combo is to try small cases, try to observe a pattern and come up with a construction / explanation, try to use induction to justify the pattern, and if this approach fails (and it often does) or trying small cases isn't doable in a problem, I give up.

by AnonymousBunny, Jul 27, 2014, 6:34 PM

Still an immature proof writer :)

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  • Good blog! I hope to see more posts here!

    by WizardMath, Feb 3, 2018, 6:57 PM

  • Thanks Harry Potter :D

    by XmL, Jul 30, 2014, 10:47 PM

  • Hey dude. Nice blog. I like your work.

    by gamjawon, Jul 30, 2014, 7:43 AM

  • Thanks...although most of the posts here won't be about geometry...rightnow I'm trying to improve my skills in other areas such as number theory and combo.

    by XmL, Jun 5, 2014, 1:11 AM

  • Hi Geometer,,nice blog..:D

    by vanu1996, Jun 4, 2014, 1:27 PM

  • Check your private massage :)

    by zabihpourmahdi, Jan 11, 2013, 5:34 PM

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