#13) Trivial Right Triangle Problem

by pro_4_ever, Feb 18, 2018, 2:58 PM

Find atleast 10 solutions to the following problem.

Problem:In $\Delta ABC, \angle A = 90^\circ$ . Let $M$ be the mid point of $BC$ . Prove that $MA=MB=MC$ .

Solutions:

Solution 1 (pro_4_ever)

Solution 2 (pro_4_ever)

Solution 3 (AnArtist)

Solution 4 (WizardMath)

Solution 5 (WizardMath)

Solution 6 (Vrangr)

Solution 7 (Vrangr)
Post your solutions in the comments and It will be added in this post as soon as I see it.
Can we reach 15 proofs?
Bashes are Permitted (Only this problem)!

This post has been edited 12 times. Last edited by pro_4_ever, Feb 21, 2018, 8:32 AM

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Tangent at $B$ makes angle with $BC$ equal to angle subtended by $BC$ = 90 deg. Hence $BC$ passes through the circumcentre, which must then be $M$ . Hence $MA = MB = MC$ .

by Learner13, Sep 4, 2018, 5:00 PM

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Tangents at $B$ , $C$ are perpendicular to $BC$ as in my other solution, so tangents at $B$ , $C$ are parallel as in Solution 5; thus $B$ and $C$ are diametrically opposite, so M is the circumcentre, so $MA = MB = MC$ .
OR
Tangents at $B$ , $C$ are parallel, so radii at $B$ and $C$ (as lines) are parallel and so coincide (having the circumvented in common), both being nothing but $BC$ , so circumcentre is midpoint of $BC$ , so $MA = MB = MC$ .

by Learner13, Sep 4, 2018, 5:08 PM

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Let $C_1$ , $B_1$ be midpoints of $AC$ , $AB$ respectively. In $MAC$ , $MC_1$ is both a median trivially and an altitude because $MC_1$ is parallel to $AB$ by Midpoint Thm. which is perpendicular to $AC$ . Thus $MAC$ is isosceles with $MA = MC$ . $MA = MB$ follows similarly, and $MB = MC$ by definition.

by Learner13, Sep 4, 2018, 5:13 PM

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Use complex co-ordinates, with circumcentre as origin, $A = a$ , $B = b$ , $C = c$ .
Then $a =$ orthocentre $= a+b+c$ , so $b+c = 0$ , so the origin, i.e., circumcentre is M. Thus $MA = MB = MC$ .

by Learner13, Sep 4, 2018, 5:17 PM

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Let $B_1$ , $C_1$ Be midpoints of $AB$ , $AC$ respectively. Then perpendicular bisector of $AC$ is through $C_1$ and parallel to $AB$ ; hence passes through $M$ by Converse Midpoint Thm. Similarly for perpendicular bisector of $AB$ . Perpendicular bisector of $BC$ passes through $M$ because it is its midpoint. Thus $M$ is the intersection of the three perpendicular bisectors and so the circumcentre by definition. So $MA = MB = MC$ .

by Learner13, Sep 4, 2018, 5:37 PM

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Hey, what a nice blog!!!!!
by Power_Set, Jun 24, 2018, 9:53 PM
Is your blog dead ? No new post in almost 2 months.

My blog has died due to malnutrition of problems.
by integrated_JRC, May 2, 2018, 5:52 AM
@ below, I will do it soon
by pro_4_ever, Mar 1, 2018, 10:37 AM
Hi, I think we should 'hide' our solutions, as it is difficult to go up and down the blog
by Drunken_Master, Mar 1, 2018, 10:36 AM
Yeah, I created a test blog, it seems to be an AoPS problem not CSS, can't do anything about it. I'll post it in site support forum later.
by Vrangr, Feb 20, 2018, 5:48 AM
The problem is not in the CSS, I believe, as the CSS only looks after the appearance of the blog, and not the $\text{\LaTeX}$ rendering.
by WizardMath, Feb 19, 2018, 11:34 PM
I'll fix the CSS after boards of need be (I'll probably be starting my own blog too then)
by Vrangr, Feb 19, 2018, 4:41 PM
@ below,
Pls ask WizardMath...
The CSS is taken from his amazing Blog, "An Olympiad Journey"...
I know nothing abt programming.
(BTW I got permission for taking the CSS. No Copyright Issues!)
by pro_4_ever, Feb 19, 2018, 4:16 PM
Huge flaw in website design: comment gets erased if while typing $\LaTeX$ has an error and needs to be fixed.
by Vrangr, Feb 19, 2018, 12:26 PM
Upon WizardMath's Request, let us all post synthetic solutions. Well, let's declare that this blog is intended to give the reader some good problems with simple synthetic solutions. All the best. Not trying to be rude,but I might delete upcoming Bashes
by pro_4_ever, Feb 13, 2018, 4:50 PM
Here's an idea. Ban all bashes on this blog from now on. Also try posting your own previous geo solutions as storage (I hope this is allowed). This would make the quality of the blog better. Bashes are basically write-once, read-never things. Why bother?
by WizardMath, Feb 13, 2018, 4:33 PM
Somebody type Egmo 2013 Problem 1.
No bash,purely troll problem.
by QWERTYphysics, Feb 13, 2018, 4:11 PM
I hate having to see a pure geometry blog being swamped to death with bashes.

Obviously a bash is a super terrible idea. Don't do one.
Unless you are bad at synthetic.
by WizardMath, Feb 13, 2018, 2:58 PM
@ayan Done:)
by pro_4_ever, Feb 13, 2018, 1:38 PM
9th shout
by MEGAKNIGHT, Feb 13, 2018, 11:23 AM
@below Pls don't post a bash if you are thinking about posting a new entry!
by pro_4_ever, Feb 13, 2018, 11:11 AM
Wait, I thought bash is frowned upon in this blog.
by WizardMath, Feb 12, 2018, 6:57 PM
Hmmm...
Combi Geometry is accepted...
by pro_4_ever, Feb 10, 2018, 7:32 AM
Can we post Combi here?
by ayan.nmath, Feb 10, 2018, 7:31 AM
Parag dey
by Paragdey12, Feb 6, 2018, 1:17 PM
More posts please
by AnArtist, Feb 6, 2018, 2:38 AM
Second Shout!
by ccx09, Feb 3, 2018, 9:20 PM
1st shout!
by AnArtist, Jan 31, 2018, 9:20 AM

23 shouts

Problems In Geometry

#13) Trivial Right Triangle Problem

by pro_4_ever, Feb 18, 2018, 2:58 PM

5 Comments