AM=BM, ABCD #, 3AM=AC, BD=BC (2013 Math. Multiathlon Reggata J 4.2)

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AM=BM, ABCD #, 3AM=AC, BD=BC (2013 Math. Multiathlon Reggata J 4.2)

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#1 h Mar 20, 2021, 9:54 PM

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In parallelogram $ABCD$ , $BD=BC$ . A point $M$ on $AC$ is such that $3AM=AC$ . Prove that $AM=BM$ .

(13th Ural Tournament, Major League)

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#2 h Mar 20, 2021, 10:39 PM

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So the shape is actually a rhombus. Take an affine transformation sending the rhombus to a square with area $9$ . Now we use coordinates. $A=(0,0),B=(3,0),C=(3,3),D=(0,3)$ . Then $M=(1,1)$ . But wait this doesn't work. Is there a typo with the problem?

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#3 h Mar 20, 2021, 10:46 PM

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@above I'm not sure how you got that it's a rhombus

This post has been edited 1 time. Last edited by natmath, Mar 20, 2021, 10:46 PM

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#4 h Mar 20, 2021, 10:49 PM

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Doesn't $BD=BC$ imply it is a rhombus?

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#5 h Mar 20, 2021, 10:53 PM

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I searched for the original wording, from the 1st source, there it was stated as:

In the parallelogram $ABCD$ , $BD = BC$ . Points $M$ and $N$ are taken on the diagonal $AC$ so that $AM = MN = NC$ . Prove that triangle $ABM$ is isosceles.

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#6 h Mar 20, 2021, 10:55 PM • 1 Y

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The wording/question is correct.
@2above Note that BD is a diagonal not a side

This post has been edited 2 times. Last edited by natmath, Mar 20, 2021, 10:56 PM

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#7 h Mar 20, 2021, 10:59 PM

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franzliszt wrote:

Doesn't $BD=BC$ imply it is a rhombus?

I don't think so. I don't know where you went wrong but $BD$ is a diagonal if you read it wrong. Think what happens when $\angle{DBC}$ is increased.

I'm solving the OP's problem below:
straightforward trig bash

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#8 h Mar 21, 2021, 1:11 AM

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$M$ is the barycentric of triangle $ABD$ . And $BD=BC=AD$ so that $AM=BM$ .

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#9 h Mar 21, 2021, 12:52 PM

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Welp, I guess I did misread the problem. To make up, here is a solution.

Apply Barycentric coordinates with reference triangle $ABD$ . Set $A=(1,0,0),B=(0,1,0),D=(0,0,1)$ . By the parallelogram lemma, $C=(-1,1,1)$ . Points on line $AC$ are parameterized by $(t,\tfrac{1-t}2,\tfrac{1-t}2)$ . Here, clearly $t=1/3$ so we find that $M=(1/3,1/3,1/3)$ and it is the centroid of $\triangle ABD$ . But recall that $BC=BD=DA$ so $\triangle ABD$ is isosceles with $DA=DB$ . So clearly $MB=MA$ and we are done.

This post has been edited 1 time. Last edited by franzliszt, Mar 21, 2021, 12:52 PM

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