[Solutions Posted]Geometry Proof Problem of the day #3 (Diagram Added)

To get a better idea of the community's current geometric proof "skill level," I will keep dropping the difficulty of these problems until we find an adequate starting point for the synthetic geometry group. I will appreciate any feedback on today's problem as you approach it.

Given ABC\triangle ABC, let D,ED,E denote the midpoints of the arc BCBC, where A,EA,E lie on the same side BCBC. Construct FF on ABAB such that CFABCF\perp AB, and GG on AEAE such that GFDFGF\perp DF. Prove that CG=ECCG=EC.

Note by Xuming Liang
3 years, 9 months ago

No vote yet
1 vote

</code>...<code></code> ... <code>.">   Easy Math Editor

MarkdownAppears as
*italics* or _italics_ italics
**bold** or __bold__ bold

- bulleted
- list

  • bulleted
  • list

1. numbered
2. list

  1. numbered
  2. list
Note: you must add a full line of space before and after lists for them to show up correctly
paragraph 1

paragraph 2

paragraph 1

paragraph 2

[example link](https://brilliant.org)example link
> This is a quote
This is a quote
    # I indented these lines
    # 4 spaces, and now they show
    # up as a code block.

    print "hello world"
# I indented these lines
# 4 spaces, and now they show
# up as a code block.

print "hello world"
MathAppears as
Remember to wrap math in </span>...<span></span> ... <span> or </span>...<span></span> ... <span> to ensure proper formatting.
2 \times 3 2×3 2 \times 3
2^{34} 234 2^{34}
a_{i-1} ai1 a_{i-1}
\frac{2}{3} 23 \frac{2}{3}
\sqrt{2} 2 \sqrt{2}
\sum_{i=1}^3 i=13 \sum_{i=1}^3
\sin \theta sinθ \sin \theta
\boxed{123} 123 \boxed{123}

Comments

Sort by:

Top Newest

Hey I got around more than half way to the problem. It's almost completed. Feeling tired. I will post it tommorow. (Feeling very excited but Tired). :P :P

Surya Prakash - 3 years, 9 months ago

Log in to reply

image image

Hurray!!! I got the solution.

Since EABCEABC is cyclic. It implies that ABC=1800AEC=CEG\angle ABC = 180^{0} - \angle AEC = \angle CEG.

We have to prove that CE=EGCE = EG i.e. CEG=CGE\angle CEG = \angle CGE. It suffices to prove that CGE=ABD\angle CGE = \angle ABD.

Let 'II' be foot of perpendicular from 'CC' onto EGEG. But, EFC=EIC=900\angle EFC = \angle EIC = 90^{0}. So, ΔBFCΔCEI\Delta BFC \sim \Delta CEI. So, BFBC=EIEC\dfrac{BF}{BC} = \dfrac{EI}{EC} (#)

Observe that DBF=1800DEA=DEG\angle DBF = 180^{0} - \angle DEA = \angle DEG (1)

Since EDED is diameter of the circle. It implies that DAE=900\angle DAE = 90^{0}. But DFG=900=DAG\angle DFG = 90^{0} = \angle DAG. So, A,G,D,FA, G, D, F are cyclic.

It implies that EGD=1800AFD=BFD\angle EGD = 180^{0} - \angle AFD = \angle BFD (2)

From (1) and (2) it implies that ΔBFDΔEGD\Delta BFD \sim \Delta EGD. So, BFBD=EGED\dfrac{BF}{BD} = \dfrac{EG}{ED}(##)

Dividing (#) by (##), we get,

BDBC=EIEC×EDEG\dfrac{BD}{BC} = \dfrac{EI}{EC} \times \dfrac{ED}{EG} EIEG=BDBC×ECED=12×BDBN×ECED\dfrac{EI}{EG} = \dfrac{BD}{BC} \times \dfrac{EC}{ED} = \dfrac{1}{2} \times \dfrac{BD}{BN} \times \dfrac{EC}{ED}

Since, ΔECD\Delta ECD is right angled at 'CC' and also CED=CBD\angle CED = \angle CBD. So, cosCED=cosCBD\cos{\angle CED} = \cos{\angle CBD}. It implies that ECED=BNBD\dfrac{EC}{ED} = \dfrac{BN}{BD}

It gives out that EIEG=12\dfrac{EI}{EG} = \dfrac{1}{2} i.e. EG=2EIEG = 2 EI. So, EI=GIEI =GI. It implies that CEG=CGE\angle CEG = \angle CGE i.e. CG=CECG = CE as required.

SPECIAL THANKS TO @Xuming Liang

Surya Prakash - 3 years, 9 months ago

Log in to reply

Nice solution! Well Done. I have understood your solution and I realize that I was close to prove it. But I think there are few trivial typo mistakes at start that you should fix. Overall your solution is quite rich! :) :) :)

Nihar Mahajan - 3 years, 9 months ago

Log in to reply

Congradulations :) You are the first Brilliant solver of my challenges. Great Job

Xuming Liang - 3 years, 9 months ago

Log in to reply

A slightly alternate solution:

Since DAAG,DFGFDA\perp AG, DF\perp GF, D,F,A,GD,F,A,G are concyclic and thus DGA=DFB\angle DGA=\angle DFB. In addition,note that DBA=DEG\angle DBA=\angle DEG, therefore DEGDBFDEG\sim DBF. At this point one can use the new-found information to ratio chase GEGE related lengths, as Surya Prakash has demonstrated, or we can continue to utilize similar (they are almost the same thing),

Since DEGDBFDEG\sim DBF, we ask what point corresponds to CC in the DBFDBF configuration, i.e. For what point is it to DBFDBF as CC is to DEGDEG(Not sure if this makes sense...) By some thought we quickly see that it is the midpoint of BCBC denoted by MM (I will elaberate: it must lie on BCBC since GEC=FBC\angle GEC=\angle FBC, and MM is the only point that satisfy BM=FMBM=FM). Clearly DBMDECDBM\sim DEC, thus DEGCDBFMDEGC\sim DBFM. SInce BM=FMBM=FM, we have CG=ECCG=EC and are done.

Similarity is powerful isn't it?

Xuming Liang - 3 years, 9 months ago

Log in to reply

Nice and easy one.

Surya Prakash - 3 years, 9 months ago

Log in to reply

Can you post the figure?

Agnishom Chattopadhyay Staff - 3 years, 9 months ago

Log in to reply

@Agnishom Chattopadhyay

01000011 01101111 01101110 01100111 01110010 01100001 01110100 01110011 00100000 01101111 01101110 00100000 00110001 00110000 00110000 00110000 00100000 01000110 01101111 01101100 01101111 01101111 01110111 01100101 01110010 01110011 00100001

Mehul Arora - 3 years, 9 months ago

Log in to reply

What's that @Mehul Arora

Surya Prakash - 3 years, 9 months ago

Log in to reply

@Surya Prakash I'm sure Agnishom will understand :)

Mehul Arora - 3 years, 9 months ago

Log in to reply

@Mehul Arora Tell me too :P @Mehul Arora

Surya Prakash - 3 years, 9 months ago

Log in to reply

@Surya Prakash Come to Slack ;)

Mehul Arora - 3 years, 9 months ago

Log in to reply

74 68 61 6e 6b 20 79 6f 75 21

Agnishom Chattopadhyay Staff - 3 years, 9 months ago

Log in to reply

Binary

Aditya Chauhan - 3 years, 9 months ago

Log in to reply

@Aditya Chauhan Yeo. Binary it is :P

Mehul Arora - 3 years, 9 months ago

Log in to reply

@Mehul Arora I guess this is a geometrical discussion forum... Or wait , are you thinking of computer scientific geometry? :O :3

Nihar Mahajan - 3 years, 9 months ago

Log in to reply

@Nihar Mahajan Yep. Computer scientific geom would be better :P

Mehul Arora - 3 years, 9 months ago

Log in to reply

@Nihar Mahajan You mean computational geometry?

Agnishom Chattopadhyay Staff - 3 years, 9 months ago

Log in to reply

@Agnishom Chattopadhyay Yeah.

Leave it , I was just joking :P

Nihar Mahajan - 3 years, 9 months ago

Log in to reply

27741264475741985937324611021968519394650689232351903626759926561\large{27741264475741985937324611021968519394650689232351903626759926561}

Satyajit Mohanty - 3 years, 9 months ago

Log in to reply

What difficulty are you encountering in drawing the diagram? May be I can save your time by helping you out :)

Nihar Mahajan - 3 years, 9 months ago

Log in to reply

Can you help me understand how I construct the arc?

Agnishom Chattopadhyay Staff - 3 years, 9 months ago

Log in to reply

@Agnishom Chattopadhyay Draw circumcircle of ΔABC\Delta ABC. Now draw perpendicular bisector of BCBC and let it intersect the circumcircle at D,ED,E and we have D,ED,E as mid-points of arc BCBC. Now suppose if AA lies on major arc BCBC , then EE must also lie on major arc and same if its case of minor arc.

Nihar Mahajan - 3 years, 9 months ago

Log in to reply

@Nihar Mahajan Did u get the problem @Nihar Mahajan

Surya Prakash - 3 years, 9 months ago

Log in to reply

@Surya Prakash Ah ,Don't expect me to solve Xuming's challenge so early :P I am not as awesome as he is ;)

Nihar Mahajan - 3 years, 9 months ago

Log in to reply

Also , Congrats for having 1000 followers! :P :) :3

Nihar Mahajan - 3 years, 9 months ago

Log in to reply

Thanks! I decided that I will be excited at first, but then again, I said, I thought I will save the excitement for 1024 followers instead

Agnishom Chattopadhyay Staff - 3 years, 9 months ago

Log in to reply

@Agnishom Chattopadhyay Why 1024 only? Wait , you planning something awesome?

Nihar Mahajan - 3 years, 9 months ago

Log in to reply

So basically, we have to prove that G and E coincide, right?

@Xuming Liang

Mehul Arora - 3 years, 9 months ago

Log in to reply

No. Not at all. In short you must prove that triangle CGE is isosceles.

Nihar Mahajan - 3 years, 9 months ago

Log in to reply

@Xuming Liang Do D and E lie on the circumcirlce of ΔABC\Delta ABC?

Surya Prakash - 3 years, 9 months ago

Log in to reply

Yes , they do.

Nihar Mahajan - 3 years, 9 months ago

Log in to reply

×

Problem Loading...

Note Loading...

Set Loading...