Geometry Problem

Level: Mathematics Education, High School, Honors Geometry, College.

Click the figure below to see the problem 741 details.

## Monday, April 16, 2012

### Problem 741: Scalene Triangle, Incenter, Circumcircle, Midpoint, Arc, Congruence

Labels:
arc,
chord,
circumcircle,
congruence,
incenter,
midpoint,
triangle

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http://img37.imageshack.us/img37/4078/problem741.png

ReplyDeleteSince D is the midpoint of arc AC , so DA=DC and ∠ ABD=∠DBC=∠CAD…. ( angles face same arc DC)

In triangle AID , ∠IAD=∠IAC+∠CAD=∠A/2+∠B/2

In triangle AIB, external angle ∠AID=∠BAI+∠IBA=∠A/2+∠B/2 => Triangle AID is isosceles

So DA=DC=DI

Problem 741 - Corollary:

ReplyDelete∆AID is equilateral ⇔ ∠C = 60°

Let /_ACB = C. Now /_ADI = /_ADB = C (angles in the same sector) whereas /_ACI = C/2 simce I is the incentre of Tr. ABC. Thus, AI subtends twice the angle at D than at C which means D is the circumcentre of Tr. AIC or DA = DI = DC

ReplyDeleteTo Ajit

ReplyDeleteIn my opinion, the statement "AI subtends twice the angle at D than at C which means D is the circumcentre of Tr. AIC or DA = DI = DC" may not true in general case.

It will be true in the case DA=DI . if DA=DI , we can draw a circle center D , radius DA=DI . Since angle(ICA)=1/2angle(ADI) => C will be on the circle and DA=DI=DC

Ref. Problem 741 (A Poser):

ReplyDeleteLet D, E, F be the midpoints of the arcs AC, AB, BC respectively

Show that

(i)I is the ortho-centre of ∆DEF

(ii)∠ODI = ∠C - ∠B

To Peter,

ReplyDeleteSee Problem #8 - Solution by Nilton Lapa.

Would this solution, in your opinion, be invalid?

To Antonio,

I'd very much like a word by way of an explanation from you.

To Ajit, I don't see the solution by Nilton of problem #8.

DeleteI meant Problem #9. Sorry abt the typo.

DeleteActually, I agree with Peter that just the fact that /_ADI=2*/_ACI is insufficient to prove that D is the circumcentre of triangle ACI but this fact coupled with the fact that D lies on the perpendicular bisector of AC since DA=DC is sufficient to say that D is the circumcentre of triangle ACI and hence etc.

ReplyDeleteTo Peter and Ajit:

ReplyDeleteLooked in another way,

∠AIC = ∠A/2 + ∠C/2 + ∠B = 90° + ∠B/2

reflex∠ADC = 360° - ∠ADC = 180° + (180° - ∠ADC) = 180° + ∠B = 2∠AIC

Follows D is the centre of the circle through A, I, C

To Pravin:

ReplyDeleteAbout your suggested problem and questions:

In my opinion, question (i) "I is the ortho-centre of ∆DEF" is OK but question (ii)

"(ii)∠ODI = ∠C - ∠B" may not correct. I got ∠ODI = 1/2(∠C - ∠A ).

Please check it

You are right Peter! Thank You.

ReplyDelete∠ODI = ∠ODC - ∠IDC = (90° - ∠B/2) - ∠A

= ∠A/2 + ∠C/2 - ∠A = (∠C - ∠A) /2

(or)

∠ODI = ∠ODB = ∠OBD = ∠ABD - ∠ABO = ∠B/2 - (90 - ∠C)

= ∠C + ∠B/2 - ∠A/2 - ∠B/2 - ∠C/2 = (∠C - ∠A) /2

To Ajit: What is the problem which solution sended by me is (perhaps) invalid? I'd like to review that solution.

ReplyDeleteThanks.

The solution is uploaded to the following link:

ReplyDeletehttps://docs.google.com/open?id=0B6XXCq92fLJJenVNQ1NYZWJKSHM

Video solution http://youtu.be/Asa1SJnv_v0

ReplyDeleteGreetings =)

Hi Alejandro, I was wander why did you put into the problem conditions that D is mid point, indeed this is obviously satisfied by any configuration.

ReplyDeleteMay be you should restating by puting "prove that: 1- D is mid piont of arc AC, 2- AD=DI=DC.

MReyes

Simply < ADC = <C + <A and so < DAC = 90 - 1/2(A+C). But < IAC = A/2 and so IAD = 90 - C/2 = < AID since < ADI = < C.

ReplyDeleteSo AD = DI = DC

This solution assumes that DIB is collinear which must be the case since AD = DC and IB bisects < ABC

As others have pointed out earlier D is hence the circumcentre of Tr. AIC

Sumith Peiris

Moratuwa

Sri Lanka

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