Geometry Problem. Post your solution in the comments box below.

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

Click the diagram below to enlarge it.

## Friday, August 14, 2015

### Geometry Problem 1146: Triangle, Interior Point, 10, 20, 30, 130 degrees, Congruence

Subscribe to:
Post Comments (Atom)

Let AD intersect BC at E; <DEB=30 ( from tr. ACE) and let F be the reflection of A in CD; tr ACF is a 80-20-80 triangle and, well known, CE=AF and tr. ABF is isosceles giving BF=AF, hence BF=CE. But easily D is the circumcenter of tr ACF, so D lies onto perpendicular bisector of CF (and BE), tr. BDE is isosceles, giving <DBE=<DEB=30, done.

ReplyDeleteHow did you know that CE=AF? What it follows from?

DeleteExtend AE to a point G such that tr AGC is a right triangle. The right triangle AGC obtained is congruent to tr CFO. Also triangle EGC is 30-60-90.

DeleteThis helps with CE=AF

In fact this is problem #2 from this site!

ReplyDeleteOops, that's another problem!

ReplyDeleteCircumcircle ABC with center O, connect OA, OB, OC,OD;

ReplyDeleteFrom symmetry, OD bisect <AOB, so <AOD=50 degree;

<OAD=50 degree; so AD=DO;

Also AD=DC, so DO=DC;

Also tr OBC is equilateral tri;

From symmetry, BD bisect <OBC, X=30 degree

How can OD bisect <AOB? It isn't even contained in that angle :\

DeleteFind E such that EA and EC are perpendicular to AB and BC respectively. Then ABCE is cyclic with M the mid point of diameter BE the centre of this circle.

ReplyDeleteThen by computing the angles we can easily show that DM=DC from which we show that Tr.s MBD and CBD are congruent SSS and so BD bisects < MBC.

Hence x=30

Sumith Peiris

Moratuwa

Sri Lanka

We can also show that

ReplyDelete1)BE is perpendicular to AD

2)and if N is the point of intersection of AC and BE that DMNC is cyclic

3)and that DN is perpendicular to AM

4) and that D is the I centre of Tr. BNC

Excellent observations, Sumith!

ReplyDeleteThanx Ajith - sorry I saw your comments only just now.

ReplyDeleteFurther

ReplyDelete5) if DN meets AE at P then P lies on circle AMC whose centre is D

6) if BD meets CE at Q then Tr. BQE is isoceles and BCQM is cyclic with diameter BQ

Another solution: Extend (AB to P so that BP=PC; <APC=80 degs, so D is the circumcenter of tr. ACP and subsequently tr. CPD is equilateral, making BP=PD, <BPD=20, thus <CBD=<PBD-<PBC=30 degs.

ReplyDeleteBest regards

Hi

DeleteHow D is the circumcenter of tr. ACP?

Hi

DeleteHow circumcenter of tr. ACP?

.Forming equilateral triangle ACE (B is interior point triangle ACE).Then tri AEB= tri ABC, so <AEB=<ACB=20.But AE=EC, AD=DC then ED is perpendicular AC and <DEC=30. Therefore

ReplyDelete<BED=<BCD=10 . So BDCE is concyclic. Then <DBC=<DEC=30.

Can you guys elaborate some more about the process you went through to figure out that these particular auxiliary constructions can help to find the solution?

ReplyDeleteIt's bad that people omit this important detail so often in their proofs, because then it is hard to learn from these solutions.

Or perhaps you can recommend some resources from which I could learn these techniques myself?

Few tips....

DeleteRead Mind Map on this site.

What we see is the end result of the Problem often much refined.

The process of course is mainly trial and error, at times using many sketches. And then suddenly it dawns....

Try to infer whatever u can from the given data, some or all of these inferences may lead u to the answer.

Many a time these problems can be deviously difficult or most obvious.

Let me explain my thought process on Antonio's latest Problem (1250) which appeared today in Gogeometry.

The crucial thing was to use the many cyclic quads on show and realize that EG//FH, then the rest was easy.

"Read Mind Map on this site."

DeleteWhich one? You mean this one?:

http://www.gogeometry.com/problem/problem_solving_techniques_mind_map.html

I read it, but it didn't much informative on the subject of solving geometry problems, because it's too general. I'd rather like to know how to figure out what auxiliary lines I need to draw and where to make the problem solvable.

_"What we see is the end result of the Problem often much refined."_

That's exactly what I have problem with, because such an result doesn't contain the path which lead to that particular solution. And I'm more interested in how can I come up with a solution myself instead of just staring in awe at how others did it and puzzling how did they figure it out.

_"The process of course is mainly trial and error"_

In my case, my every trial so far ends up with me being stuck somewhere and not knowing how to progress any further :/ There's always something missing between where I am and the statement to prove. (Even if I try going backwards from the last statement.)

I'd like to learn how to solve such problems, because I stumble at many problems in my own research in geometry where I need to calculate angles and lengths in different polygons, and it always take me hours of hopeless brooding until I find a way to calculate something I need. Learning these techniques could save me a lot of time.

_"Try to infer whatever u can from the given data, some or all of these inferences may lead u to the answer."_

That's what I do. But I always get stuck somewhere eventually.

_"Many a time these problems can be deviously difficult or most obvious."_

Is there any list of the problems on this website arranged by the level of difficulty? I'd like to start from something simple and learn gradually.

Oh, and how to see if a problem has been solved or not? Are there listed somewhere on separate lists? (solved vs. unsolved)

Sorry if my questions are dumb, but I find this website to be a real mess, lots and lots of subpages and I can't really grasp its "structure" :/

"_Let me explain my thought process on Antonio's latest Problem (1250) which appeared today in Gogeometry."_

That's another thing I'd like to learn: stating such problems. How to make sure that there is enough information in the problem that it could be solved, but not too much so that it wasn't too easy?

Can anyone post their problems here? Does one need to be the author of the problem? Or any problem will do?

_"The crucial thing was to use the many cyclic quads on show and realize that EG//FH, then the rest was easy."_

OK I'll give it a shot myself firs, and then I'll check out your solution if I get stuck.

Problem 1146 answer

ReplyDeleteIs triangleAEB=triangleABC (AB is common,AE=AC,<BAE=<BAC)

AE=EC (triangle AEC =equilateral ). ΑD=DC(<DAC=<DCA)

ED perpendicular AC (AE=EC,AD=DC perpendicular bisector )

<ΒΕD=<AED-<AEB=30-20=10

BDCE is concyclic (<BED=10=<BCD)

APOSTOLIS MANOLOUDIS 4 HIGH SHCOOL OF KORYDALLOS PIRAEUS GREECE

In fact, it is a special case of more general problem.

ReplyDeletet=10; (t, 30-2t):(2*t, 30-2t) -> (90+t, 30)

See https://output.jsbin.com/fofecum#20,10,10,10.

<DAC=<DCA

ReplyDeleteDA=DC

<ABD=130-x

In triangle ABD

sin(130-x)/AD=sin20/BD

AD/BD=sin(130-x)/sin20--------(1)

In triangle BDC

sinx/DC=sin10/BD

DC/BD=sinx/sin10-------(2)

AD=DC, (1)=(2)

sin(130-x)/sin20=sinx/sin10

sin(130-x)/sinx=2cos10

sin(130-x)=2sinxcos10

sin(130-x)=sin(x+10)+sin(x-10)

sin(130-x)-sin(x-10)=sin(x+10)

2cos60sin70=sin(x+10)

sin70=sin(x+10)

x=60