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By Xiggi (Xiggi) on Tuesday, July 20, 2004 - 03:48 pm: Edit |

This classic problem would never appear on a SAT test because it does not have a fast and easy solution. While it is not an easy problem, it can be solved using SAT geometry concepts.

**Draw an isosceles triangle ABC with Side AB = Side AC. Draw a line from C to side AB and label that line CD. Now draw a line from B to side AC. Label that line BE. Let angle EBC = 60 degrees, angle BCD equal 70 degrees, angle ABE equal 20 degrees, and angle DCE equal 10 degrees. Now draw line DE. Find what angle EDC is by using geometry only and no trigonometry.**

By Adidasty (Adidasty) on Tuesday, July 20, 2004 - 04:30 pm: Edit |

The answer is 70 degrees. Angle EDC equals 70 degrees. That problem was a b****. I suck at math, and that one took some hardcore reasoning. Thanks for the challenge Xiggi!

By Xiggi (Xiggi) on Tuesday, July 20, 2004 - 05:31 pm: Edit |

The answer is not 70 degrees. Try again.

By Feuler (Feuler) on Tuesday, July 20, 2004 - 05:34 pm: Edit |

It cannot be 70 degrees because that could make DE parellel to BC, which clearly cannot be the case.

By Jaredthegreat (Jaredthegreat) on Tuesday, July 20, 2004 - 05:45 pm: Edit |

...that's what I thought...I got 30

By Xiggi (Xiggi) on Tuesday, July 20, 2004 - 05:50 pm: Edit |

It's not 30 degrees. No guessing

By Joshjmgs (Joshjmgs) on Tuesday, July 20, 2004 - 05:52 pm: Edit |

Is it 130 degrees?

If you want an explanation, just ask.

By Optimizerdad (Optimizerdad) on Tuesday, July 20, 2004 - 05:54 pm: Edit |

Um, I don't think so. If angle EDC=70 degrees, and (input) BCD is also 70 degrees, this implies that ED is parallel to BC; since EBC BCD, ED cannot be parallel to BC.

True?

Xiggi - I got every **** angle in the resulting set of interlocking triangles except EDC ! It's staring me in the face, and I can't see it...

By Loop123 (Loop123) on Tuesday, July 20, 2004 - 05:59 pm: Edit |

I got 43 and 1/3 ....? LOL I must be a crackhead.

By Optimizerdad (Optimizerdad) on Tuesday, July 20, 2004 - 06:04 pm: Edit |

If F is the intersection of lines BE and CD, I got as far as

angle DAE = 20

angle BFC = 50 = angle DFE

angle EFC = angle DFB = 130

angle FEC = 40

angle BDF = 30

and the rest of the angles are functions of angle EDC (say x degrees)

angle ADE = 150 - x

angle AED = 10 + x

angle DEF = 130 - x

One small last step for man...

By Jblackboy05 (Jblackboy05) on Tuesday, July 20, 2004 - 06:05 pm: Edit |

Please let it be 40 degrees

By Jaredthegreat (Jaredthegreat) on Tuesday, July 20, 2004 - 06:07 pm: Edit |

Sorry--20 degrees. typo. I accidentally typed in the angle ADE instead (which I found to be 130) and just deleted the 1.

By Loop123 (Loop123) on Tuesday, July 20, 2004 - 06:07 pm: Edit |

Xiggi, end our misery soon.

My way for solving it was as follows (you might want to skip this --- may impair mathematical ability)

Since the original isosceles triangle ABC has angles 20-80-80, that means that an angle of 80 degrees would encompass one whole side, if you get one I am saying. Thus, (and here's where I'm probably wrong) an angle of 70 degrees encompasses 7/8 of side of the triangle, leaving 1/8 of the side to be the left side of the triangle with bottom side DE. Likewise for the 60 degtree angle: the segment above the intersection with AB will have 1/4 of the side above it and 3/4 below it. Then you can get the sides of triangle AED. This might be close/correct?

I'm the wrong one to try to solve this.

By Joshjmgs (Joshjmgs) on Tuesday, July 20, 2004 - 06:09 pm: Edit |

Good job optimizerdad, thats exactly what I did, you jst didn't finish ;)

BTW they would never put a problem like this on the SAT, because it would take AT LEAST 2 minutes, and youre only allotted 72 seconds.

By Inopa (Inopa) on Tuesday, July 20, 2004 - 06:10 pm: Edit |

its 130

By Ilovefood (Ilovefood) on Tuesday, July 20, 2004 - 06:14 pm: Edit |

damn....now i feel dumb when i get an angle that i havent seen yet...anyone else hve 65 degrees?

By Ilovefood (Ilovefood) on Tuesday, July 20, 2004 - 06:21 pm: Edit |

Is my reasoning solid? I'm really not a math dude...

ADE= 180-20-(140-x), so = 20+x

AED= 180-20-(150-x), so = 10+x

Now 20+x+10+x+20=180, so 50+2x=180, so x=65

Does this make sense?

By Korey (Korey) on Tuesday, July 20, 2004 - 06:24 pm: Edit |

I got 90 degrees, though, I bet that's wrong...

By Loop123 (Loop123) on Tuesday, July 20, 2004 - 06:26 pm: Edit |

LOL this is one controversial problem! Those dudes seem pretty confident about 130 though...

By Ilovefood (Ilovefood) on Tuesday, July 20, 2004 - 06:29 pm: Edit |

I want answers now, dammit!

ahhh....the sweet smell of impatience...

By Legendofmax (Legendofmax) on Tuesday, July 20, 2004 - 06:33 pm: Edit |

I think the answer is 65. Angle ADE is 85 and angle AED is 75. If you notice, 85 + 75 + 20 = 180 for triangle ADE, so this does work. Therefore, to get andle EDC you know that line AB is 180 degrees, and that EDC touches it, but you know the angles around it. Therefore, 180 - 85 - 30 is 65 degrees.

By Adidasty (Adidasty) on Tuesday, July 20, 2004 - 06:35 pm: Edit |

It is possibly 130. i redid the problem and this seems right.

By Joshjmgs (Joshjmgs) on Tuesday, July 20, 2004 - 06:36 pm: Edit |

I agree with legendofmax.

I was checking over my arithmatic and it didn't seem right.

Besides b+x = 130, so if x is 130, then b would be zero, which is impossible

By Legendofmax (Legendofmax) on Tuesday, July 20, 2004 - 06:36 pm: Edit |

*Denoting angle EDC as x, we know that line AB is 180. Angle BDC is 30 degrees (found through deduction and knowledge that angles ABC and ACB are equal), so angle ADE is 180 - 30 - x, or 150 - x. Likewise for AED, which is 140 - x (180 - BEC - x), where BEC is 40. BAC is 20, so 2- + 150 - x + 140 - x gives you 65 for x.

By Legendofmax (Legendofmax) on Tuesday, July 20, 2004 - 06:37 pm: Edit |

I don't think it is 130. What is the reasoning behind it? I made a drawing and it does not seem probably that the angle is very large. It isn't even an obtuse angle.

By Korey (Korey) on Tuesday, July 20, 2004 - 06:42 pm: Edit |

XIGGI, PLEASE TELL US THE ANSWER BEFORE WE ALL DIE!

By Ilovefood (Ilovefood) on Tuesday, July 20, 2004 - 06:42 pm: Edit |

Wow! Somebody agrees with me! Thank you, legendofmax. This means I might be bordering on the edge of being a competent math dude!

By Adidasty (Adidasty) on Tuesday, July 20, 2004 - 06:44 pm: Edit |

i dunno, i really suck at math. i like biology. biology is better.

By Optimizerdad (Optimizerdad) on Tuesday, July 20, 2004 - 06:46 pm: Edit |

1. The answer has to be < 70 degrees. Suppose we started with ED // BC, then EDC = BCD = 70 degrees. Now make D 'travel' on line BA towards A, until all of Xiggi's conditions are met. As D travels towards A, the angle EDC gets smaller & smaller, reducing from its initial value of 70.

2. Legend - I'm not sure your

AED = 180 - BEC - x

is right, since DEB is not necessarily equal to EDC=x?

By Ilovefood (Ilovefood) on Tuesday, July 20, 2004 - 06:48 pm: Edit |

hmmm...im pretty bad at bio, so thats also not my thing. my subject would involve eating and sleeping. maybe a little proofreading, but only if i couldnt fall asleep.

By Legendofmax (Legendofmax) on Tuesday, July 20, 2004 - 06:49 pm: Edit |

It's 65! I will bet my nuts on it. Another tidbit of proof:

We know that BFC is 50, yes? That means DFE is also 50 (vertical angles). Now, if EDC is 65, then this shouldmean that DEF is 180 - 50 - 65, or 65. So now we can check that measurement with triangle DEC, where we have 65 + 10 + (65 + angle FEC, or 40), which is, oh good lord and butter, 180!

optimizerdad: oh crap, you're right.

Nevermind I don't bet my nuts on it. Everything still seems to make sense though. I'll come up with something here. Luckily, if it were 65, the other side is also 65. There has to be something more than luck here, one moment

By Optimizerdad (Optimizerdad) on Tuesday, July 20, 2004 - 06:52 pm: Edit |

Legend:

Sorry, I don't think so. If EDC=65=DEF, then triangle DEF is isosceles, which would imply DE is parallel to BC, which it ain't. Unless I'm missing something...

By Optimizerdad (Optimizerdad) on Tuesday, July 20, 2004 - 06:53 pm: Edit |

...good job you withdrew your bet, or you'd be singing high soprano... :-)

By Legendofmax (Legendofmax) on Tuesday, July 20, 2004 - 06:55 pm: Edit |

I've got it optimizerdad. if you draw a line from point F as an altitude of triangle DEF you can also find x this way. By cutting that triangle into a right triangle you now have angles of x, 90, and 25 (50 cut in 2). Thus 180 - 25 - 90 is 65.

Edit: Wait this doesn't prove it either. There's no guarantee that DE will make the 50 into a 25 with an altitude if it doesn't bisect that angle with a perpedicular segment. D'oh

By Optimizerdad (Optimizerdad) on Tuesday, July 20, 2004 - 06:57 pm: Edit |

Legend:

Nope. The 'altitude line' within triangle DEF doesn't necessarily bisect angle DFE.

By Legendofmax (Legendofmax) on Tuesday, July 20, 2004 - 06:57 pm: Edit |

Optimizerdad: I don't think it's impossible to have an isoceles encapsullated within another isoceles.

By Legendofmax (Legendofmax) on Tuesday, July 20, 2004 - 06:59 pm: Edit |

I would still like to know why 65 does not work?

By Optimizerdad (Optimizerdad) on Tuesday, July 20, 2004 - 07:00 pm: Edit |

...but you just proved that angle EDC must be < 65 (tighter than 70), since if EDF=DEF=65, we would have an isosceles triangle DEF, and we know that ain't so.

By Loop123 (Loop123) on Tuesday, July 20, 2004 - 07:00 pm: Edit |

WHY did Xiggi have to take a daytrip after posting this problem? lol

By Legendofmax (Legendofmax) on Tuesday, July 20, 2004 - 07:03 pm: Edit |

Why do you know it isn't so? Why is it not possible for an isoceles to exist within another? The smaller base doesn't have to be parallel to the larger base. If you draw any arbitrary line as a "small base" in a larger triangle it is not impossible to make another triangle from that base such that it makes a set of equal angles. But, I did not yet "prove" that EDF = DEF yet because as you said, I had accidentally assumed it when deriving x

By Optimizerdad (Optimizerdad) on Tuesday, July 20, 2004 - 07:22 pm: Edit |

Curses. You're right. If EDF=DEF, then DE would have to be parallel to BC iff the line bisecting angle DFE was perpendicular to BC.

OK, let's start with placing D exactly opposite E (i.e. angle DCB=60, not Xiggi's 70), and then move point D towards A till angle DCB=70. At the start, DE // BC and angle EDC=DCB= 60 degrees. Now start moving D towards A. EDC would reduce gradually, from its initial value of 60; so our final answer has to be less than 60, no?

By Loop123 (Loop123) on Tuesday, July 20, 2004 - 07:27 pm: Edit |

Hey, just for future reference, could you explain why I'm wrong in my proportionality assumption some odd posts above?

By Legendofmax (Legendofmax) on Tuesday, July 20, 2004 - 07:27 pm: Edit |

Hmmmm that does make sense. Oy oy oy....

By Legendofmax (Legendofmax) on Tuesday, July 20, 2004 - 07:31 pm: Edit |

However, this goes by the assumption that the isoceles triangle inside ABC is parallel. The left side of this triangle is above the "would-be" parallel which does make it seem like it would be < 60...

However.

"DE would have to be parallel to BC iff the line bisecting angle DFE was perpendicular to BC. "

Triangle BFC is not equilateral; therefore bisecting angle DFE or BFC may not always be perpendicular to both DFE and BFC. Encapsulated triangles don't always have to be perpendicular by base.

By Optimizerdad (Optimizerdad) on Tuesday, July 20, 2004 - 07:38 pm: Edit |

Sorry, I'll restate. 'If DEF is itself an isosceles triangle, then DE // BC iff the line bisecting angle DFE was perpendicular to BC'.

Intuitively, this makes sense - my po' old way_older_than_college_kid brain is dissolving slowly right now, so I don't have the energy to prove it rigorously. Not to mention I'm goofing off, I should be working on designing a scheduling algorithm right now...

By Legendofmax (Legendofmax) on Tuesday, July 20, 2004 - 07:41 pm: Edit |

"iff the line bisecting angle DFE was perpendicular to BC'." Seems fine to me, but only if this bisecting line forms an altitude with DE

By Appliedmath (Appliedmath) on Tuesday, July 20, 2004 - 08:36 pm: Edit |

your retarded people the answer is 50, plug it in and test the other sides too see if they add to 180 on a supplementry line.

By Xiggi (Xiggi) on Tuesday, July 20, 2004 - 08:51 pm: Edit |

Appliedmath~

The answer is not 50. Actually, I'll go even further, the answer is smaller than 30.

By Jaredthegreat (Jaredthegreat) on Tuesday, July 20, 2004 - 08:54 pm: Edit |

I'm absolutely certain it's 20.

By Legendofmax (Legendofmax) on Tuesday, July 20, 2004 - 08:55 pm: Edit |

How did you get 20?

By Xiggi (Xiggi) on Tuesday, July 20, 2004 - 08:58 pm: Edit |

Jared~

It think that it is impossible to solve this problem in 2 minutes and have a geometry proof.

OptimizerDad~

I will give you additional hints. Some keys are the use of the isosceles triangle properties and congruency of triangles. The solution -at least mine- requires additional constructions.

By Jaredthegreat (Jaredthegreat) on Tuesday, July 20, 2004 - 09:02 pm: Edit |

I originally posted 20 a loooong time ago (top quarter of the page). And I said nothing about it taking only two minutes...

By Xiggi (Xiggi) on Tuesday, July 20, 2004 - 09:04 pm: Edit |

Jared~

The answer is indeed 20. Now, the hard part, can you demonstrate that the answer is exactly 20?

I confused your post with Josh's.

By Jaredthegreat (Jaredthegreat) on Tuesday, July 20, 2004 - 09:12 pm: Edit |

I'm working on it. But first I have to do dinner, so it'll probably be at least an hour till I can post it.

By Xiggi (Xiggi) on Tuesday, July 20, 2004 - 09:48 pm: Edit |

I am in no rush. It took me a VERY long time to finish this problem.

By Feuler (Feuler) on Tuesday, July 20, 2004 - 09:55 pm: Edit |

This is a rough problem. What's so great about it is that it looks so easy at first, but is actually so hard. Someone showed me this problem in 9th grade, and after working on it for awhile I just gave up. Maybe I can get it now though...

By Jaredthegreat (Jaredthegreat) on Tuesday, July 20, 2004 - 11:40 pm: Edit |

Haha...It doesn't look like an hour's gonna do it...Maybe tomorrow....

By Xiggi (Xiggi) on Wednesday, July 21, 2004 - 01:23 am: Edit |

I'll give you a small hint: AD = BC.

By Haithman (Haithman) on Wednesday, July 21, 2004 - 01:49 am: Edit |

Wow..I've been owned by a geometry problem!

By Jaredthegreat (Jaredthegreat) on Wednesday, July 21, 2004 - 11:43 am: Edit |

"I'll give you a small hint"

Thats not necessary...

I figured out how to do it last night, while in bed. I'll write a whole thing with pictures, and make a PDF later today.

My method probably is not the preferred method though...

By Xiggi (Xiggi) on Wednesday, July 21, 2004 - 12:04 pm: Edit |

Cool

By Tongos (Tongos) on Wednesday, July 21, 2004 - 12:13 pm: Edit |

i actually have the answer to, but before i found it, i thought it was nearly impossible to solve.

By Jaredthegreat (Jaredthegreat) on Wednesday, July 21, 2004 - 12:43 pm: Edit |

It is...haha

By Cavalier302 (Cavalier302) on Thursday, July 22, 2004 - 01:23 pm: Edit |

bump...please post a solution if you have one! lol, stupid "easy" geometry question....

By Seahorsekid (Seahorsekid) on Thursday, July 22, 2004 - 01:46 pm: Edit |

grr... i have all of the angles EXCEPT DEB and EDC... although I know they equal 130

By Xiggi (Xiggi) on Thursday, July 22, 2004 - 02:58 pm: Edit |

It's not an easy problem.

I'll wait to post the solution to give more time to people who are still working on it.

The only answer is **20 degrees**. It is not 30, 65, or 130 degrees.

By Seahorsekid (Seahorsekid) on Thursday, July 22, 2004 - 03:11 pm: Edit |

I meant that the two angles together add up to 130 degrees... they can't both be 130 degrees or line segment DE would not exist

By Apocalypse_Now (Apocalypse_Now) on Saturday, July 24, 2004 - 12:28 am: Edit |

Jared, most people have to figure out how to do a problem before they get the answer. Is it not so with you?

Anyway, can someone post a solution for us wannabe math geniuses?

By Spectzix (Spectzix) on Saturday, July 24, 2004 - 02:09 am: Edit |

AD doesnt equal BC

By Xiggi (Xiggi) on Saturday, July 24, 2004 - 03:11 am: Edit |

Spectzix, AD does equal BC, and it plays an important role in solving this problem.

By Scm007 (Scm007) on Saturday, July 24, 2004 - 03:44 am: Edit |

You guys have got to be kidding me, all you do is bisect one of the angles. Childish almost.

By Apocalypse_Now (Apocalypse_Now) on Saturday, July 24, 2004 - 01:04 pm: Edit |

Really scm007, all you do is bisect one of the angles ??? I somehow doubt that. Anyway, post your solution if it's so simple.

By Tongos (Tongos) on Saturday, July 24, 2004 - 01:32 pm: Edit |

and its not childish its very grownupy. somehow i look at my work and i just see annoying constructions, confused of where i actually am.

By Apocalypse_Now (Apocalypse_Now) on Saturday, July 24, 2004 - 02:21 pm: Edit |

Allright, Xiggi! Post the mfing solution already! If someone wants to continue working on the problem, they simply don't have to read the solution.

By Xiggi (Xiggi) on Saturday, July 24, 2004 - 03:52 pm: Edit |

Scm007, sorry to disappoint you with such a childish problem. There are times that things are not what they appear.

If needed, I'll post one solution. I still think that some of you could solve it with additional hints. Thus, I will give the basis of the solution plus the hints to finish the proof.

Make sure that your drawing is correct with angles of 80, 80, 20.

First, this is for Spectzix

Step 1: You want to establish that AD equals BC

1. You will need to construct a figure (AGC) that will end up looking like an inverted triangle ABC where the acute 20 degrees angle will be at C, the segment AC will be common, and a new segment to be labeled CG will appear and yield two new intersections to be labeled F and E.

2. Construct a congruent triangle to ADC by reflectinng ADC along the axis formed by CD. Label the new triangle DGC. The newly formed segment GC will intersect AB at F and BE at H. Join A and G.

3. Take note that triangle AGC is congruent to ABC. From there, analyze the properties of triangle AGD: demonstrate that it is an equilateral triangle and that AG = GD = AD. Using the congruency of AGC and ABC, you can then establish that AD = BC.

Step 2:

In this step, you will analyze the properties of triangle AEB (isosceles) and the properties of EF (that you can establish as being // to BC via congruency rules).

Hints:

1. Note that AE = EB

2. Establish that triangles EFH and BCH are equilateral

3. Remember conclusion of step 1.

That is 80% of the solution but still need some light work to demonstrate that CDE is equal to 20 degrees.

PS There are probably different approaches and solutions. I hope I did not miss any labels!

By Seahorsekid (Seahorsekid) on Saturday, July 24, 2004 - 04:06 pm: Edit |

Perhaps I misread, but how can angle C be 20 degrees? isn't that one of the 80 degree angles and angle A is 20 degrees?

By Xiggi (Xiggi) on Saturday, July 24, 2004 - 05:08 pm: Edit |

Sea~

It is the acute angle of ACG that is 20 degrees.

By Slipstream99 (Slipstream99) on Saturday, July 24, 2004 - 08:30 pm: Edit |

question...doesn't GDF have to 120 degrees because DGF has to be 20 and DFG has to be 40 because BFC is 40 because ABE is 20 and FHB has to be 120 since BHC is 60 because triBHC is equilateral. So if GDF is 120, then ADE is 120 because opp angles of intersecting lines are equal. Then, if ADE is 120 degrees, CDE has to be 30 degrees because BDC has to be 30 since it is given that DBC is 80 and BCD is 70. So CDE CAN'T be 20.

By Rachelvishy (Rachelvishy) on Saturday, July 24, 2004 - 09:44 pm: Edit |

whoever said it was 130, thats not possible because....

* lets label the intersection of Line BE and Line CD "F" *

-CFB must equal 50 because BCF is 70 and CBF is 60, and they must add up to 180

-By property of Vertical Angles, angle EFD should be 50

-If x is 130, E would have to be zero, since EFD is 50. EFD and EDF would add up to 180, and angle FED would be zero. *this is not possible

By Slipstream99 (Slipstream99) on Saturday, July 24, 2004 - 10:11 pm: Edit |

^^^No, BCD is 70, BCF is 60

By Xiggi (Xiggi) on Saturday, July 24, 2004 - 11:45 pm: Edit |

*So if GDF is 120, then ADE is 120 because opp angles of intersecting lines are equal.*

Slight problem: GDF and ADE are NOT opposing angles.

DH and DE are different segments that do NOT form a line. There is no line EH. ADE is NOT equal to 120 degrees.

By Jaredthegreat (Jaredthegreat) on Sunday, July 25, 2004 - 12:38 am: Edit |

It was really late when I tried to figure out how to actually get to 20 being the right answer. I drew the triangle as it was given, and found all the angles that can easily be determined from there. I also found that EDC=AED-10 and BED=ADE-20. So if I could find either AED or ADE I could find EDC..... Then I drew some parallel and perpendicular lines because that would throw in some angles that would have to add up to 90. I found a 20 degree angle that formed a vertical angle with the other base angle ... Okay, I'm losing myself here....

It comes down to me mistaking a 30 degree angle that formed an isosceles triange with angle AED for angle AED. They were equal, but since I couldn't prove that it was an isosceles triangle, this was useless.(I had previously gone through the same thing with that aforementioned 20 degree angle.)

Anyhow, I was able to prove one thing, and that was that angle EDC had to be less, significantly less than 70 degrees.

The black lines are what was given. The blue are my parallel/perpendicular lines. The red shows where I made my mistake (the top angle is what I found, the bottom is the one I needed).

I've returned to trying to draw out a solution, this time on the computer, so I don't make the same misreads. I actually think that my newest addition to the drawing, the green line, was both my original forgotten solution, and does work.

The green line (KM) is the bisector of angle DKL, and perpendicular [bisector] to line DL. Which would make triangle DKL isosceles, so angle KLD which equals 20 degrees would be equal to angle EDC, making EDC 20 degrees.

I think I'm finished with it, and I've put my drawing online, at http://www.csun.edu/~jn673464/hardtriangleproblem.html

By Xiggi (Xiggi) on Sunday, July 25, 2004 - 02:08 am: Edit |

Jared~

How do you demonstrate that the KM is the bisector of angle DKL, and perpendicular [bisector] to line DL?

By Jaredthegreat (Jaredthegreat) on Sunday, July 25, 2004 - 12:16 pm: Edit |

Wouldn't you just draw the bisector of angle DKL and then the line perpendicular to DL that goes through K? The two are the same line.

Let the point where it intersects DL be "N". Then DN would equal NL (perpendicular bisector), KN=KN, and angle KNL=DNL (they're both right angles). So using SAS triangles DKN and LKN would be congruent. and angle EDC would equal angle KLD which is 20 degrees.

Or, if you're not sure that KM bisects DL: Angle DKN=LKN (Angle bisector). KN=KN. And angle KNL=DNL (right angles). So using ASA, triangles DKN and LKN would be congruent, and angle EDC would equal angle KLD which is 20 degrees.

By Xiggi (Xiggi) on Sunday, July 25, 2004 - 12:40 pm: Edit |

I am afraid you have a circular demonstration. You assume that the bisector is perpendicular because the triangle is isosceles and vice-versa. I think that you have to show that the right angles are equal to 90 degrees and that the bisector is really one.

Did I miss something?

By Jaredthegreat (Jaredthegreat) on Sunday, July 25, 2004 - 01:02 pm: Edit |

The last two paragraphs were just showing how you get EDC=20 from there.

In your method you had to construct a congruent line. Then you had to construct a congruent reflected triangle...

Why can't you construct the bisector of angle DKL and the perpend icular line to DL through K?

By Tongos (Tongos) on Sunday, July 25, 2004 - 03:11 pm: Edit |

i have it, but my method is wayyyyyy toooo longggg. and to be honest, i really dont feel like writing the whole proof down. maybe later though, when i feel like it. my real problem is how to always keep accurate on everything, when mishap on an equation and its gone.

By Apocalypse_Now (Apocalypse_Now) on Sunday, July 25, 2004 - 03:15 pm: Edit |

No Jared, not unless the triangle is isosceles. Altitudes and angle bisectors are not always the same.

It's probably hard to see that in this case, because if the answer is indeed 20 (as Xiggi says) then that triangle IS isosceles and the the angle bisector of angle DKL IS the same as the altitude through K. However, we obviously don't know that the answer is 20 yet, so we CANNOT assume that that triagle is isosceles.

By Jaredthegreat (Jaredthegreat) on Sunday, July 25, 2004 - 03:36 pm: Edit |

You can construct an angle bisector, right?

You can also construct a line that is perpendicular to a line and goes through a certain point, right?

In this case they are the same. If you were to draw the figure accurately you would see that if you constructed the bisector of angle DKL, and then KM perpendicular to DL through K that they are the same line.

I just don't see why this construction is not acceptable, while the one Xiggi uses is.

By Apocalypse_Now (Apocalypse_Now) on Sunday, July 25, 2004 - 04:08 pm: Edit |

Jared, you are only ASSUMING that they are the same. They don't have to be the same UNLESS triangle DKL is ISOSCELES. You have no proof that triangle DKL is isosceles, so you CANNOT assume that they are the same!

"If you were to draw the figure accurately you would see that if you constructed..."

A so-called "accurate" figure, is NOT proof. Just because the bisector and the altitude LOOK the same does not mean that they ARE the same. Think about it: Suppose they differ by 1/1000000 of a freaking milimeter. To the human eye they would look identical, while in reality they would be DIFFERENT.

I understand why you're confused because, as you said, they ARE the same in this case (because angle EDC is 20 degrees). However you have no proof of that, which is essentially the whole point of the problem: to prove that angle EDC is 20 degrees.

Your method of proof is essentially: angle EDC is 20 degrees because angle EDC is 20 degrees

By Miamidude (Miamidude) on Sunday, July 25, 2004 - 04:52 pm: Edit |

according to my insane calculations using the old techniques of euclidean geometry blah blah blah, i have found the answer to be somewhere greater than 0 but less than 30........ intriguing? indeed!

By Miamidude (Miamidude) on Sunday, July 25, 2004 - 04:57 pm: Edit |

ok i just actually looked at the problem, why isnt it 70 with the alternate interior angle crap? maybe i drew wrong, dunno

By Miamidude (Miamidude) on Sunday, July 25, 2004 - 05:04 pm: Edit |

dude can this problem be solved? Angle B and C are both 80 meaning angle A has to be 20. If angle EDC is less than than 30 then ADE would be 149ish or less. If this was true angle AED would be some small retarded triangle. Man why u giving problems with some retarded looking triangles?

By Miamidude (Miamidude) on Sunday, July 25, 2004 - 05:05 pm: Edit |

lets just say the answer is infinity

By Xiggi (Xiggi) on Sunday, July 25, 2004 - 06:56 pm: Edit |

Yes, Tongos, it takes a while to write down a solution. I was in the same boat and it does take a while to write down all the steps. However, you can take a few shortcuts when providing a solution. For instance, I did not describe all the steps to demonstrate that triangle XYZ is equilateral. I think that people who have an interest in a solution can fill the blanks for intermediate steps. For instance, I could write the solution step-by-step, but I find it more interesting to have other people build on the hints I provided.

Jared, I think that there might be a way to demonstrate that the triangle is isosceles, maybe by expanding the construction of the bisector and play with the 90 degrees angle. It would be nice to find more than one solution, as I am positive that there are different ways to solve this problem.

Miamidude, it is obvious that the problem can be solved, as I posted 80% of the solution.

By Tongos (Tongos) on Sunday, July 25, 2004 - 07:26 pm: Edit |

i know there are shortcuts, but in geometry, i always seem to overlook these.

By Anijen21 (Anijen21) on Sunday, July 25, 2004 - 08:14 pm: Edit |

I feel like I did something wrong, but I got this problem in like a second...I just drew a diagram

By Tongos (Tongos) on Sunday, July 25, 2004 - 08:21 pm: Edit |

then you did. no guessing

By Jaredthegreat (Jaredthegreat) on Sunday, July 25, 2004 - 10:31 pm: Edit |

"Your method of proof is essentially: angle EDC is 20 degrees because angle EDC is 20 degrees"

Not quite, but I understand what you're saying. I was trying to use the bisector and perpendicularity relationship to prove the triangle to be isosceles.

I'll continue trying to prove that triangle isosceles. An alternate method would be to find a way to prove that the triangle in the top right of my diagram, formed by the red line, is isosceles.

By Xiggi (Xiggi) on Sunday, July 25, 2004 - 11:26 pm: Edit |

Regarding the shortcuts, I may have been unclear. You CANNOT take shortcuts when developing the solution, each step is important. I meant one could use small shortcuts to describe the solution in writing, especially when your audience is able to follow the logic.

By Tongos (Tongos) on Sunday, July 25, 2004 - 11:38 pm: Edit |

yes, i know. i know what you meant xiggi.

By Legendofmax (Legendofmax) on Friday, July 30, 2004 - 05:02 pm: Edit |

http://img48.exs.cx/my.php?loc=img48&image=geometryproof.jpg&t=y

Ok this is what I have so far. Unfortunately I still don't know what ADE is or AED, which is what I wanted initially. However, although I can't prove it yet, I think that triangles ADC and DEC are similar triangles. If I can prove this, you can prove that angle EDC is 20 degrees.

By Legendofmax (Legendofmax) on Friday, July 30, 2004 - 05:37 pm: Edit |

oh dear god i have a headache

By Xiggi (Xiggi) on Friday, July 30, 2004 - 06:50 pm: Edit |

LegendofMax:

You are very close. Pay closer attention to what I typed earlier.

Step 2:

3. Remember conclusion of step 1. (AD = BC)

That is 80% of the solution but still need some light work to demonstrate that CDE is equal to 20 degrees.

This is what I've done from here. If you can establish that triangle DEF is isosceles, you will be able to find the angles FDE and DEF. You know that angle DFE is equal to 80 degrees. If you can demonstrate that EF = DF, you are done and the triangle DEF will be a 80-50-50 one, with angle EDF being 50. This then yields EDF - FDC = 20. And a big smile!

To get this done, look at what you have so far. You have several equilateral triangles and several isosceles ones. You can easily establish that EB = AF. See that AF = AD+DF and that EB = EH+HB. You can play around with the various segments and end with demonstrrting that DF = EF.

I could have written it out step by step, but I think you'll enjoy more working through the steps.

By Legendofmax (Legendofmax) on Friday, July 30, 2004 - 07:56 pm: Edit |

Oh duh, why didn't I see this. The two triangles have the same setup through the intersection at C. Congruency of AGC and ABC...

Ok, so given that AF = EB, we know that EB is composed of BH, a side of an eq. triangle, and EH, a part of a smaller eq. triangle. Since this same equilateral triangle exists at GAD, we know that AF is also composed of the equilateral-triangle side AD and DF. If AD = BH, then EH must equal DF. Since EF also equals EH because it's part of an equilateral triangle, DF is equal to EF, and thus DFE is isoceles. So 180 - 80 leaves us with 100 degrees evenly split between FDE and DEF, meaning each is 50 degrees. The angle we want is EDC, which shares its space with FDC measured at 30 degrees, which both occupy FDE. Therefore EDC = EDF - FDC, which is the same as 50 - 30, or 20!

THEREFORE ANGLE EDC IS 20 DEGREES.

QED fo sho

By Xiggi (Xiggi) on Friday, July 30, 2004 - 08:23 pm: Edit |

Nice.

By Legendofmax (Legendofmax) on Saturday, July 31, 2004 - 01:34 am: Edit |

That would be a killer SAT problem... too bad they'd never use it

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