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Solved the World's Hardest Easy Geometry Problem

Wednesday Dec 05, 2007

There's a famous "World's Hardest Easy Geometry Problem" in http://thinkzone.wlonk.com/MathFun/Triangle.htm, It's solved now. Here is the step by step for the solution.

From Triangle

Figure 1

From Figure 1, we have

[1] AC = BC

[2] BD = CD

From Triangle

Figure 2

In Figure 2, parallel to AB, we draw DF

Because [1] and AB||DF, we get

[3] DC = FC

[4] AD = BF

From Triangle

Figure 3

In Figure 3, we draw line AF.

From [4], and AB=BA, and angle DAB = FBA (=80),

the triangle ABD equals to triangle ABF. So

[5] AF = BD

From [2], AF = CD

From [3], we get

[6] AF = CF

So CAF = 20, EAF = 10, FAB = 60,

[7] DGF = 60

[8] AB = BG = GA

From [5], [7] and [8], we get

[9] DF = FG = GD

From Triangle

Figure 4

In Figure 4, we draw FH vertical to AC.

From [6] and angle CAF = ACF, we get triangle AFH equals to triangle CFH. So

[10] AH = CH

From Triangle

Figure 5

In Figure 5, we draw CI.

From [10] and FH vertical to AC, we get

[11] angle DCI = DAI = 10

[12] CI = AI

From [11], we get

[13] angle FCI = 10 = DCI

[13'] angle FCI = FAI

From Triangle

Figure 6

In Figure 6, extend CI to AB.

From [13], and angle ABC = BAC = 80, we get

line CI vertical to AB,

[14] line CI is the same as line CG

From [12], [13], [14], we get triangle CEI equals to triangle AGI. So

[15] EI = GI

From [12], [15], we get

[16] CG = AE

From [13'], [16], we get triangle CGF equals to triangle AEF. So

[17] EF = FG

Finally

From [9], [17], we get

[18] EF = DF

Because DF || AB, so

[19] angle CFD = CBA = 80

From [18], [19], we get

[20] angle DEF = EDF = 50

So angle X = angle DEF - angle AEF

= 50 - (180 - 70 - 60 - 20)

= 50 - 30

= 20

[42] Comments

Like this post?

Posted at 12:34AM Dec 05, 2007 by Simford Dong in General
Tags: geometry

Comments:

Wow, very good work. Nicely explained.

Posted by Garble on December 06, 2007 at 10:10 AM CST #

I like your proof but it could have been much simpler if you had avoided the perpendicular construction. Triangle AGC is congruent to triagle CEA so CE equals BA. That's all you needed after constructing the little equilateral triangle.

I proved it a completely different way using the fact that EG is also equal to AB. But this is (with the modification) the slick proof.

Posted by Michael H. on December 08, 2007 at 12:35 AM CST #

I prefer that people not post solutions to this puzzle openly on the internet. In my opinion, the solution to a puzzle, like the secret of a magic trick or the ending of a movie, should not be made too easily available.

Posted by Keith Enevoldsen on December 09, 2007 at 04:41 AM CST #

Hi Michael H.,
Right, the line FH seems not needed.
For EG equals to AB, I saw similar solution, which simply draws a circle that covers point A, point B and point E.

Posted by simford on December 10, 2007 at 08:56 AM CST #

Hi Keith Enevoldsen,
To me, this geometry puzzle is a bit different from the magic trick or movie. You only have one secret or ending, but you can have multiple solutions to this single puzzle. I only provided one, and it's sure not the best one.

Posted by simford on December 10, 2007 at 11:44 AM CST #

Thank you for solving my homework.

Posted by Homework Help on December 14, 2007 at 07:22 PM CST #

Great problem
I find a more straight forward proof,
after u proved the so-called "second hardest problem" (angle BAE, EAD, ABD, DBE) = (60, 20 , 50, 30) in ur link, then there is a very simple method to prove the "hardest problem" (60, 20, 70, 10)

the proof I use also gives the angle BDE for the combo (60, 20, 30, 50)

I dont want to spoil the fun for those trying, so I only give a brief description:
step 1, use equilateral triangle to prove "second hardest problem"
step 2, now we know the answer of "second hardest problem", we add the "50, 30" line to the "hardest problem", then use concyclic

Posted by Or2 on December 18, 2007 at 07:32 PM CST #

My solution is conceptually similar, but I think simpler. (See my blog, referenced below.)

Posted by Mike Coffin on December 28, 2007 at 11:27 AM CST #

There's a problem in

[9] DF = FG = GD

because if that's true, then each of the angle in that triangle should be 60 degrees.

From Angle GAB and Angle GBA, Angle AGB is equal to 50.

60 <> 50

Therefore, this solution might not work.

Posted by Someone on January 16, 2008 at 08:56 PM CST #

Sorry

Mistake ;)

Posted by Someone on January 16, 2008 at 09:01 PM CST #

very nice

Posted by very nice on February 11, 2008 at 06:39 AM CST #

it's interesting how the first and 2nd problem are different... I basically substitute them with a,b,c,d and tries to solve them -.-
my epic fail...

Posted by Mgccl on February 23, 2008 at 05:51 AM CST #

And THAT'S why we invented trig. :)

Posted by matt on March 31, 2008 at 02:04 PM CST #

Thx 4 helping me solve this

Posted by johnny on April 01, 2008 at 08:04 PM CST #

How to prove lines HF, CI and AE intersect at point I?

Posted by 76.30.92.53 on May 12, 2008 at 05:03 AM CST #

20 is correct answer. I used sin. cos laws to calculate it. just don't know how do you prove lines HF, CI and AE intersect at point I.

Posted by 76.30.92.53 on May 12, 2008 at 05:06 AM CST #

Thanks alot for your generous donations. I have spent considerable amount of time trying to figure out the logical way of proving x= 20 degrees, but kept going aound the circle. It was an eye-opening experience. Thank you again.

Posted by calculuslover on May 20, 2008 at 05:10 AM CST #

How can you make the assumption (read: prove) that D is the midpoint of AC? I realize you could measure it, but that's not really proving anything..

Posted by Some Guy on June 07, 2008 at 11:04 AM CST #

@ Some guy

Because angle C and B are equal.

That means wherever they meet is the midpoint of 'that' triangle.

I think that hsould make sense

Posted by Brian on August 24, 2008 at 01:16 AM CST #

everybody here who asks questions like why is d midpoint knows nothing about geometry... So gtfo
This proof is very well done!! And those points intersect because of properties of isosceles triangles!!!!!!! Gtfo!!!!

Posted by Wow@ on October 24, 2008 at 12:49 PM CST #

Am I the only the one that just used the properties
- the corners of a triangle add up to 180
- a straight line makes a 180 angle
- a corner can be partitioned (like the corner in A is partitioned into 10 and 70)

And then just made it into a linear system of equations that was pretty much trivial to solve (4 equations, 4 unknowns). Or was this not allowed or something?

Posted by firlitor on November 11, 2008 at 07:54 AM CST #

I did the same, firlitor, but it's not enough. Call

angle EDB x
angle CDE y
angle CED z
angle DEA w

the equations are

z + w = 150
x + y = 140
x + w = 130
y + z = 160

This is four equations in four unknowns, but they're not independent :-(

Posted by Mike on November 11, 2008 at 08:50 AM CST #

nice work

Posted by danniella on December 08, 2008 at 11:35 AM CST #

Your solution is way too complicated. Don't need to draw FH, only CG. Then notice that triangles ACE and ACG are equal (side and two angles). This CE = AG, but CF = AF, hence GF=EF, so the triangle DEF is isosceles and thus x + 30 = 50.

Posted by Bill on December 21, 2008 at 01:02 PM CST #

wow. this problem has been rediculous. i worked on it for about 2 hours and then decided to look for the answer. but that is amazing that you found the answer.

Posted by Kk on December 23, 2008 at 11:00 AM CST #

But who really cares...because I'm never going to need to do this in my life.

Posted by Mitch on January 11, 2009 at 05:02 AM CST #

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Posted by 123.230.162.211 on March 20, 2009 at 09:16 AM CST #

Very nice proof. I will use this as my geometry project.

Posted by Michael on May 02, 2009 at 06:36 AM CST #

You guys should all know the easiest way to solve this problem (Triangle 1); it took me about two minutes and no constructions.

1. In triangle ABE, there is a 70 degree angle and a 60 degree angle so the third angle (AEB) must be 50 degrees.

2. In triangle ACE, there is a 20 degree angle and a 10 degree angle so the third angle (AEC) must be 150 degrees.

3. Since both angles are on the same line, they should add up to 180 degrees. But wait! 150 + 50 = 200, not 180. This is because there is an overlapping section of both angles (AED). To make the measures correct at 180, you must subtract 200 - 180, and that is 20 degrees.

Posted by Leonard on May 31, 2009 at 12:11 PM CST #

leonard u did urs wrong just lucky it came out the same way

Posted by vbc on June 07, 2009 at 05:42 AM CST #

How is this information acheived in step 1:
[2] BD = CD

???

Posted by Daniel on June 24, 2009 at 02:01 PM CST #

Ah, DBC = DCB... Sorry fo rdouble posting.

Posted by Daniel on June 24, 2009 at 02:02 PM CST #

Hi,
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Posted by 498a on July 21, 2009 at 04:20 PM CST #

This is the simplest solution to a complex geometry problem.
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Posted by Tax Filing on July 30, 2009 at 07:13 PM CST #