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Simplest known solution to “World's Hardest Easy Geometry Problem”?

Finding an angle within an 80-80-20 isosceles triangleDetermine angle $x$ using only elementary geometry“World's Hardest Easy Geometry Problem”Finding the angle $x$.Simpler solution to this geometry/trig problem?Simpler solution to a geometry problemPlane Geometry problem“World's Hardest Easy Geometry Problem”Easy little geometry probGeometry problem, find lengthsTiling the unit square with right trianglesInteresting Geometry Problem. The Converse is easy.A geometry problem - easy with trigonometry, harder without itSimplest way to solve this trapezoid geometry problem?

4

0

$begingroup$

I've seen multiple solutions to this well-known problem (posed, among other sites, at http://thinkzone.wlonk.com/MathFun/Triangle.htm). None of the solutions are particularly simple. I was wondering if trigonometry (which the site I mentioned instructs you not to use) would yield an easier solution. Can anybody provide one?

EDIT: Multiple users have marked this as a duplicate question. The question they are referring to is far more general and probably far more difficult. All the answers that were given to it either lacked proofs or were incorrect. I'm looking for a specific, detailed, simple proof for this particular problem that takes advantage of trigonometry instead of elementary geometry.

geometry trigonometry

share|cite|improve this question

edited Feb 10 at 22:00

Glorfindel

3,41581830

asked May 18 '14 at 2:59

user151686user151686

21113

$endgroup$

• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  $begingroup$
  I checked out that link and I wouldn't call this a duplicate. That one was more of a generalized problem, with vague, very general, and possibly incomplete answers. I'm interested in a simple solution for this particular problem.
  $endgroup$
  – user151686
  May 18 '14 at 3:38
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  On other hand, without trigonometry this math.stackexchange.com/questions/63819/… is a duplicate/related. And also the generalisation has a non-trig related :-D math.stackexchange.com/questions/6942/…
  $endgroup$
  – arivero
  Aug 18 '15 at 11:14
  
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Lots of these problems (google "adventitious angle problems") can be solved by identifying the picture as part of a regular $n$-gon (for some $n$) with its diagonals. I take it this is one such case? In that case, that's probably the simplest solution.
  $endgroup$
  – darij grinberg
  Mar 13 at 6:15
  

  
  
  
  

add a comment | 

4

0

$begingroup$

I've seen multiple solutions to this well-known problem (posed, among other sites, at http://thinkzone.wlonk.com/MathFun/Triangle.htm). None of the solutions are particularly simple. I was wondering if trigonometry (which the site I mentioned instructs you not to use) would yield an easier solution. Can anybody provide one?

EDIT: Multiple users have marked this as a duplicate question. The question they are referring to is far more general and probably far more difficult. All the answers that were given to it either lacked proofs or were incorrect. I'm looking for a specific, detailed, simple proof for this particular problem that takes advantage of trigonometry instead of elementary geometry.

geometry trigonometry

share|cite|improve this question

edited Feb 10 at 22:00

Glorfindel

3,41581830

asked May 18 '14 at 2:59

user151686user151686

21113

$endgroup$

• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  $begingroup$
  I checked out that link and I wouldn't call this a duplicate. That one was more of a generalized problem, with vague, very general, and possibly incomplete answers. I'm interested in a simple solution for this particular problem.
  $endgroup$
  – user151686
  May 18 '14 at 3:38
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  On other hand, without trigonometry this math.stackexchange.com/questions/63819/… is a duplicate/related. And also the generalisation has a non-trig related :-D math.stackexchange.com/questions/6942/…
  $endgroup$
  – arivero
  Aug 18 '15 at 11:14
  
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Lots of these problems (google "adventitious angle problems") can be solved by identifying the picture as part of a regular $n$-gon (for some $n$) with its diagonals. I take it this is one such case? In that case, that's probably the simplest solution.
  $endgroup$
  – darij grinberg
  Mar 13 at 6:15
  

  
  
  
  

add a comment | 

$begingroup$

I've seen multiple solutions to this well-known problem (posed, among other sites, at http://thinkzone.wlonk.com/MathFun/Triangle.htm). None of the solutions are particularly simple. I was wondering if trigonometry (which the site I mentioned instructs you not to use) would yield an easier solution. Can anybody provide one?

EDIT: Multiple users have marked this as a duplicate question. The question they are referring to is far more general and probably far more difficult. All the answers that were given to it either lacked proofs or were incorrect. I'm looking for a specific, detailed, simple proof for this particular problem that takes advantage of trigonometry instead of elementary geometry.

geometry trigonometry

share|cite|improve this question

edited Feb 10 at 22:00

Glorfindel

3,41581830

asked May 18 '14 at 2:59

user151686user151686

21113

$endgroup$

share|cite|improve this question

edited Feb 10 at 22:00

Glorfindel

3,41581830

asked May 18 '14 at 2:59

user151686user151686

21113

share|cite|improve this question

edited Feb 10 at 22:00

Glorfindel

3,41581830

asked May 18 '14 at 2:59

user151686user151686

21113

edited Feb 10 at 22:00

Glorfindel

3,41581830

edited Feb 10 at 22:00

Glorfindel

3,41581830

asked May 18 '14 at 2:59

user151686user151686

21113

asked May 18 '14 at 2:59

user151686user151686

21113

1 Answer
1

active

oldest

votes

0

$begingroup$

First, take note that $triangle BCD$ is isosceles, and that $angle BDC$ is 140 degrees.

Assuming arbitrarily sides $BD$ and $CD$ are of unit length, we can use the law of cosines with $angle BDC$ to obtain that the length of $CB$ is $approx 1.87939$.

Similarly, notice that triangle ABC is isosceles. $AC$ must then be congruent to $CB$, and have the same measure of $approx 1.87939$.

Using law of cosines with $AC$, $CB$, and $angle ACB$, we determine that $AB$ has measure $approx 0.652705$.

To determine the length of $AD$, we use the law of cosines with $AB$, $BD$, and $angle ABD$. We learn that $AD$ has measure $approx 0.879385$.

Using the law of sines with $angle BAE$, $angle BEA$, and $AB$, we can determine that the length of $BE$ is $approx 1.22668$.

Using the law of cosines with $BD$, $BE$, and $angle DBE$, we can determine the length of $DE$, which is $approx 0.446475$.

Finally, using the law of sines, we can calculate the angle measure of $angle BED$ using all three sides of the involved triangle, $BE$, $DE$, and $BD$. This angle is 50 degrees. Since $angle BEA$ was 30 degrees, we can finally conclude that angle $angle DEA$ is 20 degrees!

share|cite|improve this answer

answered May 18 '14 at 4:21

beanshadowbeanshadow

180213

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  This took me a long time to write and I probably made mistakes. Correct me if I goofed.
  $endgroup$
  – beanshadow
  May 18 '14 at 4:24
  

  
  
  
  

add a comment | 

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0

$begingroup$

First, take note that $triangle BCD$ is isosceles, and that $angle BDC$ is 140 degrees.

Assuming arbitrarily sides $BD$ and $CD$ are of unit length, we can use the law of cosines with $angle BDC$ to obtain that the length of $CB$ is $approx 1.87939$.

Similarly, notice that triangle ABC is isosceles. $AC$ must then be congruent to $CB$, and have the same measure of $approx 1.87939$.

Using law of cosines with $AC$, $CB$, and $angle ACB$, we determine that $AB$ has measure $approx 0.652705$.

To determine the length of $AD$, we use the law of cosines with $AB$, $BD$, and $angle ABD$. We learn that $AD$ has measure $approx 0.879385$.

Using the law of sines with $angle BAE$, $angle BEA$, and $AB$, we can determine that the length of $BE$ is $approx 1.22668$.

Using the law of cosines with $BD$, $BE$, and $angle DBE$, we can determine the length of $DE$, which is $approx 0.446475$.

Finally, using the law of sines, we can calculate the angle measure of $angle BED$ using all three sides of the involved triangle, $BE$, $DE$, and $BD$. This angle is 50 degrees. Since $angle BEA$ was 30 degrees, we can finally conclude that angle $angle DEA$ is 20 degrees!

share|cite|improve this answer

answered May 18 '14 at 4:21

beanshadowbeanshadow

180213

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  This took me a long time to write and I probably made mistakes. Correct me if I goofed.
  $endgroup$
  – beanshadow
  May 18 '14 at 4:24
  

  
  
  
  

add a comment | 

0

$begingroup$

First, take note that $triangle BCD$ is isosceles, and that $angle BDC$ is 140 degrees.

Assuming arbitrarily sides $BD$ and $CD$ are of unit length, we can use the law of cosines with $angle BDC$ to obtain that the length of $CB$ is $approx 1.87939$.

Similarly, notice that triangle ABC is isosceles. $AC$ must then be congruent to $CB$, and have the same measure of $approx 1.87939$.

Using law of cosines with $AC$, $CB$, and $angle ACB$, we determine that $AB$ has measure $approx 0.652705$.

To determine the length of $AD$, we use the law of cosines with $AB$, $BD$, and $angle ABD$. We learn that $AD$ has measure $approx 0.879385$.

Using the law of sines with $angle BAE$, $angle BEA$, and $AB$, we can determine that the length of $BE$ is $approx 1.22668$.

Using the law of cosines with $BD$, $BE$, and $angle DBE$, we can determine the length of $DE$, which is $approx 0.446475$.

Finally, using the law of sines, we can calculate the angle measure of $angle BED$ using all three sides of the involved triangle, $BE$, $DE$, and $BD$. This angle is 50 degrees. Since $angle BEA$ was 30 degrees, we can finally conclude that angle $angle DEA$ is 20 degrees!

share|cite|improve this answer

answered May 18 '14 at 4:21

beanshadowbeanshadow

180213

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  This took me a long time to write and I probably made mistakes. Correct me if I goofed.
  $endgroup$
  – beanshadow
  May 18 '14 at 4:24
  

  
  
  
  

add a comment | 

$begingroup$

First, take note that $triangle BCD$ is isosceles, and that $angle BDC$ is 140 degrees.

Assuming arbitrarily sides $BD$ and $CD$ are of unit length, we can use the law of cosines with $angle BDC$ to obtain that the length of $CB$ is $approx 1.87939$.

Similarly, notice that triangle ABC is isosceles. $AC$ must then be congruent to $CB$, and have the same measure of $approx 1.87939$.

Using law of cosines with $AC$, $CB$, and $angle ACB$, we determine that $AB$ has measure $approx 0.652705$.

To determine the length of $AD$, we use the law of cosines with $AB$, $BD$, and $angle ABD$. We learn that $AD$ has measure $approx 0.879385$.

Using the law of sines with $angle BAE$, $angle BEA$, and $AB$, we can determine that the length of $BE$ is $approx 1.22668$.

Using the law of cosines with $BD$, $BE$, and $angle DBE$, we can determine the length of $DE$, which is $approx 0.446475$.

Finally, using the law of sines, we can calculate the angle measure of $angle BED$ using all three sides of the involved triangle, $BE$, $DE$, and $BD$. This angle is 50 degrees. Since $angle BEA$ was 30 degrees, we can finally conclude that angle $angle DEA$ is 20 degrees!

share|cite|improve this answer

answered May 18 '14 at 4:21

beanshadowbeanshadow

180213

$endgroup$

share|cite|improve this answer

answered May 18 '14 at 4:21

beanshadowbeanshadow

180213

answered May 18 '14 at 4:21

beanshadowbeanshadow

180213

answered May 18 '14 at 4:21

beanshadowbeanshadow

180213