# Looking easy but hard question on geometry

#### Masaru

##### New member
Please find the attached file and look at the diagram of a triangle.

As you can see, we need to find the size of the angle marked "x" (i.e. angle DEF).

You can identify the sizes of surrounding angles using your basic knowledge of geometry as follows:

Angle ACD = 20 (Angle sum of triangle = 180)
Angle AFB = 50
(Angle sum of triangle = 180)
Angle DFE = 50 (Opposite angles are equal in size)
Angle AFD = 130 (Angles on a straight line add up to 180)
Angle BFE = 130
(Opposite angles are equal in size)
Angle ADB = 40
(Angle sum of triangle = 180)
Angle AEB = 30
(Angle sum of triangle = 180)

But how can it possible to find out the size of the angle DEF?

Honestly, I am stuck and cannot see any way to find it out.

The diagram on the file is drawn to scale, so I actuallly measured the angle DEF to see the size of it, then I found that it is exactly 20 degrees, therefore, the answer must be 20.

If so, the triangles EDC and FEB are similar, however, I cannot find any way to prove that they are similar only with these pieces of information on the diagram without measuring the angle using a protractor.

I would much appreciate it if someone can help me with this question.

Thank you.

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#### stapel

##### Super Moderator
Staff member
Please find the attached file and look at the diagram of a triangle.

As you can see, we need to find the size of the angle marked "x" (i.e. angle DEF).

You can identify the sizes of surrounding angles using your basic knowledge of geometry as follows:

Angle ACD = 20 (Angle sum of triangle = 180)
Angle AFB = 50
(Angle sum of triangle = 180)
Angle DFE = 50 (Opposite angles are equal in size)
Angle AFD = 130 (Angles on a straight line add up to 180)
Angle BFE = 130
(Opposite angles are equal in size)
Angle ADB = 40
(Angle sum of triangle = 180)
Angle AEB = 30
(Angle sum of triangle = 180)

But how can it possible to find out the size of the angle DEF?
For ease of computation, I have labelled angle EDC as "a", DEC as "b", and EDF as "c". Including "x" and looking at the straight lines ADC and BEC and triangles CDE and DEF, we get the following equations:

. . . . .$$\displaystyle 40\, +\, c\, +\, a\, =\, 180$$

. . . . .$$\displaystyle 30\, +\, x\, +\, b\, =\, 180$$

. . . . .$$\displaystyle x\, +\, c\, +\, 50\, =\, 180$$

. . . . .$$\displaystyle a\, +\, b\, +\, 20\, =\, 180$$

This is a linear system of four equations in four unknowns. Thus, it is solvable algebraically. After some rearranging, we get:

. . . . .$$\displaystyle a\, \hphantom{+\, b}\, +\, c\, \hphantom{+\, x}\, =\, 140$$

. . . . .$$\displaystyle \hphantom{a\, +}\, b\, \hphantom{+\, c}\, +\, x\, =\, 150$$

. . . . .$$\displaystyle \hphantom{a\,+}\, \hphantom{b\, +}\, c\, +\, x\, =\, 130$$

. . . . .$$\displaystyle a\, +\, b\, \hphantom{+\, c}\, \hphantom{+\, x}\, =\, 160$$

However, when I solve this system, I get:

. . . . .$$\displaystyle a\, =\, 10\, +\, x$$

. . . . .$$\displaystyle b\, =\, 150\, -\, x$$

. . . . .$$\displaystyle c\, =\, 130\, -\, x$$

...which isn't entirely helpful.

I'm wondering why they specified that the picture is to scale (other than to allow you to measure the angle directly). Clearly, the equations we've come up with are missing something, but I'm not seeing what that is right now. I'll keep working on it....

#### Dr.Peterson

##### Elite Member
Please find the attached file and look at the diagram of a triangle.

As you can see, we need to find the size of the angle marked "x" (i.e. angle DEF).

You can identify the sizes of surrounding angles using your basic knowledge of geometry as follows:

Angle ACD = 20 (Angle sum of triangle = 180)
Angle AFB = 50
(Angle sum of triangle = 180)
Angle DFE = 50 (Opposite angles are equal in size)
Angle AFD = 130 (Angles on a straight line add up to 180)
Angle BFE = 130
(Opposite angles are equal in size)
Angle ADB = 40
(Angle sum of triangle = 180)
Angle AEB = 30
(Angle sum of triangle = 180)

But how can it possible to find out the size of the angle DEF?

Honestly, I am stuck and cannot see any way to find it out.

The diagram on the file is drawn to scale, so I actuallly measured the angle DEF to see the size of it, then I found that it is exactly 20 degrees, therefore, the answer must be 20.

If so, the triangles EDC and FEB are similar, however, I cannot find any way to prove that they are similar only with these pieces of information on the diagram without measuring the angle using a protractor.
This is a classic problem that requires fairly subtle tricks. In fact, your picture could come from the "very small hint" given here: World's Hardest Easy Geometry Problem. Solutions are not given there, but can be found elsewhere if you searched appropriately (which I am not suggesting).

I have seen many solutions to similar problems; they typically involve constructing at least one extra line, perhaps with an angle the same as another in the figure so that you have an isosceles triangle or similar triangles; some have involved reflections of one line in another, etc. One involved showing that the figure is equivalent to part of a figure composed of diagonals of an 18-gon. I haven't put any effort into this one just yet; I just know not to expect it to be either easy or impossible. Be creative.

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#### Masaru

##### New member
This is a classic problem that requires fairly subtle tricks. In fact, your picture could come from the "very small hint" given here: World's Hardest Easy Geometry Problem. Solutions are not given there, but can be found elsewhere if you searched appropriately (which I am not suggesting).
Thank you for your help. I did not know that this has come from a website. With the help of
yma16, I was able to figure out how to work it out, but this is a really hard question because you need to draw even a circle!

#### Masaru

##### New member
"because", the three dots mean "View attachment 9179View attachment 9180therefore"

On the second page, I should have said that name the three base angles to be S,Q, and R. Also, I should have said angle ODY=60=angle DYO.
Thank you very much for your help.
Initially, I found your explanation on your notes hard to understand, but eventually I was able to follow everything. Thank you very much, indeed.

#### Masaru

##### New member
I'm wondering why they specified that the picture is to scale (other than to allow you to measure the angle directly). Clearly, the equations we've come up with are missing something, but I'm not seeing what that is right now. I'll keep working on it....
Thank you so much for spending lots of time trying to work out this hard question for me.
But now I see how to work it out with the help of
yma16.
This is, indeed, a hard question though it looks simple...

#### Dr.Peterson

##### Elite Member
Thank you for your help. I did not know that this has come from a website. With the help of
yma16, I was able to figure out how to work it out, but this is a really hard question because you need to draw even a circle!
As I mentioned, there are many ways to solve these problems; some involve circles, some don't. Actually, it is not so much a matter of solving as of making and proving a conjecture: this is why the hint was to draw it to scale, so that you can not only guess that the answer is 20, but also experiment with the figure to find interesting features to use. yma16's method probably started with drawing lines and observing that O appeared to be the center of the circumcircle of ABE.

Here is one source I know for various solutions to these problems, which all depend on interesting properties of the 80-80-20 triangle: Cut-the-knot. (I'm only showing you this because a solution has already been given. Perhaps you can see why it was impossible to guide you to discover a solution of your own, which I would have preferred; I have no idea what hints might lead to a solution you could discover, unless I knew a lot more about you.)

yma16: Was this your own discovery, or did you, like me, have a source? If it's your own, perhaps you can tell us if I'm right about how you found it. The process of discovery is more interesting than the solution!

#### yma16

##### Junior Member
As I mentioned, there are many ways to solve these problems; some involve circles, some don't. Actually, it is not so much a matter of solving as of making and proving a conjecture: this is why the hint was to draw it to scale, so that you can not only guess that the answer is 20, but also experiment with the figure to find interesting features to use. yma16's method probably started with drawing lines and observing that O appeared to be the center of the circumcircle of ABE.

Here is one source I know for various solutions to these problems, which all depend on interesting properties of the 80-80-20 triangle: Cut-the-knot. (I'm only showing you this because a solution has already been given. Perhaps you can see why it was impossible to guide you to discover a solution of your own, which I would have preferred; I have no idea what hints might lead to a solution you could discover, unless I knew a lot more about you.)

yma16: Was this your own discovery, or did you, like me, have a source? If it's your own, perhaps you can tell us if I'm right about how you found it. The process of discovery is more interesting than the solution!
This is not an isosceles triangle. I think it is harder than the isosceles triangle problems. I do not have source from the web. It was an old problem and was solved by a few people together through discussion and I was one of them.

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#### Dr.Peterson

##### Elite Member
This is not an isosceles triangle. I think it is harder than the isosceles triangle problems. I do not have source from the web. It was an old problem and was solved by a few people together through discussion and I was one of them.
It IS an isosceles triangle: both base angles are 80 degrees.

But it's great that your group worked it out. It's a big challenge.

#### yma16

##### Junior Member
You are right. Thank you.

It IS an isosceles triangle: both base angles are 80 degrees.

But it's great that your group worked it out. It's a big challenge.
I'll try to do it without the circle. I thought it over and recalled that we needed the circle to solve it.

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#### Dr.Peterson

##### Elite Member
I'll try to do it without the circle. I thought it over and recalled that we needed the circle to solve it.
I think the circle was a wonderful idea; I wasn't saying there's any reason to try for another proof.

But this problem is the type where people can enjoy the challenge of finding as many different proofs as they can, so have fun with it if you wish.

#### Dr.Peterson

##### Elite Member
Interesting. I think the last expression was copied wrong, but the result is correct. I'd be happier with a method that gave an exact value (not one that you get as a decimal approximation); it might be fun to try to show that it is exactly 20.

#### Dr.Peterson

##### Elite Member
You're my kind of person! You took the challenge.

I'm sure that work can be shortened a bit, but it's definitely a proof.