Can I prove if this trapezoid is possible/impossible?

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pineapplewithmouse

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So I had a math test today, and there was a trig question with a trapezoid.
I couldn't figure out the answer for the question, so after the test I asked my friend what he got, and I used their way of solving the problem to try and solve it, and I got a different answer from theirs. I tried to draw the trapezoid in geogebra, to see which one of us is right, but not only that I couldn't draw the trapezoid with our solutions, I couldn't draw it at all. So I thought, maybe this trapezoid is impossible (or I just can't draw properly), but if so, how can I prove it?
1681658631312.png
AD=BC=3b
DE=2b
EC=4b
The angle BAD is equal to α (degrees)

And then there are 2 properties that just look like they can not exist together:
AB=1.25EC=5b
The angle AEB is equal to 90 degrees.

This is what my friend did and it seems to work theoretically:
By using the law of cosines in triangles ADE and BCE, you can see the AE is equal to sqrt(13b^2+12b^2*cosα) and BE is equal to sqrt(25b^2+24b^2*cosα).
Then by using the pythagorean theorem in triangle AEB, you get that α=111.17 degrees.
After you find α, the problem tells you that the perimiter of the escribed circle of the trapezoid is equal to 25.
With that information I got that b=1.312, and my friend got b=1.107.
Both options don't seem to work in geogebra as you can see:
b=1.107
1681660220354.png
b=1.312
1681660267567.png
In both cases AB is equal to approximately 3.8b, and AEB is not equal to 90 degrees.
Can I somehow prove that this trapezoid is possible/impossible?
I will really appreciate any help.
 
Here is my drawing, using only the side lengths given (with b=1), and then measuring the angles.

1681663076376.png​

Clearly the data are inconsistent.

Are you sure you have stated the problem exactly as given?
 
khansaheb said:
Was this property GIVEN to you?
Click to expand...
Yes, it was GIVEN, not a conclusion from calculations, GIVEN
Dr.Peterson said:
Here is my drawing, using only the side lengths given (with b=1), and then measuring the angles.


Clearly the data are inconsistent.

Are you sure you have stated the problem exactly as given?
Click to expand...
The problem stated:
AD=BC=3b
DC=6b (Or it said that DC=2AD? I do not remember clearly, but probably it's that)
The area of BCE is twice as big as the area of ADE (from here I concluded that DE=2b and EC=4b)
Angle BAD = α
Question 1: Express the length of AE and BE using α and b.
After that, the problem stated: AB=1.25CE and that AE is perpendicular to BE.
Question 2: Find α.
This is where I was originally stuck, and my friend used the pythagorean theorem to figure out that α=111.17 degrees.
Then the problem stated: The perimeter of the escribed circle of the trapezoid is equal to 25 (Although, because I was stuck in the previous question, I didn't read the later part, so it's based on my friend's memory)
Question 3: Find b.
My friend got b=1.107, I got b=1.312 if I use the apparent fact that α=111.17.
 
Why do you say that the measure of angle BAD is equal to the measure of angle ABC?

That is the basis for your splitting the length of line DC into 2b and 4b. And in your outline of the text of the problem, that does not seem to be specified.
 
NotJeffM said:
Why do you say that the measure of angle BAD is equal to the measure of angle ABC?
Click to expand...
Each set of an isosceles trapezoid's base angles are equal to one another.
NotJeffM said:
That is the basis for your splitting the length of line DC into 2b and 4b.
Click to expand...
No it's not? I got it because the problem stated that "The area of BCE is twice as big as the area of ADE".
The heights of the bases are equal to each other, so it means that EC=2DE.
 
pineapplewithmouse said:
The problem stated:
AD=BC=3b
DC=6b (Or it said that DC=2AD? I do not remember clearly, but probably it's that)
The area of BCE is twice as big as the area of ADE (from here I concluded that DE=2b and EC=4b) [this is a valid conclusion]
Angle BAD = α
Question 1: Express the length of AE and BE using α and b.
After that, the problem stated: AB=1.25CE and that AE is perpendicular to BE.
Question 2: Find α.
This is where I was originally stuck, and my friend used the pythagorean theorem to figure out that α=111.17 degrees.
Then the problem stated: The perimeter of the escribed circle of the trapezoid is equal to 25 (Although, because I was stuck in the previous question, I didn't read the later part, so it's based on my friend's memory)
Question 3: Find b.
My friend got b=1.107, I got b=1.312 if I use the apparent fact that α=111.17.
Click to expand...
I've shown that the description as you understand it is not consistent; AEB is not a right angle (in fact can't be, if the sides of ABCD are as stated; there is no location of E that would work).

Possibly what is said about DC is not either of the things you say. I'd be much more confident if I saw an image of the actual problem. We may need to wait until you get the test back.
 
Can I ask what app you are using in drawing the sketches? Thanks.
 
pineapplewithmouse said:
Question 1: Express the length of AE and BE using α and b.
After that, the problem stated: AB=1.25CE and that AE is perpendicular to BE.
Question 2: Find α.
Click to expand...
Thank you for supplying us with the question. It really is hard to offer help without it.
 
Dr.Peterson said:
As the OP said, we are both using GeoGebra (geogebra.org).
Click to expand...
Thanks. I did have a quick read to see if it was mentioned. But simply I missed that bit.

This name certainly rings a bell with me. Now with a check of its online version, it's have gone through improvements!
 
Dr.Peterson said:
I've shown that the description as you understand it is not consistent; AEB is not a right angle (in fact can't be, if the sides of ABCD are as stated; there is no location of E that would work).

Possibly what is said about DC is not either of the things you say. I'd be much more confident if I saw an image of the actual problem. We may need to wait until you get the test back
Click to expand...
I am sure I remember those parts of the question, but of course it will be much better if I had the question itself.
I will try getting the questions from the test.
 
Update 1: The teacher said that she will send me the test later today.
Update 2: I thought it's a good idea to see if the areas of ADE, ABE and BEC will match the area of ABCD, because I thought, if the trapezoid is impossible, the areas won't match, and the equation won't have a real solution.
ADE: [math]\frac{1}{2} * 3b*2b*sin(180-α) = 3b^2sinα[/math]ABE: [math]\frac{1}{2}*\sqrt{13b^2+12b^2cosα}*\sqrt{25b^2+24b^2cosα}*sin90=\frac{b^2\sqrt{612cosα+144cos2α+469}}{2}[/math]BEC: [math]\frac{1}{2}*3b*4b*sin(180-α)=6b^2sinα[/math]
Height of DC:
[math]\frac{4b*h}{2}=6b^2sinα[/math][math]h=3bsinα[/math]ABCD: [math]\frac{3bsinα(6b+5b)}{2}=16.5b^2sinα[/math]So now:
[math]3b^2sinα+\frac{b^2\sqrt{612cosα+144cos2α+469}}{2}+6b^2sinα=16.5b^2sinα[/math]And after a lot of Wolfram Alpha'ing and thinking, I got the answer:
[math]α=2πn±arccos(\frac{1}{48}(\sqrt{1801}-51))[/math]Which is of course in radians, so after converting it to degrees, we get [math]α = 100.2748915°...[/math]Which is also, guess what? Not working!
1681723026966.png
 
You're missing the point. There is too much information, which is incompatible; so if you just keep doing different things with different combinations of the data, you will keep getting different answers.

If you use only the length information, and consider a trapezoid with bases 5 and 6, and sides 3 and 3, then its shape is fully determined. You can calculate alpha, and it work out to the 99.5941 degrees that my drawing showed. So that angle is not really a variable, and the angle at E can't be made to be 90 degrees without changing something in the problem.
 
Dr.Peterson said:
You're missing the point. There is too much information, which is incompatible; so if you just keep doing different things with different combinations of the data, you will keep getting different answers.

If you use only the length information, and consider a trapezoid with bases 5 and 6, and sides 3 and 3, then its shape is fully determined. You can calculate alpha, and it work out to the 99.5941 degrees that my drawing showed. So that angle is not really a variable, and the angle at E can't be made to be 90 degrees without changing something in the problem.
Click to expand...
But I can't just say to my teacher: "The information is contradicting itself", I need to prove why is that.
So how can I say it? Because of the given lengths, the angle can not be 90 degrees?
 
pineapplewithmouse said:
But I can't just say to my teacher: "The information is contradicting itself", I need to prove why is that.
So how can I say it? Because of the given lengths, the angle can not be 90 degrees?
Click to expand...
First, wait until you have a copy of the problem so we can make sure it is really wrong.

Next, do what I suggested: Use only the given lengths and determine what the angles ARE. You will be able to show that alpha has a fixed value, and that AEB is not a right angle. I've given you all the information needed to prove your (my) claim; so do so, in order to earn the credit. It is not hard.
 
I got the original problem, and it goes like this:
In the isosceles trapezoid ABCD the point E is on the base CD.
Given that: CD=2AD. We will call AD=3b, and angle BAD = α.
The area of triangle BCE is twice as big as the area of triangle ADE.
A) Express with α and b the lengths of BE and AE.
Given that: AE ⊥ BE and AB=1.25CE
B1) Find α.
B2) Express with b the diagonal of the trapezoid, BD.
The perimeter of the circumscribed circle of the trapezoid ABCD is 25cm.
C) Find b.
1681753652876.png
This is the drawing attached to the problem.
 
Dr.Peterson said:
Next, do what I suggested: Use only the given lengths and determine what the angles ARE. You will be able to show that alpha has a fixed value, and that AEB is not a right angle. I've given you all the information needed to prove your (my) claim; so do so, in order to earn the credit. It is not hard.
Click to expand...
I read the problem again, I think I understood it very well, and as I thought the trapezoid does not exist.
So unless there is something I missed, there is my proof that AEB can't be a right angle:
1681759475387.png
Do you have a shorter proof in your head?
 
pineapplewithmouse said:
I read the problem again, I think I understood it very well, and as I thought the trapezoid does not exist.
So unless there is something I missed, there is my proof that AEB can't be a right angle:
View attachment 35572
Do you have a shorter proof in your head?
Click to expand...
I would have done much of the same; I just would have ended by showing that with legs of [imath]\sqrt{11}[/imath] and [imath]\sqrt{21}[/imath], the hypotenuse of a right triangle would not be 5. And rather than actually finding [imath]\alpha[/imath] (though I'd done that previously), I would just find the height of the trapezoid and use the Pythagorean Theorem to find AE and BE. That is, I'd keep the proof that the problem is bad as simple as possible, avoiding the trig.
 
Dr.Peterson said:
I would have done much of the same; I just would have ended by showing that with legs of [imath]\sqrt{11}[/imath] and [imath]\sqrt{21}[/imath], the hypotenuse of a right triangle would not be 5. And rather than actually finding [imath]\alpha[/imath] (though I'd done that previously), I would just find the height of the trapezoid and use the Pythagorean Theorem to find AE and BE. That is, I'd keep the proof that the problem is bad as simple as possible, avoiding the trig.
Click to expand...
Maybe I could avoid the trig, but I do not think it's existence makes the proof significantly more complex than without it.
Anyway, thanks for the help, I will show it to my teacher tomorrow, and I'll update.
It's a very important test, so I guess she should know.
 
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