Math Question

[ColdFlare]

Math problem

Can anyone by looking at the picture prove that DEC is a equalatoral triangle?

Here is the link

http://www.theforumisdown.com/uploadfiles/0702/math.JPG

since theforumisdown is refusing to show the picture

 

[Mday]

is abcd a square?

 

[ColdFlare]

yes it is, sorry left that out, yes ABCD is a square

 

[blahblah99]

I know da answer but ain't telling cuz im in da same class as u

 

[ColdFlare]

i'm 35 and unemployed, i dont thinkwe know each other

plus i solved it muahah

 

[Fandu]

If you know it's a square, then give each side an arbitrary length, say 10. Now you have enough information to work out the length of every segment. cheers

 

[Mday]

since it's a square, you can use complementary and supplementary angles, the laws of sine and cosine to figure out what's what...

 

[heliomphalodon]

OK, here's one way...

We will show that angle(ADE) = 30 from which it follows immediately that DEC is equilateral.

Let z = length(AB).

Erect a perpendicular on AD passing through E. Call it FE. Clearly length(FE) = z/2.
Let x = length(AF) and y = length(FD). Clearly x+y = z and angle(AEF) = angle(BAE) = 15.

Now, x/(z/2) = 2x/z = tan(15). By the half-angle formula tan(a/2) = (1-cos(a))/sin(a) we have
2x/z = (1-cos(30))/sin(30). But these values are known exactly, so we substitute and obtain
2x/z = (1-sqrt(3)/2)/(1/2) which simplifies to
2x/z = 2-sqrt(3) which is easily solved for x to give
x = z*(1-sqrt(3)/2)

Now, we have x+y = z and x = z*(1-sqrt(3)/2) so we substitute for x to obtain
z*(1-sqrt(3)/2)+y = z which is easily solved for y to give
y = z*sqrt(3)/2 or, equivalently, 2y = z*sqrt(3).

Let T = angle(ADE). Then tan(T) = (z/2)/y = z/(2y) so we substitute for 2y to obtain
tan(T) = z/(z*sqrt(3)) which simplifies to
tan(T) = 1/sqrt(3) so we have
T = arctan(1/sqrt(3)). But this value is known exactly, so we substitute and obtain
T = 30 QED

I know, it's ugly and there's probably an elegant 3-line pure-geometric proof...

 

[Arrgghh]

Well, here's my go at it:
Given that angle BAE = angle ABE, it follows that triangle AEB is isoceles, and therefore segement AE = segement BE. Also, it is given that segement AD = BC, and by angle subtraction angle DAE = angle CBE.

Using a SAS correspondence, we can find that triangle DAE = triangle BEC. It follows then that segments DE and CE are congruent. Also, segment DA congruent to segment DE, and segment DA congruent to segment DC implies that DE conguent to DC. Also, since DE and CE are congruent, the triangle is equilateral.

I did fudge a bit in there, but i'm tired... sorry.

 

[heliomphalodon]

Originally posted by: Arrgghh
Well, here's my go at it:
Given that angle BAE = angle ABE, it follows that triangle AEB is isoceles, and therefore segement AE = segement BE. Also, it is given that segement AD = BC, and by angle subtraction angle DAE = angle CBE.

Using a SAS correspondence, we can find that triangle DAE = triangle BEC. It follows then that segments DE and CE are congruent. Also, segment DA congruent to segment DE, and segment DA congruent to segment DC implies that DE conguent to DC. Also, since DE and CE are congruent, the triangle is equilateral.

I did fudge a bit in there, but i'm tired... sorry.

You did more than "fudge" -- your statement "segment DA congruent to segment DE" is not justified and essentially assumes the result! :Q Still, I think you're on the right track (and I suspect that your approach is closer to what was intended). The "clean" way to do this geometrically might be to show that angle(ADE) = 2*angle(ABE), or else to show (not assume) that triangle ADE is isosceles.

C'mon folks, someone must see an elegant proof for this!

 

[Arrgghh]

Well, like I said I was tired, and that was what came to mind when I looked at the picture. Proving DEC is isosceles is easy, an easy way to get that last segment (DC) just didn't spring to mind.

It looks to me like you could use angle and segment constructions to draw triangle DAE outside the square with segment DA in common, and then using congurences it should be possible to show that angle DAE and angle DEA are congurent, thus making the triangle isosceles. That would justify my huge leap. I leave it to someone else to fill in the holes.

 

[BumJCRules]

Since ABCD is a Square,

Postulate #1 - Side A, B, C, D all equal 10 units in length. This is just a number that I am using.

Postulate #2 - Singe all the angle in a square are 90 degree angles, all of the side are equal lengths.

With that being said...


Use the law of sines to figure out the length of segments AE and EB.

Law of Sines is ... (I have it set up for this problem)

(sin BAE / EB) = (sin ABE / AE) = (sin AEB / AB) So...

EB = (sin BAE x AB)/sin AEB

EB = (sin 15 x 10) / sin 150

EB = ~5.17638


So...

AB = 10
AE = 5.17638
EB = 5.17638

Since ABCD is a square, the angle ABD is 90 degrees. The angle BAE is 15 degrees. That makes angle EAD 75 degrees.

90 - 15 = 75



Now you have to use the Pythagorem Identity along with side-side-side to solve for the length of DE.


(DE^2) = (AD^2) + (AE^2) - (2 x AD x AE) cos EAD

(DE^2) = 10^2 + 5.17638^2 - 2(10)(5.17638)(cos 75)

(DE^2) = 100 + 26.79491 - 26.79491

(DE^2) = ~100

DE = 10

So...

Yes, triangle EDC is an Equilateral Triangle.

 

[heliomphalodon]

Originally posted by: BumJCRules
...

EB = ~5.17638

...

So...

Yes, triangle EDC is an Equilateral Triangle.

Close, and probably the right approach, but once you introduce an approximation you have lost rigor. A correct proof must be exact (or symbolic). I think your use of the Law of Sines is probably the key to a clean result, but we're not there yet...

 

[rbhawcroft]

you dont need figures to prove it. just say that angles EAB and EBA are equal and therefore their lines (A...E and B...E) intersect at point E, therefore E MUST be half way across the diameter of the square, equally a similar thing from the bottom base line with lines D to E and C to E must be equal and their angles EDC and ECD must be equal, any triangle with two equal angles must be an EQUILateral triangle.

 

[ColdFlare]

Wait you guys are proving it with all these sine forumlas and etc. Someone ask me this question and he said this was a 9th grade math question, meaning 9th grade geometry without any of these calc functions. Is there any other way or proving it? Since i am really baffled

 

[Sahakiel]

Draw a line (plane) down the middle and prove it bisects both triangles at right angles (not hard, ABCD is a square). Then prove doing so splits both into two pairs of identical triangles. Use Pythagoras. This proves angles EDC and ECD are equal, thus triangle CDE is isosceles.

*edit*
Whoops. Forgot it's to prove equilateral. Let me think some more; it's hard without sines and cosines.

 

[Arrgghh]

Originally posted by: rbhawcroft
any triangle with two equal angles must be an EQUILateral triangle.

Close, but not quite. Any triangle with two equal angles is isosceles. That's the easy part of the proof...

 

[heliomphalodon]

Try this on for size:

Let z = length(AB).

It is clear through the use of similar triangles that DEC is isosceles.

Choose F to bisect DC and erect the perpendicular, which intersects AB at G.
Since ABCD is a square, FG is perpendicular to AB.
Since DEC is isosceles, E is on FG.

Draw the circular arc of radius z, centered at C, from B to D.
Draw the circular arc of radius z, centered at D, from A to C.

These arcs obviously intersect at some point E', and moreover it is clear that DE'C is equilateral.
We will show that E' = E.

Since DE'C is equilateral, it is isosceles. Thus E' is on FG.
Since DE'C is equilateral, angle(E'CB) = 30.
Since triangle E'CB is isosceles, we have
angle(CE'B) = angle(CBE') = 75. Thus angle(E'BA) = 15.
Recall that we were given angle(EBA) = 15.

Consider the two triangles E'GB and EGB.
By the angle-side-angle rule, these triangles are congruent.
Then since E' and E are both on segment FG, we have
E' = E QED

 

[bizmark]

okay, so this uses sine/cosine/tangent, and the half-angle formula to arrive at the exact value for tan(15), but other than that it's very simple, straightforward and 'geometric'.

Say the length of the side of the square is 10. We want to prove that DE also has length 10. Then the same could be said of CE by symmetry and voila, all 3 sides are 10.

Draw a straight line vertically through the middle of the square. We'll call the segment above the cross-point 'x', and the segment below that point 'y'. x=5*tan(15) ==>y=10-5*tan(15).

Now look at the triangle formed by D, E, and the point at the bottom of the square where your vertical line intersects. We know that the bottom part is length 5, and the vertical part is length y=10-5*10(15). It's a right triangle, so to find the length of the final side (DE) we use the Pythagorean Theorem.

The square of the length of the hypotenuse is then
5^2 + (10 - 5*tan(15))^2
=25 + 100 - 100*tan(15)+ 25*(tan(15))^2.

tan(15) is 2-sqrt(3), following heliomphalodon. So we have

=125 - 100*(2 - sqrt(3)) + 25*(2 - sqrt(3))^2
=125 - 200 + 100*sqrt(3) + 100 - 100*sqrt(3) + 75
=100

==> DE has length 10.

 

[bizmark]

heliomphalodon

Assuming we had a compass/straightedge, we could just confirm that CD=DE by any number of ways, including putting the non-drawing point of the compass on D, putting the drawing point on C, and making an arc to the left which should intersect E perfectly.

In other words, I don't think that the second half of your proof is really necessary since it's sort of assumed when using straightedge and compass that everything's exact. Like if you construct a bisection, you don't have to prove that your constructed line intersects the midpoint of the original line; it is so, by construction, even though there's a chance that it could be off (assuming a bad compass or whatever). But then, your proof does leave no doubt

 

[rbhawcroft]

Originally posted by: Arrgghh

Originally posted by: rbhawcroftany triangle with two equal angles must be an EQUILateral triangle.

Close, but not quite. Any triangle with two equal angles is isosceles. That's the easy part of the proof...

maybe im confused, doh, so they are three equal sides are they? just use either squaring or SOH, CAH, TOA (Sin equals Opposite over the Hypoteneuse etc...). I forget the ules of squaring but its basically Hyp(squared) = Adj (sq) times Opp (sq), and there maybe a root in there.

 

[rbhawcroft]

right SOH CAh TOA only works with rightangles triangle to you would have to do a few more calculations to use them.

 

[heliomphalodon]

Well done bizmark. I like your proof, although a purist would say that you should have used a symbol like 'z' for the length of the side of the square, rather than choosing a number, however arbitrary.

As to your point about compass and straightedge, I didn't really mean to imply their use. I should have said "consider" the circular arcs, rather than "draw" the circular arcs. And even in a classical compass-and-straightedge construction, one can never merely observe that two points coincide -- it must be proved.

I still think there's a similar-triangles argument out there somewhere that we haven't seen yet.

 

[bizmark]

Originally posted by: heliomphalodon
Well done bizmark. I like your proof, although a purist would say that you should have used a symbol like 'z' for the length of the side of the square, rather than choosing a number, however arbitrary.

yeah, I know... it's trivial, though and the equations are much easier to follow with actual numbers IMO.

As to your point about compass and straightedge, I didn't really mean to imply their use. I should have said "consider" the circular arcs, rather than "draw" the circular arcs. And even in a classical compass-and-straightedge construction, one can never merely observe that two points coincide -- it must be proved.

Ah, I didn't know that -- I haven't done compass-and-straightedge stuff since 9th grade and we certainly weren't very rigorous

I still think there's a similar-triangles argument out there somewhere that we haven't seen yet.

I thought about similar triangles for a while, but I don't think it's really possible. Look at the triangles you can form (post-hoc, knowing that DEC is an equilateral triangle):

150-15-15, 90-75-15, 90-60-30, 60-60-60, 75-75-30.... you want to show a 90-60-30, but all you start off with is a 150-15-15 or a 90-75-15... what's similar to either of those?

 

[heliomphalodon]

Originally posted by: bizmark
I thought about similar triangles for a while, but I don't think it's really possible. Look at the triangles you can form (post-hoc, knowing that DEC is an equilateral triangle):

150-15-15, 90-75-15, 90-60-30, 60-60-60, 75-75-30.... you want to show a 90-60-30, but all you start off with is a 150-15-15 or a 90-75-15... what's similar to either of those?

Well, if you take the AEB triangle and copy it down to the bottom of the square so that A->D, B->C and E gets copied to a new point E" below the bottom of the square, then AEE"D forms a parallelogram and triangles DAE and EE"D are congruent. But I still can't get DA = DE which would make DAE isosceles. Then again, if we bisect AE at H and look at triangles AHD and EHD, they're congruent and also similar to triangles EGA and EGB in my construction.

Ooh, I just got an idea.

We're given a square ABCD together with a point E in the square such that angle(EAB) = angle(EBA) = 15.
To prove: triangle DEC is equilateral.

Consider the triangle AEB. Reflect it in the line AC so that A->A, B->D and E->E'.
Now consider triangle AEE'. It is equilateral.
Thus triangles AE'D and DE'E are congruent, so
triangle ADE is isosceles and thus
triangle DEC is equilateral. QED

Now that's the sort of proof I like!