# Thread: Simple but difficult geometry question (picture included) - First post here :)

1. ## Simple but difficult geometry question (picture included) - First post here :)

Hi all, first of all - my first post here. Nice to meet you all. I'm active on mymathforum if you guys post there from time to time. My name is Oria, I live in israel, I'm a math private teacher for pre uni students, and I study applied mathematics in holon institute of technology.

Anyway here's my question. A student of mine asked me to solve it and I was stumped, was actually quite embarrasing since it is so simple.
23.png picture by oriagr - Photobucket

ABCD is a diamond (AD = AB = BC = CD, AD ||BC, AB ||CD), The angle A is 60 degrees. The points M and N are on the sides AB and BC accordingly.
We are given: BN+BM = AB

Prove that the triangle DMN is equilateral.

Frankly I think something is wrong with the question and we are missing a detail. But I would like a second opinion.

Thank you!

2. ## Re: Simple but difficult geometry question (picture included) - First post here :)

What you want to prove is obviously NOT true!

3. ## Re: Simple but difficult geometry question (picture included) - First post here :)

Show me why not. Yeah I know my drawing is pretty bad, but that triangle is equilateral. just got to prove how and I can't come up with any idea. Just ignore the bad drawing

4. ## Re: Simple but difficult geometry question (picture included) - First post here :)

It's routine cosine rule isn't it ?
In England, we call such a figure a rhombus, suppose that its length of side is 1 and suppose that BM is of length $a,$ then AM will be of length $1-a.$

Since $BN + BM = AB = 1$,

$BN = 1 - BM = 1-a,$

and therefore

$CN = 1 - BN = a.$

Now use the cosine rule in the two triangles $DAM$ and $DCN.$

We have,

$DM^{2} = 1^{2} + (1-a)^{2} - 2(1-a)\cos 60,$
and
$DN^{2} = 1^{2} + a^{2} - 2a\cos 60.$

It's easy to show that the two are equal, just substitute $\cos 60 = 1/2.$

5. ## Re: Simple but difficult geometry question (picture included) - First post here :)

Forgot the last line !

$MN^{2} = a^{2}+(1-a)^{2} - 2a(1-a)\cos 120 = a^{2} - a + 1 = DM^{2} = DN^{2}.$

6. ## Re: Simple but difficult geometry question (picture included) - First post here :)

Its simple geometry,ABCD is a rhombus consisting of two base to base equilateral triangles i.e. ABD and DCB. DM and DN are perpendicular bisectors of AB and BC.
If the sides of rhombus are lenght 2, BM =1, DM=DN = rad3.Half of MN = rad3/2 MN=rad3.DMN is equilateral.All of the right triangles in the diamond are 3060-90 triangles

7. ## Re: Simple but difficult geometry question (picture included) - First post here :)

That only covers the particular case where M and N are the mid-points of AB and BC ?
All that we are told is that BM + BN = AB.

8. ## Re: Simple but difficult geometry question (picture included) - First post here :)

AB=AC BM=BN=1 BM+BN =2=AC and AB

9. ## Re: Simple but difficult geometry question (picture included) - First post here :)

BM is not necessarily equal to BN, all that we have is that BM + BN = AB.

We can get there geometrically.
As earlier, let BM = a, then MA = 1 - a, BN = 1 - a and CN = a.
The angles DBA and DBC will both be 60.

There are two pairs of congruent triangles.

DBM is congruent to DCN, (sides 1 and a with included angle 60) and from which it follows that DN = DM.

DAM is congruent to DBN, (sides 1 and 1-a with included angle 60) and from which follows that the angles MDA and NDB are equal.

Then, it follows that angle NDM = NDB + BDM = MDA + BDM = BDA = 60.

So, triangle NDM is isosceles with a top angle of 60, in which case it must be equilateral.

10. ## Re: Simple but difficult geometry question (picture included) - First post here :)

Hello BobP,
Your rhombus has a side length 0f 1
DMBN is a right kite with 90 angles DMB and DNB
Triangles DMB and DNB are congruent
BM =1/2 BN=1/2
BM +BN =1=AB=AC

11. ## Re: Simple but difficult geometry question (picture included) - First post here :)

Hello bjhopper,
One of us is misreading the question.
The question does not say that angles DMB and DNB are right-angles.
It does not say that M and N are the midpoints of AB and BC
M can be any point on AB, and then, N will be a point on BC such that BM + BN = AB = 1
We could have BM = 1/3, BN = 2/3 or BM = 3/4, BN = 1/4 or ........ .

12. ## Re: Simple but difficult geometry question (picture included) - First post here :)

Hello again BobP,
In solving geometry problems the solver is allowed to draw construction lines.My construction lines were the two diagonals of the rhombus,the perpendicular bisectors of AB and BC,DM and DN.and the segment MN It was then evident that BM + BN = AC.I do not have to directly prove that.It was also evident that the sides of triangle DMN were each equal to rad 3.( 2 = sides of rhombus).If the construction shows that BM+BN =AC then I do not need to try to find where Mand N are located before p roceeding

13. ## Re: Simple but difficult geometry question (picture included) - First post here :)

Hello again bjhopper
Of course you are allowed to draw in whatever construction lines you like. Yes you are allowed to draw in the perpendicular bisectors of AB and BC. What you can't do is to call those lines DM and DN without admitting that that fixes the positions of M and N. If you do call those lines DM and DN then you are proving that DMN is equilateral only for that specific case. The point M can lie anywhere along the line AB. Suppose I say, (for example), that M is a quarter of the way from A to B. How does your proof deal with that situation ?

14. ## Re: Simple but difficult geometry question (picture included) - First post here :)

I understand your logic so tell me the length of the sides of your inner equilateral triangle and the location of M and N. I assume that your rhombus has a side length of 1

15. ## Re: Simple but difficult geometry question (picture included) - First post here :)

I picked the point M as 0.2 and then N .8 from B. Using the cosine law I found an equilateral triangle in the diamond side length 0.91. Your method must apply to any chose for M consistent with BM + BN =AB.There are then many solutions.

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