Difficult geometry problem

[Stevelau]

In a quadrilateral ABCD, AB = BC = CD and M is the midpoint of AD.
Angle BMC = 90 degrees.
AC intersects BD at E.
Find angle AEB.

 

[Stevelau]

In a quadrilateral ABCD, AB = BC = CD and M is the midpoint of AD.
Angle BMC = 90 degrees.
AC intersects BD at E.
Find angle AEB.

I know that triangles ABC and BCD are isosceles, and so the base angles are the same.
The required angle is the sum of the two base angles. And it seems that there are some more
isosceles triangles. I don't know exactly how to use the info that AM = MD.

 

[soroban]

Hello, Stevelau!

$\displaystyle \text{In a quadrilateral }ABCD,\:AB = BC = CD.$
$\displaystyle M\text{ is the midpoint of }AD.\;\;\angle BMC = 90^o.\;\;AC\text{ intersects }BD\text{ at }E.$
$\displaystyle \text{Find }\angle AEB.$

We have an isosceles trapezoid.

                  B     a     C
                  *  *  *  *  *
                 * *         * *
              a *   *       *   * a
               *     *     *     *
              *       *90d*       *
             *      45d* *45d      *
          A *  *  *  *  *  *  *  *  * D
                        M

Draw $\displaystyle AC$ and $\displaystyle BD.\;\;AC \cap BD\,=\,E$

Let $\displaystyle a = AB = BC = CD.$
We have: $\displaystyle \angle BMC = 90^o$
. . Hence: .$\displaystyle \angle AMB = \angle DMC = \angle BCM = \angle CBM = 45^o.$

In isosceles right triangle $\displaystyle BMC\!:\;BM = CM = \dfrac{a}{\sqrt{2}}$
In $\displaystyle \Delta ABM$, Law of Sines: .$\displaystyle \dfrac{\sin A}{\frac{a}{\sqrt{2}}} \:=\:\dfrac{\sin45^o}{a}$
. . Hence: .$\displaystyle \sin A \,=\,\frac{1}{2} \quad\Rightarrow\quad A \,=\,30^o$

And we have: .$\displaystyle \angle ABM = 105^o \quad\Rightarrow\quad \angle ABC = 150^o.$

In isosceles triangle $\displaystyle ABC\!:\;\angle BAC = 15^o$

In $\displaystyle \Delta ABE\!:\;\angle AEB \:=\:180^o - \angle BAE - \angle ABE \:=\:180^o - 15^o - 150^o$

. . Therefore: .$\displaystyle \angle AEB \:=\:15^o$

 

[Stevelau]

thanks Soroban.
But ABCD is just a quadrilateral, it is not an isosceles trapezium.
We cannot assume that BC is // to AD and AM equals AB.

 

[Stevelau]

A quadrilateral may not be a square, rectangle, and so on.
The question only says ABCD is a quadrilateral, so we cannot assume any particular type of shape for ABCD.
I actually constructed just a quadrilateral satisfying all the measurements in the question, but it is not of any
specified shape.
In your method, angle ABE (the second last line) is not 150 degrees.
Thanks

 

[wjm11]

angle ABE (the second last line) is not 150 degrees.

You are correct; it is 135 degrees.

...it is not an isosceles trapezium.

The problem only states that the figure is a quadrilateral -- but it is up to us to discover that it is indeed an isosceles trapezoid.

 

[Stevelau]

If we only consider an isosceles trapezoid, we are looking at a very special (easy) case.
The answer is then limited to that particular case.
The difficult part is to find the angle for a general case.

 

[mmm4444bot]

If we only consider an isosceles trapezoid, we are looking at a very special (easy) case.

The difficult part is to find the angle for a general case.

I'm thinking that only an isosceles trapezoid has the given right angle at BMC. This angle will not be 90 degrees, with any quadrilateral that is not also an isosceles trapezoid. Am I correct?

After you've worked this exercise, writing a general solution (in terms of some other angle) should be easier. :cool:

 

[Stevelau]

I'm thinking that only an isosceles trapezoid has the given right angle at BMC. This angle will not be 90 degrees, with any quadrilateral that is not also an isosceles trapezoid. Am I correct?

After you've worked this exercise, writing a general solution (in terms of some other angle) should be easier. :cool:

It is possible to construct a quadrilateral (I tried to attach the diagram, but it didn't work. I also tried to attached the file, but the file size was too big.) in which angle BMC = 90 degrees,
but ABCD is not an isosceles trapezoid.

 

[Stevelau]

Well, I take back what I said; you're correct: lots of "non isosceles" possibilities.

                                C
                        (1)
               B
                             c         (1)
        (1)       b   
                      [90]
A         a            M         a             D

Let AB = BC = CD = 1; assign a size to angle CBM; let's take 50; so angle BCM = 40.

a = AM = BM, b = BM, c = CM.
So b = SIN(40) = ~.643, c = SIN(50) = ~.766

Let angle CMD = u degrees; then angle BMA = 90-u degrees.

j = -2cCOS(u) : DM = {SQRT[j^2 - 4(c^2-1)] - j}/ 2
k = -2bCOS(90-u) : AM = {SQRT[k^2 - 4(b^2-1)] - k} / 2

So SQRT[j^2 - 4(c^2-1)] - j = SQRT[k^2 - 4(b^2-1)] - k
Solve to get u = ~46.4 degrees.
Makes AD = ~2.724, angle BAM = ~26.3 and angle CDM = ~33.5

For any values assigned to angle CBM where SQRT[j^2 - 4(c^2-1)] - j
and SQRT[k^2 - 4(b^2-1)] - k are both positive, you get a solution,
and in ALL cases I believe angle AEB = 30 degrees...did no work on that!

Thanks for agreeing with me.
I was actually trying to use simple geometry (angle sum of a triangle, isosceles triangles, etc) to solve this difficult problem --- it's more challenging and fun. There should be some similar triangles and isosceles triangles. I suspect the two triangles (with E as their common point) within triangle BMC are isosceles and similar as well.
I got stuck in using the info AM = MD

 

[mmm4444bot]

Well, I take back what I said; you're correct: lots of "non isosceles" possibilities.

                                C
                        (1)
               B
                             c         (1)
        (1)       b   
                      [90]
A         a            M         a             D

Let AB = BC = CD = 1; assign a size to angle CBM; let's take 50; so angle BCM = 40.

a = AM = BM, b = BM, c = CM.
So b = SIN(40) = ~.643, c = SIN(50) = ~.766

Let angle CMD = u degrees; then angle BMA = 90-u degrees.

j = -2cCOS(u) : DM = {SQRT[j^2 - 4(c^2-1)] - j}/ 2
k = -2bCOS(90-u) : AM = {SQRT[k^2 - 4(b^2-1)] - k} / 2

So SQRT[j^2 - 4(c^2-1)] - j = SQRT[k^2 - 4(b^2-1)] - k
Solve to get u = ~46.4 degrees.
Makes AD = ~2.724, angle BAM = ~26.3 and angle CDM = ~33.5

For any values assigned to angle CBM where SQRT[j^2 - 4(c^2-1)] - j
and SQRT[k^2 - 4(b^2-1)] - k are both positive, you get a solution,
and in ALL cases I believe angle AEB = 30 degrees...did no work on that!

Has my supposition been proven false? Could be, but I'm not sure, yet!



Starting with 50° for angle CBM and using your algorithm, I get u = 46.3429°.

Using the Law of Sines, I find angle CDM to be 31.9245°. This value is not close enough to your 33.5° approximation for me to be comfortable.

Likewise, I find angle BAM (27.7138°) to differ too much from your result (26.3°).

These different measures cause my calculated values for AM and DM to differ (~1.77 and ~1.97, respectively), which means that M is not the midpoint of AD or that maybe I made a mistake.



Perhaps round-off error explains the differing measures for angle u (46.4 versus 46.3), or maybe I made a mistake there, as well.

Will you please double-check your calculations for angles BAM and CDM? I would like to confirm all of the angles and line-segment lengths involved.



I'm also playing around with the following, letting x equal the measure of angle CBM.

a = sec(u) times the average of sin(x) and cos(x)

 

[Stevelau]

Has my supposition been proven false? Could be, but I'm not sure, yet!



Starting with 50° for angle CBM and using your algorithm, I get u = 46.3429°.

Using the Law of Sines, I find angle CDM to be 31.9245°. This value is not close enough to your 33.5° approximation for me to be comfortable.

Likewise, I find angle BAM (27.7138°) to differ too much from your result (26.3°).

These different measures cause my calculated values for AM and DM to differ (~1.77 and ~1.97, respectively), which means that M is not the midpoint of AD or that maybe I made a mistake.



Perhaps round-off error explains the differing measures for angle u (46.4 versus 46.3), or maybe I made a mistake there, as well.

Will you please double-check your calculations for angles BAM and CDM? I would like to confirm all of the angles and line-segment lengths involved.



I'm also playing around with the following, letting x equal the measure of angle CBM.

a = sec(u) times the average of sin(x) and cos(x)

You may like to try the following construction:

Draw a line segment BC = 5 cm
Using B as centre, draw a circle of radius 4 cm
Using C as centre, draw a circle of radius 3 cm
The two circles intersect at M at which angle M = 90 degrees.
(M is clearly off centre of BC)
Using B as centre, draw a circle (C1) of radius 5 cm
Using C as centre, draw a circle (C2) of radius 5 cm
Draw a line segment passing through M with endpoints A and D on C1 and C2 respectively such that AM = MD
(Or Using M as centre, draw circles of varying radii until AMD are collinear)
Then we have ABCD as a simple quadrilateral.

By changing the right-angled sizes (4 cm and 3 cm), we can construct many many non-symmetric quadrilaterals
satisfying the conditions in the questions.

 

[Stevelau]

You may like to try the following construction:

Draw a line segment BC = 5 cm
Using B as centre, draw a circle of radius 4 cm
Using C as centre, draw a circle of radius 3 cm
The two circles intersect at M at which angle M = 90 degrees.
(M is clearly off centre of BC)
Using B as centre, draw a circle (C1) of radius 5 cm
Using C as centre, draw a circle (C2) of radius 5 cm
Draw a line segment passing through M with endpoints A and D on C1 and C2 respectively such that AM = MD
(Or Using M as centre, draw circles of varying radii until AMD are collinear)
Then we have ABCD as a simple quadrilateral.

By changing the right-angled sizes (4 cm and 3 cm), we can construct many many non-symmetric quadrilaterals
satisfying the conditions in the questions.

For a more accurate construction, please visit

http://home.scarlet.be/j-willekens/exchange/prob912291.png

The diagram shows the answer, but with no working.

 

[mmm4444bot]

I wasn't worrying much about a half degree or so.

Is this what's meant by "fuzzy math" ?

That's okay. I assumed some round-off error.

The concerning issue for me was that some of the numbers did not jive (i.e., off by too much). However, I found my errors.

In triangle BCM, for some odd reason, you named the side opposite the vertex at point B using symbol c and you named the side opposite the vertex at point C using symbol b.

It took me awhile to figure out that these values were transposed in my calculations because I kept associating b with B and c with C.

Doh!


I wanted to confirm the results from your example, before attempting to understand what you did. I was also curious about your statement concerning the two expressions that must be positive, in order to arrive at a solution, because it seems that they are both positive for any acute starting angle chosen for CBM.


I'm now satisfied that my prior claim has been proven false. (Wait, that did not come out quite right...) :lol:

 

[Stevelau]

That's okay. I assumed some round-off error.

The concerning issue for me was that some of the numbers did not jive (i.e., off by too much). However, I found my errors.

In triangle BCM, for some odd reason, you named the side opposite the vertex at point B using symbol c and you named the side opposite the vertex at point C using symbol b.

It took me awhile to figure out that these values were transposed in my calculations because I kept associating b with B and c with C.

Doh!


I wanted to confirm the results from your example, before attempting to understand what you did. I was also curious about your statement concerning the two expressions that must be positive, in order to arrive at a solution, because it seems that they are both positive for any acute starting angle chosen for CBM.


I'm now satisfied that my prior claim has been proven false. (Wait, that did not come out quite right...) :lol:

Hope you enjoy the following method.

Please look at the drawing I (Doctor Jacques, from Ask Dr Math) made on:


http://mathforum.org/dr.math/gifs/prob912291.gif


I used your notations: the quadrilateral is ABCD, the midpoint of AD is M, and the intersection of the diagonals is E. I will write a for the common length of the sides AB, BC, and CD. We must compute the size of the angle AEB.


As you correctly wrote, AEB = BCE + CBE = BAC + BDC.


On the other hand, looking at the triangle AED, we find that AEB = EAD + EDA. Adding these two relations, we get:


AEB = (BAD + CDA)/2 [1]


If H is the intersection of AB and CD, we have therefore:


AEB = (180 - AHD)/2 [2]


Now, it is illuminating to consider how we can construct a quadrilateral ABCD that satisfies the hypothesis.


We start by drawing the points A, B, and C such that AB = BC = a.


Because of the right angle at M, the point M must lie on the circle c1 with diameter BC and center G.


Because M is the midpoint of AD, D must lie on the circle c2, image of c1 by a homothety of center A and ratio 2; let F be the center of that circle. Note that the radius of c2 is equal to 2*GB = a.


As G is the midpoint of both BC and AF, the quadrilateral ACFB is a parallelogram. In particular, this shows that CF = a and the point C lies on c2.


Because CD = a, D is the intersection of c2 with a circle c3 of center C and radius a. The circles c2 and c3 share the radius CF (=a); this means that the triangle CFD is equilateral and the angle FCD is equal to 60 degrees.


As CF is parallel to AH, the angle AHD is equal to 120 degrees, and [2] shows that AEB is equal to 30 degrees.


This elegant and clean method shows the beauty of Mathematics.

 

[Stevelau]

In his example, Steve really did this, and assigned the widely
known "3-4-5"er to ABC. Right (no pun intended!), Steve?

Sure, I did.