Another Sporadic Problem-of-the-Week

[mmm4444bot]

ABCD is a unit square.

E is a point along AB.

F is a point along AD.

Triangle AEF has perimeter 2.

(1) Find angle ECF (exact radian or degree measure).

(2) Find exact length of line segment AE such that the perimeter of triangle CEF is minimized.



(I enjoy analytical geometry, so I wrote functions that input length AE and output the measure of angle ECF and the sides of triangle CEF.

I would like to see other approaches.)

 

[Denis]

Hmmm...I'll bite!

Angle ECF is obviously 45 degrees; easily shown by making AE = AF; from the problem statement:
> "(1) Find angle ECF (exact radian or degree measure)."
we're "indirectly told" there is only one size.

> "(2) Find exact length of line segment AE such that the perimeter of triangle CEF is minimized."
AE = AF = 2 / [2 + sqrt(2)]

> "(I enjoy analytical geometry, so I wrote functions that input length AE and output the measure of
> angle ECF and the sides of triangle CEF. I would like to see other approaches.)"

Angle ECF always = 45

x = AE, a = EF, b = EC, c = FC

a = sqrt(x^2 + y^2)
b = sqrt[1 + (1 - x)^2]
c = sqrt[1 + (1 - y)^2]
where y = (2 - 2x) / (2 - x)

Angle ECF = ACOS[(b^2 + c^2 - a^2) / (2bd)] = 45

 

[mmm4444bot]

… Angle ECF is obviously 45 degrees; easily shown by making AE = AF … we're "indirectly told" there is only one size.

Yes, when triangle AEF is isoceles, then angle ECF is 45 degrees.

You "indirectly perceived" being told that the measure of angle ECF is constant. (Nice try.)

You need to explain all cases.

-

> "(2) Find exact length of line segment AE such that the perimeter of triangle CEF is minimized."

AE = AF = 2 / [2 + sqrt(2)]

How do you know that the minimum occurs when AE = AF?

You need to explain this.

-

x = AE, a = EF, b = EC, c = FC

a = sqrt(x^2 + y^2)
b = sqrt[1 + (1 - x)^2]
c = sqrt[1 + (1 - y)^2]
where y = (2 - 2x) / (2 - x)

Angle ECF = ACOS[(b^2 + c^2 - a^2) / (2bc)] = 45

Okay, Denis, I'll buy this explanation for part (1), since we can examine this function for 0 < x < 1.

(It's basically the same as what I did.)

I then began finding the minimum perimeter by writing a function for the perimeter, too.

My result 2 - sqrt(2) matches yours above.

I would still like to see your explanation for part (2), before giving you full credit. :twisted:

 

[Denis]

Re:

I would still like to see your explanation for part (2), before giving you full credit. :twisted:

Perimeter is minimized when area is maximized; area is maximized when triangle is half a square,
thus when AE = AF = 2 / [2 + sqrt(2)]

[Hmmm......] :?

 

[mmm4444bot]

… area is maximized when triangle is half a square, thus when AE = AF …

The maximum area of triangle CEF is half of what square?

 

[Denis]

Square with sides 2 / [2 + sqrt(2)], when AE = AF of course.

 

[mmm4444bot]

Perimeter is minimized when area is maximized …

I understand why this perimeter shortens as we let this triangle's area increase, but how did you determine that triangle CEF's maximum area is half of a square with side 2 - sqrt(2)?

Or, did you reason instead that the area of triangle CEF is maximized when the area of triangle AEF is maximized (i.e., when triangle AEF is an isoceles right-triangle)?

 

[Denis]

Something like that :wink:

 

[mmm4444bot]

Something like that

Oh, I see.

You guessed, right? (I mean, an educated guess, followed by some kind of confirmation.)

Going back to your initial response to part (1), where you read one angle measure for ECF, I subsequently thought that one could put point E at B, and put point F at A.

Then, CF is the diagonal of the unit square, and it's pretty obvious that angle ECF bisects a right-angle. Perhaps, one could get away with that.

So, I should give you full credit (for that part).

I'm working on another method involving extending AB past point B to a point G, where BG = DF. Maybe I can get by without Pythagorus.

 

[Denis]

I refuse to divulge my math-shattering method; will submit such to the Newfoundland Mathematical Disassociation :idea:

 

[Denis]

Had a "slower" look at this thing...

a = AE, b = AF ; so BE = 1-a, DF = 1-b
c = EF, e = EC, f = FC

So problem is: find value of a that creates minimum c + e + f

Since c = 2 - a - b and c = sqrt(a^2 + b^2), easy to get b,c in terms of a:
b = (2 - 2a) / (2 - a)
c = (a^2 - 2a + 2) / (2 - a)

Since e = sqrt[1 + (1 - a)^2], then:
e = sqrt(a^2 - 2a + 2)

Since f = sqrt[1 + (1 - b)^2] and b = (2 - 2a) / (2 - a), then:
f = sqrt[(2a^2 - 4a + 4) / (a^2 - 4a + 4)]

I verified those (program) and they come out ok.
And I see no way to simplify them.

SO:
c+e+f = (a^2 - 2a + 2) / (2 - a) + sqrt(a^2 - 2a + 2) + sqrt[(2a^2 - 4a + 4) / (a^2 - 4a + 4)]
Set to 0, get 1st derivative:
(a^2 - 2a + 2) / (2 - a) + sqrt(a^2 - 2a + 2) + sqrt[(2a^2 - 4a + 4) / (a^2 - 4a + 4)] = 0
sqrt(a^2 - 2a + 2) + sqrt[(2a^2 - 4a + 4) / (a^2 - 4a + 4)] = -[(a^2 - 2a + 2) / (2 - a)]

Now, who in his/her right mind wants to square both sides, then square both sides again :shock:

 

[mmm4444bot]

… Now, who in his/her right mind wants to square both sides, then square both sides again

(Why does Denis exclude me from his questions?)

My second approach is to extend AB past B to a point G such that BG = FD

Therefore, angle DCF = BCG

Note that angle DCF is complementary to BCF.

Since angle BCG is adjacent to BCF, angle FCG must be 90 degrees.

Note that angle FCG is the sum of FCE and ECG.

Note that triangles FCE and ECG share a common side EC and that FC = CG.

Therefore, triangles FCE and ECG are congruent and angles FCE and ECG are equal.

The only two equal angles that sum to 90 degrees are two 45 degree angles.

Angle ECF must be 45 degrees.

I think that my second approach is, in general, easier than analytical geometry. (I most always end up in the slow lane at the grocery store.)

 

[Denis]

Re:

(I most always end up in the slow lane at the grocery store.)

In my case, if I line up at MacDonald's in the line where there's only 1 guy, he turns out
to be a little league baseball coach ordering different meals for his 12 players