A Somewhat Large Cake

[Lepton]

Puzzle
That's no mean cake...

[Dr. Borodog]

I don't even understand the statement of the problem.

The sums of the areas of opposite sides are equal?

------------------
You will respect my philosophai.

[Quagmire]

Yes, mathgrant didn't specify it was length, so I'm guessing that there's some kind of play on words there and you should do something other than length to finish the problem.

[Lepton]

It should read as follows:

[Dr. Borodog]

I think the problem is under-specified. Assuming that the sides of the cake are vertical, and that the edges where the sides meet are similarly vertical (the problem is totally impossible without those restrictions), then I get 2 equations for 4 unknowns (the heights of the edges where the sides meet).

This is because the vertical height of the cake can be scaled by any arbitrary factor without changing the equality of the areas of the opposing sides.

Am I missing something obvious?


------------------
You will respect my philosophai.

[Dr. Borodog]

Oh! Never mind. That makes it completely different.



------------------
You will respect my philosophai.

[Dr. Borodog]

But that still doesn't address the problem of the vertical scale of the cake. It's impossible to calculate the cake's surface area without it.



------------------
You will respect my philosophai.

[Dr. Borodog]

Ok, I now have 6 equations for 6 unknowns, but it doesn't matter because I don't know the height of the cake.



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You will respect my philosophai.

[Lepton]

A valid point. (usually, eh)
I just want the area of the top of the cake. Pretend it's flat or something. I'll mention that in the puzzle.
Thanks

[Lepton]

code:

---

C +-------+ A


  |       |


E +-------+ K

---

Angle C + Angle K = Angle A + Angle E
(the shape isn't a rectangle, though)

[This message has been edited by Lepton (edited 05-07-2004 01:55 PM).]

[Dr. Borodog]

OK. Since I have the equations I'll consider the problem solved.



------------------
You will respect my philosophai.

[Lepton]

Needless to say, there are a few elegant solutions.

[Hypnosis]

132 or did I miss something? Seems too easy...

[mathgrant]

There was an error in the original posting. Two pairs of ADJACENT angles, not opposite angles, add up to the same. But the old version is interesting.

[kevinatilusa]

It still feels like there's something strange. Adjacent angles adding up to 180 degrees implies that the quadrilateral is a parallelogram, but opposite sides of a parallelogram are congruent, which is not the case here.

[ZutAlors!]

An example of two adjacent sides adding to 180 degrees:

code:

---

 


      /|


     / |


    /  |


   /   |


  /    |


 /     |


/      |


|      |


|      |


|      |


|      |

---

[shadyforce]

Answer: Total area = 155.12 square inches.???

I'd post how I got it but I'm afraid if I do, I'll give away the whole thing, assuming I'm even right.

[Dr. Borodog]

Um, that's slightly larger than 1 square foot. Perhaps you should check your math.


------------------
You will respect my philosophai.

[kevinatilusa]

Oh, right. Only one set of adjacent angles (C/A vs K/E) add up to the same amount, and parallelograms require every pair of adjacent angles sum to 180.

I get an area of 165sqrt(7)/4=109.13... square feet

[Bicho the Inhaler]

I think I remember when mathgrant posted this originally. IIRC, a couple of good solutions surfaced.

[Lepton]

Kevin is correct. Bicho is also correct. In fact, I used his solution from when the puzzle was originally posted. (proper credit is given - hope this is okay?)

[shadyforce]

Sorry, I meant feet instead of inches. I just got the units mixed up. Bah, I wish you guys would convert to metric already .

[Bicho the Inhaler]

Originally posted by Lepton:

---

Kevin is correct. Bicho is also correct. In fact, I used his solution from when the puzzle was originally posted. (proper credit is given - hope this is okay?)

---

HELL YEAH!!!! WHO DA MAN, HUH? WHO D--

I mean, yes, that's okay.

[IQ221]

The fact that C+A = K+E implies that C+A = K+E = 180 degrees (since C+A+K+E = 180 degrees), which then implies that the straight line segments CE and AK are parallel. Given that fact, it is simple geometry to deduce the area of CAKE... just apply Pythagoras theorem a few time! Hint: you should end-up with 3 equations and 3 unknowns.

[IQ221]

Sorry... I meant C+A+K+E = 360 degrees!

[chewie]

So being new to this board how do you submit your answers without revealing them to everybody?

[Quailman]

you may use spoiler codes. just type [SP]enter your answer here[/SP].

It will look like this: I think the answer is [December 7, 1941].

highlight the area in the brackets to see what appears there, or [ctrl]A to see all hidden text on the page.

[IQ221]

The area of the cake is m * k / 2 + l * m / 2 + m * (15 - k) where [ k * k + m * m = 8 * 8 and m * m + l * l = 10 * 10 and 15 - k = 18 - l].

[This message has been edited by Lepton (edited 05-13-2004 12:24 PM).]

[shadyforce]

Well, this is how I worked out my answer. It's wrong, but I'm not sure where:

[Kinda Spoiler]


I think we have a quadrilateral, where the tob and base are parallel. The base is 18, top 15. Also, we have 2 sides of lengths 8 and 10, with the 10 being more sloped than the 8 (naturally).

Let the perpindicular height be h.

Taking out the central square, leaves 2 right angled triangles on either side, (which for the purposes of calculation, are assumed to have hypotenuses of slopeing in opposite directions. One with hypotenuse 8, height h, and base x, the other with hypoteneuse 10, height h and base (3 - x).

code:

---

A Rather crude diagram:


            15


      ----------------


   8 /|             h|\ 10


    / |              | \


    --------------------


     x      18        3-x

---

=> h^2 + x^2 = 8^2,
=> h^2 + (3-x)^2 = 10^2.
(Simultanious equations)

Solving for h and x: h works out as 9.179, but x is (-27/6 = 4.5), meaning the shape has a parallel base and top, but the slope of the 2 sides are the same.

code:

---

Result:


      4.5    10.5


      ----------------


      \ |    h=9.179 |\ 10


     8 \|            | \


        ----------------


              10.5   7.5

---

A1 = 10.5 * 9.179 = 100.05
A2 = 0.5 * 4.5 * 9.179 = 20.65
A3 = 0.5 * 7.5 * 9.179 = 34.42

=> Total area = 155.12 square feet.


[End Spoiler]

[This message has been edited by shadyforce (edited 05-13-2004 08:55 AM).]

[Quailman]

The right triangle on the left side of your diagram has a side of 9.179 and a hypotenuse of 8. That's impossible.

[What you should have is a parallelogram with base and top = 15 and height = H (to be determined in the next step); and a triangle with base of 3 and sides of 8 and 10. This gives you two equations:

8^2 = x^2 + H^2
10^2 = (x+3)^2 + H^2

Note that the top of the triangle is not above the 3" base. The distance to one side is x. substituting for H^2 gives you

8^2 - x^2 = 10^2 - (x^2 + 6x + 9)
6x = 100 - 9 - 64
x = 4.5
H = ~6.614

the total area is 6.614*15 + 6.614*3/2
or 109.1372 square inches]

[This message has been edited by Quailman (edited 05-13-2004 09:59 AM).]

[Quailman]

I see now that you just confused that factors in the pythagorean theorem after you found the 4.5. Try sqrt(8^2 - 4.5^2) in stead of sqrt(8^2 + 4.5^2)

[mikegoo]

A slightly different approach:
[First I realized it didn't matter exactly how the bases of the trapezium lined up I could use the 3,8,10 triangle (a,b,c) to find the height. To do this I used the law of cosines to find angle C. I then used 2 triangle area formulas .5*a*b*sin C & .5*a*h to find h. Finish with the area of a trapezium .5*(b1+b2)*h and get the same answer as the rest of the universe.]

[Mathtech]

I drew in the diagonals and used the Law of Cosines in order to find each angle measure. I then used area of a triangle to find the areas of the two triangles and got the same answer when doing it from both sides. I got an answer of [ 146.154 ]. Sorry if the answer is showing, this is my first time.

[Quailman]

That's not it. Look at Shadyforce's second diagram. In spite of his having said the slope of the two sides are the same, they are not. If you draw an 8 inch line parallel to the 8 inch line on the left, you'll have a parallelogram with sides of 15 and 8 (and as yet unknown height H), with a triangle with a base of 3 and sides of 8 and 10.

code:

---

Pretend the 10 inch line is connected to the end of the top corner


            15


      ----------------


      \ |            |\8\ 10


     8 \|            | \  \


        ----------------\--


              15         3

---

I don't know the law of cosines, but the only way to get an area of > 146 sq. in. is to have a height greater than 8 inches, which is the length of the hypotenuse of a right triangle. It does not compute. I used simultaneous equations to get the height. Hit [ctrl]A to highlight the concealed text in the above solutions.

I know you think it's longer than eight inches, but it's really just over six.

[shadyforce]

When I said that the slope of the 2 sides are the same, I meant the slope of the 2 sides are in the same direction (as indicated in the diagram), but not the same value (as was impossible to draw on a diagram). Sorry for the confusion.

[Cello_dude]

It seems to me that everyone is making too many assumptions... I believe the only rational way to answer this is to write a guess-and-check program that finds the areas of the four separate triangles using the law of cosine. The difficult part is having the four randomly generated thetas (based upon C+A= K+E) working simultaneously; based on this, each of the four hypot. must by figured simultaneously... with the hypot. defined, it's only matter of simple trig. to figure out the rest... too bad my VB skills are not current, nor sufficient in the first place.

[Quailman]

No - there are not too many assumptions. You point out that C+A=K+E (given). Since it's a quadrilateral, C+A+K+E=360, so C+A=K+E=180. You can reason from there that two opposite sides are parallel. Since it cannot be the 8 & 10 sides (try to make this work with the other sides = 15 & 18), it must be the long sides that are parallel. There's only one layout that fits, as shown in shadyforce's diagram above. Then work it out as several people demonstrated. Just don't accept a solution that has a right triangle with a leg longer than it's hypotenuse.

[Lambert]

Make CE the base and then extend the shorter sides (CA and EK) to form a big triangle, call this triangle CBE. Using similar triangles CBE and ABK, we find that AB = 40 and KB = 50. Now we know the 3 sides of triangle ABK are 40, 50 15 and of triangle CBK are 48, 60, 18 and by Hero's formula we can evaluate the areas of these two triangles as sq rt 61523.4375 and sq rt 127575. The area of the cake is the difference: 109.137 sq ft.

[Cello_dude]

Sorry all...
I read my late-night/early-morning post today and discovered its harsh, frustrated tone... I am a person that makes no assumptions, but fails to realize the truth(s) on a regular basis. Thanks for setting me straight. It seems Lambert has the answer and rightfully so. Good work... maybe.

[ccw]

All the solutions I have seen posted are more complicated
than necessary. Why?: |[You don't need to find the height. ]|
Just Hero(n)'s Formula and a little arithmetic.

|[1. Disect trapazoid into a parallelagram and 10-8-3 triangle.
(2 different parallelagrams are possible, but both have a side of 15)
2. disect parallelagram into triangles of base 3 (along original 15 and 18)
with the same height as trapezoid. (they could all be 10-8-3)
3. There are 10 of these. All have the same area as 10-8-3 triangle.
4. Area=11* A(10,8,3)
5. Applying Hero gives Area=165*sqrt(7)/4
]|