Teepees and Wigwams

Let’s see, your problem as written has too much information of one kind and not enough info of other kinds.

The answer is anywhere from zero to one hundred feet, depending on how you want to cut it.
Assuming (a) that the two areas are meant to be equal, and (b) that the field is right-angled, and (c) that the trapezoid is parallel to one of the hundred-foot legs, the height of the trapezoid is 100 - 10root50 = 100 - 50root2, or approximately 29.2 feet.
On the other hand, assuming the same field but cutting the trapezoid parallel to the hypotenuse, the height of the trapezoid is 50root2 - 50, or approximately 20.7 feet.
If the original field is not right-angled, we can calculate the height on any base you want …
Diagrams available if you want.

Permanent hall pass: Please allow Victoria to go to the library at lunch hour any time she wants to. (This was a genuine hall pass which I had all year when the principal was trying to protect the books from the students, and my home teacher finally got tired of writing the same thing every day.)

Then his neighbor, the lovely widow Helgamar, invited him to build a yurt together. How many extra yurts did they need for the children and their cows?

I had not thought about dividing the triangle with a line parallel to the third side, but I think that’s a clever idea. And if one intends to use this for first or second year high school algebra or geometry students, then one should indeed say that the farmer’s pasture is a right triangle.

But if I may ask, what happened to summarize, visualize, relate, name, equation, solve, and check? (G)

Sara McName

Yes, a yurt would be good. Many of my LD students were not aware that a teepee is a wigwam. I bet that most won’t know about yurts. The convesation might be:

“Mrs. McName, how are you today?”

“Well, I can’t decide if I’m a wigwam or a yurt.”

“Mrs. McName? What’s a yurt?”

Sara McName

Gee, I did all of that — have three pages off my notepad here if you want me to fax them to you — I *did* say diagrams available on request …

True story: in Grade 12 art class, our young teacher was trying to do up-to-date and relevant projects (this was the late sixties …). One of the projects, I don’t know why, was to build a scale model of some other culture’s living space. I had read somewhere or other about yurts and for some reason the idea caught my fancy. So I decided to build a yurt. Research in the encyclopedia and dictionary turned up nothing. The teacher had never heard of it; she also asked “what’s a yurt?”. This was long before the internet. Finally I came up with a one-or-two sentence description somewhere, maybe National Geographic. And I built a yurt. Bent popsicle stick frame (you have to boil them first to make them flexible), white felt cover, cotton canvas over-cover.
The space for your yurts depends on the size of the yurts and how many you want …

You seem to imply that you’d give this problem to a class — well, *with* a class discussion as a model of too much/too little/misleading info it could be fun, but I would never assign a problem like this cold to anyone below advanced Grade 12/college, and even then as a puzzle problem with some lead-up.

BTW, I solved this for fun because I like puzzle problems; the first time I made some kind of assumption that didn’t work out and proved that h = 0, so I knew that was wrong, turned the page and tried again; the second time tried a few approximations to see what came up; the third time clunked through the algebra for two small note pages and got a reasonable answer which also agreed with the approximations; the fourth time tried the alternate cut and did some quite tricky algebra, minimum Grade 12 level, and got a reasonable (and similar) answer; the *fifth* time looked at the general case (any triangle at all, not necessarily a right triangle, not even required to be isosceles) and looked for a simpler way through the brush, found a neat simple generalization that solves all such problems — the particular answers from tries #3 and 4 gave a hint towards the general result.
Now, in far too many textbooks, the neat simple generalization is printed as the “right” answer, and no mention at all is made of the other approaches. In fact, this problem can be approached at least five different ways. The neat generalization is *not* obvious and clear — it requires a difficult change in point of view plus a strong knowledge of fairly advanced geometry.
I’d recommend thinking twice before asking young students to solve this kind of problem!

Now I get to go back to the notepad and work out three, four, n equal pieces for those yurts … results upcoming later after my ESL student.

THat problem is the kind of problem that doesn’t lend itself to the process — since it doesn’t have enough information to do it and starts off with the completely absurd sentence about a person *being* a wigwam or a teepee. Once you establish with that sentence that any word can mean anything, things like “divide it into a trapezoid and a triangle” could also mean just about anything, too.

Gee whiz…I have a diminished lexicon..I had no idea what a yurt was, teepee and wigman…thank heavens for Sheriff John, Captain Kanagaroo and Monty Montana….

I am not a math person either as this problem flew right over my ADD head…glad to see that some people think the way you three do…As for me…I think and dream in pictures…if you ask me to give measurements…and mathmatical analysis of what I am visualizng we are in big trouble.. :-P

Yes, the part about being a teepee (or yurt) and a wigwam is a pun. Puns are good for LD students because LD students usually don’t understand metaphores. At the school where I tutor, the speach therapists use puns as a teaching tool.

Dividing a triangle into a triangle and a trapezoid is the sort of thing that mathematicians do. Before I posted the problem, I knew of one solution. Victoria found a second. We should challenge you to find a third. Or maybe we could find a proof that only two solutions exist.

The problem has no numerical solution unless one assumes that that the larger triangle is a right triangle and that the areas or perimeters area equal. But problems need not have numerical solutions. This is a hard fact for LD students to absorb. The algebra 1A-1B course at my school has a test with the following question.

solve for w: 8y + w = -8.

Many students resist the notion that the answer is not a number.

Sara McName

Now what did I do with that hall pass.

Sheriff John was my son’s favorite television show. And when did I learn about yurts? I think it may have been in a book about Ghingis Khan. It was one on a series of books, popular in the 1950’s about the Wright Brothers, the Marine Corp, the Chilsom Trail, Alexander the Great. I haven’t seen them in a long time.

Sara McName

I agree, that was my first point, that the problem has too much info of one kind (names and locations and teepees and wigwams) and not enough info of another kind (goal of the division — I *assumed* equal area to make the problem solublem but nothing says this has to be so; shape of the triangle which I *assumed* to be right, etcetera.)

.

Sara, I beg to differ. This problem DOES have a solution if the triangle isn’t a right triangle, and even if it is scalene (all three sides unequal). And yes it DOES have a numerical solution, if the triangle is known/soluble. It also has a general solution, as outlined below.

Take any triangle. Choose one side (any side) as the base b. Choose a second side as leg a, and the vertex where a and b meet to be C, and name the angle between b and a (angle C) to be theta. Call the third side c if you want to (not used in calculations). Draw the altitude from the other end of b perpendicular to a (If theta is obtuse, the altitude falls outside the triangle, but that’s OK, just extend b to meet it.) Call the altitude H. (capital letter for a reason, to follow).

H = a* sine of theta if you want to calculate it, or it can be measured from a scale drawing if you haven’t done sines yet.
Sine of theta can be found using cosine and sine laws if you have at least three pieces of info including at least one side, ie the triangle is soluble.

Now, you want to draw a line parallel to b, dividing the triangle into a triangle and a trapezoid, so that the areas of the triangle and trapezoid are equal. Sketch in this line parallel to b (and at right angles to H) to show yourself it’s possible for any triangle. (Trapezoids do *not* have to be symmetric, just have two parallel sides.)

Now, the altitude of the triangle is h1 = ( half root2) times H, and the altitude of the remaining trapezoid is therefore h2 = H - (half root2) * H = (1 - (half root2))H, exactly as found in the particular cases previously.

This can be proved easily (this is the cute trick I mentioned in my other post) by noting we have similar triangles because of the parallel bases, and if h1 = (half root2) * H, then base of smaller triangle = (half root2) * b by similarity, so area of smaller triangle = 1/2 * (half root2)b * (half root2)H = 1/2 * (half root2)squared *bH = 1/2 *2/4 * bH = 1/2 * 1/2 * bH = half of area of larger triangle; and if the small triangle is half the area of the larger triangle, then the trapezoid is the othe half, and thus equal. Phew.

Now, for the yurts.
If we want to divide the triangle into three equal parts, one triangle and two trapezoids, with parallel lines (and this is *not* all so silly — look up the early history of Quebec, my home, and the land division poblem, some time — interesting problem and real solution), we can use the same argument to show that the altitude of the triangle, h1, is (1/root3) * H = (one third root 3)H.
Then the area of the next part has to be equal, so we can apply the result of the previous problem (a favouite mathematician trick!) We note above our *other* h1.1 was (halfroot2) * H and h2.1 was (1 -halfroot2)H so h2.1 = [(1 - halfroot2) / (halfroot2)] * h1.1, simplifying h2.1 = (root2 - 1) h1.1.
Going back to our threeway division, this means h2 = (root2 - 1)(onethird root3) * H.
And then by subtraction
h3 = H - (onethirdroot3)H - (root2 - 1)(onethirdroot3)H, and noting simplification,
h3 = (1 - root2 * onethirdroot3)H

The divisions for four, five, and n spaces are left as exercises for the student.

How in heck do you think I solved that thing? First I imagined a triangle, then I drew some lines across it, then I played around with it a little in my head, THEN I started playing with the calculations.
This is one of the biggest *problems* in American education now, particularly in math but also in other subjects, that students are presented with verbal strings and this is called “teaching”. Verbnal strings that don’t attach to a concrete object are just a bunch of noise.
Too bad you didn’t learn to tie your visualization to some real math, although it’s never too late — read Sheila Tobias, “Overcoming Math Anxiety” for her own adult trek. And do try to teach your kids to use their visualization, which is vital to math beyond the basics.

One student’s solution is at:

http://www.geocities.com/sara216b/math/tritrap/tritrap.html

Sara McName

Here is a general solution. It follows Victoria’s argument.

http://www.geocities.com/sara216b/math/tritrapgen/tritrapgen.html

Your posted solution follows my fourth version, using some fairly fancy algebra, quadratics with variable coefficients, etc. Good exercise but tedious.

You can get a neater solution by putting
(area of top triangle) = 1/2 (area of original triangle) (equ 1)
(base of top triangle) / (base of large triangle) = (height of top triangle) / (height of original triangle) (equ 2)

Call the original triangle ABC, with A at the top
Draw DE parallel to BC so that D is on AB and E is on AC and area ADE = 1/2 area ABC (so of course area of trapezoid DECB = 1/2 area ABC = area ADE, as required)
Draw altitude H of ABC, from A perpendicular to BC. Let h be the part of H that is the altitude of ADE (if scalene/obtuse triangle, extend BC and DE as needed)
area ADE = 1/2 * DE * h = 1/2 area ABC = 1/2( 1/2 * BC * H)
so DE * h = 1/2 * BC * H (equ 1)
Then by similar triangles, h/H = DE/BC (equ 2)
From equ 2, BC = (H/h) * DE
Subbing into equ 1, DE * h = 1/2 * ((H/h) *DE) * H
Divide both sides by DE and multiply both sides by h
(h squared) = (1/2 H squared)
so h = (1/root2) H or = (root2/2) H
Altitude of trapezoid = H - h = H - (root2/2)H = H(1 - root2/2)
= H((2-root2)/2) [choose your preferred format]

A lot shorter, and avoids that pesky quadratic

http://www.geocities.com/sara216b/math/tritrapsimp/tritrapsimp.html

Here is a problem that is the first problem in an ancient Egyptian manuscript. It may be too difficult for beginning algebra students, but it is less difficult than the teepee-wigwam problem.

A bamboo pole is nine cubits long. It is broken so that one end forms a right angle with the ground and the second end touches the ground three cubits from the first end. Find the height of the right triangle.

Using x for horizontal, y for vertical, r for hypotenuse
using x^2 for x squared

x^2 + y^2 = r^2 (Pythagoras) equ. 1
x = 3 (given) equ. 2
y + r = 9 (given) equ 3

from equ 3, r = 9 - y
Sub this and equ 2 into equ 1

3^2 + y^2 = (9 - y)^2

9 + y^2 = 81 - 18y + y^2
subtract y^2 from both sides, problem simplifies fast

9 = 81 - 18y
9 - 81 = -18y
-72 = -18y
-72/-18 = y

4 = y = height of vertical part
9-4 = 5 = length of hypotenuse

and lo and behold, we have the classic 3,4,5 right triangle!

This is just barely within the level of an advanced algebra 1 student, but I’d expect it to be soluble by a good Algebra 2 student.

Several problems of varying difficulty, from elementary logic that can be solved by a bright child, a trick question, an advanced high school/college geometry question, up to adult geometry challenge.

While in the car I remembered another old classic to add, but it’s a tricky setup so will have to wait until I can find the reference book.

I had thought that I had read about this problem in Carl B. Boyer, History of Mathematics, John Wiley and Sons Inc, New York, 1968. So far in my rereading of Boyer, I haven’t found the problem, so maybe it isn’t an example of ancient Egyptian math.

Let’s see, your problem as written has too much information of one kind and not enough info of other kinds.

The answer is anywhere from zero to one hundred feet, depending on how you want to cut it.
Assuming (a) that the two areas are meant to be equal, and (b) that the field is right-angled, and (c) that the trapezoid is parallel to one of the hundred-foot legs, the height of the trapezoid is 100 - 10root50 = 100 - 50root2, or approximately 29.2 feet.
On the other hand, assuming the same field but cutting the trapezoid parallel to the hypotenuse, the height of the trapezoid is 50root2 - 50, or approximately 20.7 feet.
If the original field is not right-angled, we can calculate the height on any base you want …
Diagrams available if you want.

Permanent hall pass: Please allow Victoria to go to the library at lunch hour any time she wants to. (This was a genuine hall pass which I had all year when the principal was trying to protect the books from the students, and my home teacher finally got tired of writing the same thing every day.)

Then his neighbor, the lovely widow Helgamar, invited him to build a yurt together. How many extra yurts did they need for the children and their cows?

I had not thought about dividing the triangle with a line parallel to the third side, but I think that’s a clever idea. And if one intends to use this for first or second year high school algebra or geometry students, then one should indeed say that the farmer’s pasture is a right triangle.

But if I may ask, what happened to summarize, visualize, relate, name, equation, solve, and check? (G)

Sara McName

Yes, a yurt would be good. Many of my LD students were not aware that a teepee is a wigwam. I bet that most won’t know about yurts. The convesation might be:

“Mrs. McName, how are you today?”

“Well, I can’t decide if I’m a wigwam or a yurt.”

“Mrs. McName? What’s a yurt?”

Sara McName

Gee, I did all of that — have three pages off my notepad here if you want me to fax them to you — I *did* say diagrams available on request …

True story: in Grade 12 art class, our young teacher was trying to do up-to-date and relevant projects (this was the late sixties …). One of the projects, I don’t know why, was to build a scale model of some other culture’s living space. I had read somewhere or other about yurts and for some reason the idea caught my fancy. So I decided to build a yurt. Research in the encyclopedia and dictionary turned up nothing. The teacher had never heard of it; she also asked “what’s a yurt?”. This was long before the internet. Finally I came up with a one-or-two sentence description somewhere, maybe National Geographic. And I built a yurt. Bent popsicle stick frame (you have to boil them first to make them flexible), white felt cover, cotton canvas over-cover.
The space for your yurts depends on the size of the yurts and how many you want …

You seem to imply that you’d give this problem to a class — well, *with* a class discussion as a model of too much/too little/misleading info it could be fun, but I would never assign a problem like this cold to anyone below advanced Grade 12/college, and even then as a puzzle problem with some lead-up.

BTW, I solved this for fun because I like puzzle problems; the first time I made some kind of assumption that didn’t work out and proved that h = 0, so I knew that was wrong, turned the page and tried again; the second time tried a few approximations to see what came up; the third time clunked through the algebra for two small note pages and got a reasonable answer which also agreed with the approximations; the fourth time tried the alternate cut and did some quite tricky algebra, minimum Grade 12 level, and got a reasonable (and similar) answer; the *fifth* time looked at the general case (any triangle at all, not necessarily a right triangle, not even required to be isosceles) and looked for a simpler way through the brush, found a neat simple generalization that solves all such problems — the particular answers from tries #3 and 4 gave a hint towards the general result.
Now, in far too many textbooks, the neat simple generalization is printed as the “right” answer, and no mention at all is made of the other approaches. In fact, this problem can be approached at least five different ways. The neat generalization is *not* obvious and clear — it requires a difficult change in point of view plus a strong knowledge of fairly advanced geometry.
I’d recommend thinking twice before asking young students to solve this kind of problem!

Now I get to go back to the notepad and work out three, four, n equal pieces for those yurts … results upcoming later after my ESL student.

THat problem is the kind of problem that doesn’t lend itself to the process — since it doesn’t have enough information to do it and starts off with the completely absurd sentence about a person *being* a wigwam or a teepee. Once you establish with that sentence that any word can mean anything, things like “divide it into a trapezoid and a triangle” could also mean just about anything, too.

Gee whiz…I have a diminished lexicon..I had no idea what a yurt was, teepee and wigman…thank heavens for Sheriff John, Captain Kanagaroo and Monty Montana….

I am not a math person either as this problem flew right over my ADD head…glad to see that some people think the way you three do…As for me…I think and dream in pictures…if you ask me to give measurements…and mathmatical analysis of what I am visualizng we are in big trouble.. :-P

Yes, the part about being a teepee (or yurt) and a wigwam is a pun. Puns are good for LD students because LD students usually don’t understand metaphores. At the school where I tutor, the speach therapists use puns as a teaching tool.

Dividing a triangle into a triangle and a trapezoid is the sort of thing that mathematicians do. Before I posted the problem, I knew of one solution. Victoria found a second. We should challenge you to find a third. Or maybe we could find a proof that only two solutions exist.

The problem has no numerical solution unless one assumes that that the larger triangle is a right triangle and that the areas or perimeters area equal. But problems need not have numerical solutions. This is a hard fact for LD students to absorb. The algebra 1A-1B course at my school has a test with the following question.

solve for w: 8y + w = -8.

Many students resist the notion that the answer is not a number.

Sara McName

Now what did I do with that hall pass.

Sheriff John was my son’s favorite television show. And when did I learn about yurts? I think it may have been in a book about Ghingis Khan. It was one on a series of books, popular in the 1950’s about the Wright Brothers, the Marine Corp, the Chilsom Trail, Alexander the Great. I haven’t seen them in a long time.

Sara McName

I agree, that was my first point, that the problem has too much info of one kind (names and locations and teepees and wigwams) and not enough info of another kind (goal of the division — I *assumed* equal area to make the problem solublem but nothing says this has to be so; shape of the triangle which I *assumed* to be right, etcetera.)

.

Sara, I beg to differ. This problem DOES have a solution if the triangle isn’t a right triangle, and even if it is scalene (all three sides unequal). And yes it DOES have a numerical solution, if the triangle is known/soluble. It also has a general solution, as outlined below.

Take any triangle. Choose one side (any side) as the base b. Choose a second side as leg a, and the vertex where a and b meet to be C, and name the angle between b and a (angle C) to be theta. Call the third side c if you want to (not used in calculations). Draw the altitude from the other end of b perpendicular to a (If theta is obtuse, the altitude falls outside the triangle, but that’s OK, just extend b to meet it.) Call the altitude H. (capital letter for a reason, to follow).

H = a* sine of theta if you want to calculate it, or it can be measured from a scale drawing if you haven’t done sines yet.
Sine of theta can be found using cosine and sine laws if you have at least three pieces of info including at least one side, ie the triangle is soluble.

Now, you want to draw a line parallel to b, dividing the triangle into a triangle and a trapezoid, so that the areas of the triangle and trapezoid are equal. Sketch in this line parallel to b (and at right angles to H) to show yourself it’s possible for any triangle. (Trapezoids do *not* have to be symmetric, just have two parallel sides.)

Now, the altitude of the triangle is h1 = ( half root2) times H, and the altitude of the remaining trapezoid is therefore h2 = H - (half root2) * H = (1 - (half root2))H, exactly as found in the particular cases previously.

This can be proved easily (this is the cute trick I mentioned in my other post) by noting we have similar triangles because of the parallel bases, and if h1 = (half root2) * H, then base of smaller triangle = (half root2) * b by similarity, so area of smaller triangle = 1/2 * (half root2)b * (half root2)H = 1/2 * (half root2)squared *bH = 1/2 *2/4 * bH = 1/2 * 1/2 * bH = half of area of larger triangle; and if the small triangle is half the area of the larger triangle, then the trapezoid is the othe half, and thus equal. Phew.

Now, for the yurts.
If we want to divide the triangle into three equal parts, one triangle and two trapezoids, with parallel lines (and this is *not* all so silly — look up the early history of Quebec, my home, and the land division poblem, some time — interesting problem and real solution), we can use the same argument to show that the altitude of the triangle, h1, is (1/root3) * H = (one third root 3)H.
Then the area of the next part has to be equal, so we can apply the result of the previous problem (a favouite mathematician trick!) We note above our *other* h1.1 was (halfroot2) * H and h2.1 was (1 -halfroot2)H so h2.1 = [(1 - halfroot2) / (halfroot2)] * h1.1, simplifying h2.1 = (root2 - 1) h1.1.
Going back to our threeway division, this means h2 = (root2 - 1)(onethird root3) * H.
And then by subtraction
h3 = H - (onethirdroot3)H - (root2 - 1)(onethirdroot3)H, and noting simplification,
h3 = (1 - root2 * onethirdroot3)H

The divisions for four, five, and n spaces are left as exercises for the student.

How in heck do you think I solved that thing? First I imagined a triangle, then I drew some lines across it, then I played around with it a little in my head, THEN I started playing with the calculations.
This is one of the biggest *problems* in American education now, particularly in math but also in other subjects, that students are presented with verbal strings and this is called “teaching”. Verbnal strings that don’t attach to a concrete object are just a bunch of noise.
Too bad you didn’t learn to tie your visualization to some real math, although it’s never too late — read Sheila Tobias, “Overcoming Math Anxiety” for her own adult trek. And do try to teach your kids to use their visualization, which is vital to math beyond the basics.

One student’s solution is at:

http://www.geocities.com/sara216b/math/tritrap/tritrap.html

Sara McName

Here is a general solution. It follows Victoria’s argument.

http://www.geocities.com/sara216b/math/tritrapgen/tritrapgen.html

Your posted solution follows my fourth version, using some fairly fancy algebra, quadratics with variable coefficients, etc. Good exercise but tedious.

You can get a neater solution by putting
(area of top triangle) = 1/2 (area of original triangle) (equ 1)
(base of top triangle) / (base of large triangle) = (height of top triangle) / (height of original triangle) (equ 2)

Call the original triangle ABC, with A at the top
Draw DE parallel to BC so that D is on AB and E is on AC and area ADE = 1/2 area ABC (so of course area of trapezoid DECB = 1/2 area ABC = area ADE, as required)
Draw altitude H of ABC, from A perpendicular to BC. Let h be the part of H that is the altitude of ADE (if scalene/obtuse triangle, extend BC and DE as needed)
area ADE = 1/2 * DE * h = 1/2 area ABC = 1/2( 1/2 * BC * H)
so DE * h = 1/2 * BC * H (equ 1)
Then by similar triangles, h/H = DE/BC (equ 2)
From equ 2, BC = (H/h) * DE
Subbing into equ 1, DE * h = 1/2 * ((H/h) *DE) * H
Divide both sides by DE and multiply both sides by h
(h squared) = (1/2 H squared)
so h = (1/root2) H or = (root2/2) H
Altitude of trapezoid = H - h = H - (root2/2)H = H(1 - root2/2)
= H((2-root2)/2) [choose your preferred format]

A lot shorter, and avoids that pesky quadratic

http://www.geocities.com/sara216b/math/tritrapsimp/tritrapsimp.html

Here is a problem that is the first problem in an ancient Egyptian manuscript. It may be too difficult for beginning algebra students, but it is less difficult than the teepee-wigwam problem.

A bamboo pole is nine cubits long. It is broken so that one end forms a right angle with the ground and the second end touches the ground three cubits from the first end. Find the height of the right triangle.

Using x for horizontal, y for vertical, r for hypotenuse
using x^2 for x squared

x^2 + y^2 = r^2 (Pythagoras) equ. 1
x = 3 (given) equ. 2
y + r = 9 (given) equ 3

from equ 3, r = 9 - y
Sub this and equ 2 into equ 1

3^2 + y^2 = (9 - y)^2

9 + y^2 = 81 - 18y + y^2
subtract y^2 from both sides, problem simplifies fast

9 = 81 - 18y
9 - 81 = -18y
-72 = -18y
-72/-18 = y

4 = y = height of vertical part
9-4 = 5 = length of hypotenuse

and lo and behold, we have the classic 3,4,5 right triangle!

This is just barely within the level of an advanced algebra 1 student, but I’d expect it to be soluble by a good Algebra 2 student.

Several problems of varying difficulty, from elementary logic that can be solved by a bright child, a trick question, an advanced high school/college geometry question, up to adult geometry challenge.

While in the car I remembered another old classic to add, but it’s a tricky setup so will have to wait until I can find the reference book.

I had thought that I had read about this problem in Carl B. Boyer, History of Mathematics, John Wiley and Sons Inc, New York, 1968. So far in my rereading of Boyer, I haven’t found the problem, so maybe it isn’t an example of ancient Egyptian math.