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Making triangles out of strawsby Rebecca Maddox & Ian ZukowskiIn the mind of the average American adult, high school math classes are all about the mindless tasks of taking notes and completing book problems for homework. However, modern math classrooms strive to be remembered for different reasons. Today, teachers are striving to find engaging tasks that challenge students to think critically, apply previous knowledge and explore mathematics in real world situations. The use of technology in the math classroom is becoming increasingly encouraged. According to the Principles and Standards for School Mathematics, published by the National Council of Teachers of Mathematics, “technology is essential in teaching and learning mathematics; it influences the mathematics that is taught and enhances students' learning (NCTM, 2000-2004).” In this article we will explore a problem where students will be expected to apply previous knowledge, to use technology, and to apply conceptual understanding of mathematical concepts to solve real world situations. This problem could be used in multiple math classes and different grade levels with some minor adjustments. One standard provided by NCTM that the lesson covers is “grades 9–12 expectations: in grades 9–12 all students should–analyze properties and determine attributes of two- and three-dimensional objects (NCTM, 2000-2004).” The problem presented will focus on properties of triangles; more specifically properties of acute, obtuse and right triangles. Another standard from the Common Core that this lesson addresses is “CCSS.MATH.CONTENT.HSS.CP.B.9, use permutations and combinations to compute probabilities of compound events and solve problems (CCSS, 2014).” The problem presented will not touch on compound events; however, it does require the knowledge of permutations and combinations from the students. Some technology that will aid students in the problem solving process include but are not limited to modeling software such as Geogebra/Geometer’s Sketchpad and a graphing calculator. The following problem is thought provoking and will not only challenge students but teachers as well.Frederick W. Stevenson presented this problem in his book titled “Twenty-one Mathematical Excursions” as a game in which the student is a participant. There are 50 total straws available, of integral lengths 1, 2, … , 50. Each participant chooses three straws and attempts to form a triangle as shown in Figure 1. Whomever can form a triangle wins. If the straws form a triangle then right triangles beat all other triangles and acute triangles beat obtuse triangles. Stevenson then asked, what is the probability you win the game?Figure 1. The straws of lengths 1, 2, 3 do NOT form a triangle, while straws of lengths 3, 5, 6 DO form a triangle.Stevenson posed a series of questions that involved the game and the main question as described above. These questions can be answered using concepts of probability and geometry. The first task asks us to find the number of “hands” that can be drawn, that is how many different combinations of 3 straws exist. The second task asks us to find the total amount of triangles which can be formed using any 3 of these 50 straws. Then using that information, the next task asks us to identify which triangles are acute, obtuse, or right triangles. After completing these three tasks, we can determine the probability of certain events. Among the variety of methods to solve this problem, one approach uses knowledge of combinations, triangle inequality, properties of acute and obtuse triangles and the Pythagorean Theorem.How many different “hands” can I draw?Before solving this problem, some base information needs to be obtained. One question to consider prior to a deeper exploration of the problem is: “How many combinations of straws can I make?” When presented with this problem, students may make a list of every possible triangle. They may begin with [1, 2, 3], then [1, 2, 4], and further to [1, 3, 4], until they realize that this will take much longer than they originally thought. In search for a more efficient method of computation, they may reason that there are 50 possible choices for the first value, 49 choices for the second value, and 48 choices for the third value. Using this logic, it could make sense to take the product 50*49*48=117600 . Some students might use their statistical knowledge and realize that they used the formula for permutations; however, the question asks how many combinations of straws exist? When calculating permutations we find all sets of three numbers that appear in a particular order. So for every set of numbers [a, b, c], there are five other permutations ([a, c, b], [b, a, c], [c, a, b], [b, c, a], [c, b, a]) which are inadvertently included. When calculating combinations we find all sets of three numbers where the order of the numbers does not matter. The statistical expression to find the combinations is 50 choose 3, or 50C3=19,600.General CaseThroughout the process of solving the problem using 50 straws, we will relate general formulas to consider any number of straws. Consider 10,000 straws of lengths 1, 2, 3, ... , 10,000; how can we compute the number of combinations when we draw 3 straws? Using the same formula as above, the number of combinations equals 10,000C3. To find the total combinations for any number of straws, n, with given lengths of 1, 2, 3, ... , n use the formula nC3.How many different triangles can be formed?Prior to finding which triples form a triangle we wanted to narrow down the number of triples. Rather than sorting through 19,600 triples, we focused on all the triples with a given value as the shortest side length. Originally we began our exploration with 1 as the shortest side length. We found 1,176 triples that are possible using the remaining 49 straws. We decided that it would be most efficient to begin with 48 as the smallest possible side. This yielded only one possible triple using the 2 remaining straws: [48, 49, 50]. Moving on to 47 as the smallest side we found 3 possible triples, and with 46 as the smallest side we found 6 possible triples. A pattern began to emerge; the summation of n numbers, also known as Triangular Numbers. The reasoning behind this pattern uses the following logic. In Table 1, the numbers represented in red have (y,z)-combinations that have already appeared in the previous sets of triples (i.e. 47, 49, 50 in lieu of 48, 49, 50).Table 1. List of triples for a given shortest side lengthThe non-red triples are those using a previously unseen y-value=(x+1). The number of triples in each set that are not in red represents the nth number in the summation. Thus, there will be n=49-x new triples added to the previous amount of triples. As x decreases by 1, n increases by 1, thus adding the next number in the summation to the previous total=i=0ni. Using the summation formula, we were able to find the values shown in Table 2.Table 2. Number of triples for a given shortest side length Returning to the original question, if 3 straws are drawn of lengths 3, 20, and 30, can a triangle be formed using these three straws? Here students may use modeling software to explore what properties create a triangle in regards to side length. Using the Triangle Inequality Theorem, we can decide whether or not the straws will form a triangle. The Triangle Inequality Theorem states that any side of a triangle is always shorter than the sum of the other two sides as shown in Figure 2. Considering all of the possible triples formed by three straws, the goal is to find which satisfy Triangle Inequality. Figure 2. The Triangle Inequality Theorem.For all triangles with a shortest side length x, the difference between z and y must be less than x in order for the Triangle Inequality to hold. The largest possible difference occurs when y=x+1, z=50. For example, when the shortest side x has a length of 47, the largest possible difference between z and y occurs when y=48, z=50. The difference is z-y=2, and 2<47, so all triples with a shortest side of length of 47 will form triangles. As x decreases by one, what is the first x-value where the Triangle Inequality fails? To find this value, we took the maximum length of 50 and divided by 2 to get 25. When x=25 we know that the largest difference of 24 occurs when y=26, z=50. Since 24<x, then every triple with x≥25will satisfy the Triangle Inequality. Now, let side x have a length of 24. The largest possible difference of 25 occurs when y=25 and z=50. This shows that not every triple made with a shortest side length of 24 forms a triangle, giving us the first x-value that fails the Triangle Inequality. The three triples that fail to form a triangle are (24, 25, 49), (24, 25, 50), (24, 26, 50). In Table 3, the x-values in red are those where all the triples form triangles.Table 3. Number of triples, number of triangles and the number of non triangles for a given shortest side lengthWhen x<25 we know that at least 3 triples do not form triangles. We want to find the exact number of triples that fail to form triangles for a specific x-value. We have stated that the triangle inequality fails when z-y≥x; thus we can rewrite the equation so that the inequality fails when z-x≥y. When considering x=24, we began with the maximum z-value to find the number of y-valuesthat will fail the Triangle Inequality. The Triangle Inequality gives: 50-x=26=(x+2). This gives us 2 numbers (x+1), (x+2) which form non-triangular triples with x=24, z=50. Now decreasing the maximum value by one, we will use x=24 and z=49 to get:49-x=25=(x+1). Thus, there is 1 number, (x+1), which forms a non-triangular triple with x=24, z=49. There are a total of 3 non-triangular triples, listed in red in Table 4 when x=24.Table 4. List of non-triangular triples for a given smallest side lengthNext we considered the case when x=23. Similarly, we consider the following equations:50-x=27=(x+4)49-x=26=(x+3)48-x=25=(x+2)47-x=24=(x+1)This shows that there are 4 non-triangular triples when x=23 and z=50, that there are 3 when z=49, there are 2 when z=48 and 1 when z=47 giving a total of 10 non-triangular triples. As we decrease our shortest side length x by 1, the pattern that arises is the summation of n numbers, except the summation takes 2 numbers at a time. Looking below at Table 5, we see that the pattern described above does hold, as the nthred number in the “# of triples” column corresponds with the nth blue number in the “# of Non-triangles” column.Table 5. Number of triples related to the number of non-triangles In total, we found 10,100 Non-Triangular triples. Finding the difference between the total triples and the Non-Triangular triples, gave us 9,500 Triangular triples.General CaseHow many triangles can be formed if we had 10,000 straws? To solve this problem we could do the same process as above; however, this would be a tedious process. Instead we want to find a general formula to represent the number of triples that will not form triangles. In Table 3, we can see the number of tripIes for each given smallest side length. Using a graphing calculator, plot these points and use regression to find an equation that best fits the pattern. Referring to Figure 3 below for help, on the graphing calculator click stat→enter (edit). Under L1 type in numbers 1-25 and under L2 type in 3, 3+10, 3+10+21 etc. This will give us the total number of non-triangular triples rather than the number of triples for a given shortest side length. Once the data is entered into the calculator go to stat→calc→cubicreg→enter. We will see that the R2value is 1. This means that the equation fits our data perfectly. With 50 straws we entered in 24 data points, and when x=(n2-1)=24 the formula yielded the total number of non-triangular numbers we found above. The general formula is 23(n2-1)3+32(n2-1)2+56(n2-1). To check this formula, we considered the case of 9 straws, with lengths 1,2,3,...,9. The formula says that there should be 49.875 non-acute triangles, when in fact we found there to be 25. When investigating this discrepancy we created a table for the 9 straws, similar to the table for 50 straws. Table 6. shows that the pattern for non-triangular triples is similar to the pattern seen with 50 straws only slightly shifted. The pattern still takes two numbers in the summation but instead of starting with 3, we start with 1. Using the cubic regression function for this new data we were able to find the formula of 23(n-12)3+12(n-12)2-16(n-12). This new formula can be used when n is an odd number. To find the number of triangular triples when n is even use the formula: nC3-23(n2-1)3+32(n2-1)2+56(n2-1). To find the number of triangular triples when nis odd use the formula: nC3-23(n-12)3+12(n-12)2-16(n-12) Figure 3. Calculator steps.Table 6. Non-Triangular Triples for 9 StrawsFinding Acute trianglesNow that we have found that 9,500 triangles can be formed, how many of them are acute triangles? To find the acute triangles we must use a form of the Pythagorean Theorem. The Pythagorean Theorem is more commonly known for its use with right triangles. The theorem states that if we have a right triangle then a2+b2=c2where c is the length of the hypotenuse and a and b are the leg lengths. By using the inverse of the theorem we can determine if the triangle is obtuse or acute. Figure 4 shows which formula to use when solving for acute or obtuse triangles. We use this formula to determine the number of acute triangles in each set of triples for a given shortest side length. Figure 4. Acute and Obtuse triangles and equations.Rather than finding the number of acute triangles we found the number of non-acute triangles. We know that 502=2500and 25002=1250. So we know that if x2>1250 then x2+y2>2500because x2<y2. To find the first shortest side length that will make a non-acute triangle take2500-1250=35.35. Thus the first shortest side length that will make a non-acute triangle is 34.Since x2+y2>z2 gives an acute triangle, we can manipulate the formula to get y2>z2-x2. Consider when x=34, and z=50:y2>2500-1156=1344.This implies that y>36.6 will form an acute triangle, so non-acute triangles are formed when y=35, 36.Now consider the case when x=34, and z=49y2>2401-1156=1245This implies that y>35.2 will form an acute triangle, so non-acute triangles are formed when y=35. So in total we have 3 non acute triangles when x=34. This method worked well for all of the shortest side values until we reached x=24. Once we reach this side length, we begin to get an overlap of triples which satisfy y2>z2-x2 to form an acute triangle, but fail to satisfy the Triangle Inequality Theorem. We had to eliminate the triples which do not form triangles. In order to do this, we used the following formula: (number of non-acute triangular triples) - (number of non-triangles) = (number of non-acute triangles). We found the number of non-acute triangular triples shown in Table 7.Table 7. Number of Non-Acute triangles and Acute Triangles General CaseHow many acute triangles can be formed if we have 10,000 straws? Similar to the previous general cases, we want to find a general formula that represents the pattern of non-acute triangles. Much to our disappointment, we were not able to find a perfect formula or pattern to represent these numbers; however, we found an equation that could approximate these values. Again, we used the graphing calculator and entered our data points into "stat→edit". After entering the data we found that the graph formed, as shown in Figure. 5, does not fit a quadratic or cubic; the closest fit is a quartic function. After entering stat→calc→quartregwe see that R2=.99999 which is close to a perfect fit. The resulting equation is as follows, -.00820965402629(n2-52)2+.21180862366927(n2-52)3+8.426965670564(n2-52)2-45.39163192244(n2-52)+69.68371571647)Figure 5. Graph of non acute triangular numbers.Finding Right and Obtuse trianglesIn order to find the number of right triangles, we found all of the Pythagorean Triples which use numbers less than or equal to 50.To find the Pythagorean Triples we used Euclid’s formula: (2mn, m2-n2, m2+n2), when m>n, GCD(m,n)=1. We know that the largest number in the triple will be the value for m2+n2, when m>2, n>1, and reasoned that m2+n2≤50. Similar to finding the non-Acute triangles, this means that n<25 or n<5. By checking coprime numbers using the formula, we are able to see 20 distincts triples using numbers less than or equal to 50.Now that the number of acute triangles and the number of right triangles have been found we can easily find the number of obtuse triangles. To find the number of obtuse triangles, we took the total number of triangles formed and subtracted the number of acute triangles and the number of right triangles. This yielded a total of 5607 obtuse triangles. Generalized CaseIn order to find the number of right triangles for any given number of straws we concluded that they would have to be found by hand using the same process as above. The obtuse triangles can be found by the process of elimination. If we know the total number of triangles, the number of acute triangles, and the number of right triangles, then the number of obtuse triangles is easily obtainable.Applying StatisticsAt this point you just want to know if you won the game! To find the probability that you will lose on your first draw of three straws brings us back to the first question. We know from earlier that there are 50 chose 3 combinations of straws, or 19600 sets of triples. We also know that there are a total of 10100 triples that do not form triangles, thus leaving 9500 triples that do form a triangle. There is a 1010019600=.5153probability of not forming a triangle and a 950019600=.4847 probability of forming a triangle.What if you and your opponent both can form a triangle? Now who wins? As stated earlier, right triangles beat any other triangle, acute triangles beat obtuse triangles. We found that there are 20 right triangles, 3873 acute triangles and 5607 obtuse triangles. The probability of forming a right triangle is 2019600=.0010. The probability of forming an acute triangle is 387319600=.1976 and the probability of forming an obtuse triangle is 560719600=.2861. So if your opponent formed an obtuse triangle the probability of you winning is .1976+.0010=.1986; however, if your opponent can form an acute triangle then your chances of winning is only .1%.Generalized CaseIf we wanted to find these probabilities for any total number of straws other than 50 we would have to use our generalized formulas. The probability of forming any triangle would be # of triples - # of non-traingular triples# of triples. The Probability for acute triangles can be estimated by # of triangular triples-# of non-acute triangles# of triples=The right triangles would have to be found by hand using the primitive triples. Once that number is found then the probability would be calculated by # of right trianglesn chose 3. The probability for obtuse triangles can be found by # of triangular triples-#of non acute triangles-# of right trianglesn chose 3.Conclusion With an ever increasing emphasis on standards and technology, this math excursion should fit right into the curriculum of a classroom that is prepared to discuss geometry and/or statistics. The journey to answer the simple question of “What are the chances I win this game” provided the opportunities to learn new information and apply previous knowledge about different types of triangles, combinations, and even included a different perspective of how to use the Pythagorean Theorem. The knowledge of Triangle Inequality, acute, obtuse, and right triangles easily allows students to “analyze properties and determine attributes of two- and three-dimensional objects (NCTM, 2000-2004)”. Furthermore, students “use permutations and combinations to compute probabilities of compound events and solve problems (CCSS, 2014)” while they figure out how many triples can be formed, and then use those triangular triple numbers throughout the rest of the problem. While we have all of these fantastic theorems to help us out, the use of technology is crucial in this problem. Technology helps to provide statistical regression to find formulas and analyze data. The modeling software assists in the process of figuring out which triples form which types of triangles, and making accurate shapes quickly instead of measuring out and drawing the shapes which could take much more time. Aside from the mathematical content and standards that this problem covers, it makes students critically think and apply their knowledge to a real-world situation that is actually relatable. Rather than doing trivial problems that are purely procedural, this problem is engaging and students can conceptually understand the content. Our hope is that students will walk away from this problem and begin to ask questions about other life events or relate mathematics outside of the classroom.References:Common Core State Standards Initiative. (2014). High School: Statistics & Probability ? Conditional Probability & the Rules of Probability. Retrieved from Council of Teachers of Mathematics. (2004). Principles for School Mathematics. Retrieved from , F.W. (1992). Exploratory problems in mathematics. Reston, VA: National Council of Teachers of Mathematics. ................
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