Math problem musings from a 57-year-old
If it looks familiar to some of you, it's quite similar to a problem on this year's AMC 12 exam (formerly known as the American High School Mathematical Exam).
ABC is a triangle
The cosine of the angle at A is 11/16
The cosine of the angle at B is 7/8
The shortest side has length 1
What is the length of the longest side?
Now if you'll excuse me, I'm going to spend what is also Valentine's Day (Vegas-related trivia: Teller the magician was also born on 2/14) having some homemade pot roast for dinner - a lesson learned a long time ago; no more "birthday dinners" in restaurants, especially at places that don't take reservations. My father sold alcohol to bars & restaurants for a living, and every year, on my birthday, we went to a restaurant that was one of his customers, and as it didn't take reservations, we spent something like two hours waiting in the bar - every year. Eventually, it got to the point where I started a new "tradition" - every year, on my birthday, I would order a pizza from someplace I had never been before.
Quote: ThatDonGuyToday, being my 57th birthday, since Heinz won't send me a box of its products (which, apparently, it did at one point), instead I'll post this math problem.
If it looks familiar to some of you, it's quite similar to a problem on this year's AMC 12 exam (formerly known as the American High School Mathematical Exam).
ABC is a triangle
The cosine of the angle at A is 11/16
The cosine of the angle at B is 7/8
The shortest side has length 1
What is the length of the longest side?
Now if you'll excuse me, I'm going to spend what is also Valentine's Day (Vegas-related trivia: Teller the magician was also born on 2/14) having some homemade pot roast for dinner - a lesson learned a long time ago; no more "birthday dinners" in restaurants, especially at places that don't take reservations. My father sold alcohol to bars & restaurants for a living, and every year, on my birthday, we went to a restaurant that was one of his customers, and as it didn't take reservations, we spent something like two hours waiting in the bar - every year. Eventually, it got to the point where I started a new "tradition" - every year, on my birthday, I would order a pizza from someplace I had never been before.
So then, happy birthday!
When we were kids, Heinz 57 was what you called a dog when you had no idea what breeds had gone into its birth. So perhaps you're in your mongrel year? Not a bad thing to be, these days.
You select a horse in each of six different races, and bet:
A six-horse parlay
6 5-horse parlays
15 4-horse parlays
20 3-horse parlays
15 2-horse parlays
Quote: ThatDonGuyToday, being my 57th birthday, since Heinz won't send me a box of its products (which, apparently, it did at one point), instead I'll post this math problem.
If it looks familiar to some of you, it's quite similar to a problem on this year's AMC 12 exam (formerly known as the American High School Mathematical Exam).
ABC is a triangle
The cosine of the angle at A is 11/16
The cosine of the angle at B is 7/8
The shortest side has length 1
What is the length of the longest side?
Now if you'll excuse me, I'm going to spend what is also Valentine's Day (Vegas-related trivia: Teller the magician was also born on 2/14) having some homemade pot roast for dinner - a lesson learned a long time ago; no more "birthday dinners" in restaurants, especially at places that don't take reservations. My father sold alcohol to bars & restaurants for a living, and every year, on my birthday, we went to a restaurant that was one of his customers, and as it didn't take reservations, we spent something like two hours waiting in the bar - every year. Eventually, it got to the point where I started a new "tradition" - every year, on my birthday, I would order a pizza from someplace I had never been before.
cos(a+b) = cos(a).cos(b) - sin(a).sin(b) = 0.25, cos(b) > cos(a), then shortest side(1) is opposite angle b.
So, (a+b) < 90 degree, c is a obtuse angle, longest length is opposite the angle c, and 180 - c = a+b,
L = sin(c)/sin(b) = sin(a)/ tan(b) + cos(a) = [(135)^0.5/16] * [7/(15)^0.5] +11/16 = 21/16 + 11/16 = 32/16 = 2
Quote: ssho88cos(a) and cos(b) both are positive, then a and b both are Acute angles,
So, 180 - c = a+b,
L = sin(c)/sin(b) = sin(a)/ tan(b) + cos(a) = [(135)^0.5/16] * [7/(15)^0.5] +11/16 = 21/16 + 11/16 = 32/16 = 2
I'm not sure how you get from "a and b are both acute" to "L = sin c / sin b," but 2 is the correct answer.
Here's how I get it:
sin A = sqrt(1 - (11/16)2) = sqrt(15) * 3/16
sin B = sqrt(1 - (7/8)2) = sqrt(15) * 1/8
A + B + C = 180, so cos C = cos (180 - (A + B)) = -cos (A + B) = sin A sin B - cos A cos B = 15 * 3/128 - 77/128 = -32/128 = -1/4
sin C = sqrt(1 - (-1/4)2) = sqrt(15) * 1/4
Let X, Y, and Z be the lengths of the sides opposite angles A, B, and C, respectively
By the Law of Sines, sin A / X = sin B / Y = sin C / Z
The ratio X:Y:Z = the ratio sin A : sin B : sin C = 3 : 2 : 4 (otherwise the fractions aren't equal)
The side opposite B is shortest: since it is 1, the other two sides have lengths 3/2 and 2, so the longest side has length 2
The original problem included the fact that cos C = -1/4, and asked for the smallest perimeter of the triangle that had all three side lengths as integers
Quote: ThatDonGuyQuote: ssho88cos(a) and cos(b) both are positive, then a and b both are Acute angles,
So, 180 - c = a+b,
L = sin(c)/sin(b) = sin(a)/ tan(b) + cos(a) = [(135)^0.5/16] * [7/(15)^0.5] +11/16 = 21/16 + 11/16 = 32/16 = 2
I'm not sure how you get from "a and b are both acute" to "L = sin c / sin b," but 2 is the correct answer.
Here's how I get it:
sin A = sqrt(1 - (11/16)2) = sqrt(15) * 3/16
sin B = sqrt(1 - (7/8)2) = sqrt(15) * 1/8
A + B + C = 180, so cos C = cos (180 - (A + B)) = -cos (A + B) = sin A sin B - cos A cos B = 15 * 3/128 - 77/128 = -32/128 = -1/4
sin C = sqrt(1 - (-1/4)2) = sqrt(15) * 1/4
Let X, Y, and Z be the lengths of the sides opposite angles A, B, and C, respectively
By the Law of Sines, sin A / X = sin B / Y = sin C / Z
The ratio X:Y:Z = the ratio sin A : sin B : sin C = 3 : 2 : 4 (otherwise the fractions aren't equal)
The side opposite B is shortest: since it is 1, the other two sides have lengths 3/2 and 2, so the longest side has length 2
The original problem included the fact that cos C = -1/4, and asked for the smallest perimeter of the triangle that had all three side lengths as integers
cos(a+b) = cos(a).cos(b) - sin(a).sin(b) = 0.25, So, (a+b) < 90 degree, c is a obtuse angle, longest length is opposite the angle c.
cos(b) > cos(a), then shortest side(1) is opposite angle b.
Quote: ssho88cos(a+b) = cos(a).cos(b) - sin(a).sin(b) = 0.25, So, (a+b) < 90 degree, c is a obtuse angle, longest length is opposite the angle c.
cos(b) > cos(a), then shortest side(1) is opposite angle b.
And sin c / L = sin b / 1 (Law of Sines), so L = sin c / sin b = sin (180-c) / sin b = sin (a + b) / sin b = (sin a cos b + cos a sin b) / sin b = sin a / tan b + cos a. Got it.
