# Two math problems I made up for my school's math league team

• 7 years ago · Quote · #1

I originally posted this in a chess.com group, but since it is a small group, I thought more people might appreciate it here.

I made up both of these problems on my sabbath, when I can't write/use a calculator, so both of them can (and should!) be done in one's head.  Scroll down for the solutions.

1.  The CIA is using sensory deprivation chambers as a new form of "enhanced" interrogation.  They want you to tell them what the area of a triangle is if the triangle has side lengths of 5 in, 9 in, and sqrt(34) in.  Since you are in a sensory deprivation chamber, you cannot use writing materials, counting materials, or a calculator.  In addition, it is against your unbreakable moral principles to use Heron's formula.  What is the area of the triangle?

2.  There is a 20% probability that an event will occur at least once in a year.  What is the approximate probability (accurate to 1/2 a percent) that the event will occur at least once in 10 years?  The hard part about this problem is probably doing it in one's head.

Solutions

1.

Envision the triangle above.  You need to solve for h so as to use the formula A = 1/2bh.  From the above diagram, you can derive the system of equations x + y = 9; x^2 + h^2 = 25; y^2 + h^2 = 34.  Fortunately, it is not particularly difficult to solve this system for h in one's head, since many squared variables "drop out."  x + y = 9 so x = 9 - y.  Substitute that into x^2 + h^2 = 25 and you get (9 - y)^2 + h^2 = 25 so y^2 - 18y + 81 + h^2 = 25 so y^2 + h^2 - 18y + 56 = 0.  The third equation, y^2 + h^2 = 34, can be changed to y^2 + h^2 - 34 = 0 and then substituted into y^2 + h^2 - 18y + 56 = 0 to get y^2 + h^2 - 18y + 56 = y^2 + h^2 -34 so -18y + 56 = -34 so -18y = -90 so y = 5.  Substitute that into y^2 + h^2 = 34 to get 25 + h^2 = 34 so h^2 = 9 so h = 3.  A = 1/2bh = 1/2(9)3 = 13.5 sq. in.

2. P(event occurs at least once) = 1 - P(event does not occur) = 1 - 0.8^10 = 1 - (4/5)^10 = 1 - 4^10/5^10 = 1 - (2 * 2)^10/(10/2)^10 = 1 - (2^10 * 2^10)/(10^10/2^10) = 1 - (1024 * 1024)/(10^10/1024) = 1 - (a little more than 10^6)/(a little less than 10^7) = 1 - a little more than 0.1 = a little less than 0.9.  The probability that the event will occur at least once in 10 years is about 89% (the probability as determined by a calculator is 89.26258176...%).

Don't be intimidated by the number of equations in each of the solutions; trust me, it really isn't difficult to do in one's head.  I carefully explained the solution to #1 orally to a pretty good math student, and he was, with some thought, then able to solve for a the area of a different triangle in his head, given their side lengths.  I actually got the idea for #2 when I was reading an article that predicted the probability of a nuclear attack against a major city occurring in any given year to be 20% (scary thought!  I'm hoping that's an overestimate!).

• 7 years ago · Quote · #2

Well yes you're right.  I've only studied basic trig, so I wouldn't know... but could this be a proof for the law of cosines?  I've tried to turn it into a proof for Heron's formula, and it gets messy... really messy... it makes me make squealing noises.