*Congratulations, Michael Grandel and Timothy White, Our Math Whiz Competition Winners!*

*Congratulations, Michael Grandel and Timothy White, Our Math Whiz Competition Winners!*

September 7, 2015

Parker Academics would like to thank all of you who attempted our Math Whiz Challenge. You put up a most valiant fight. Yet, in the end, two have prevailed! We would like to congratulate Michael Grandel of Midwood High School in Brooklyn, New York and Timothy White who, at the time of the competition, was a recent graduate of Sir Wilifrid Laurier Secondary School in Ottawa, Ontario! You are the winners of Parker Academics' inaugural SAT Math Whiz Competition! Scroll down to learn more about these super brilliant young men, and take a look at their explanations. Stay tuned for our next competition and our tournament, which has a cash prize!

Michael Grandel

Midwood High School

Brooklyn, New York

My name is Michael Grandel and I am a high school student in Midwood High school in Brooklyn, New York. I will be graduating in the year of 2017 and definitely plan on pursuing college. My plan is to pursue a profession within the field of math and physics, such as astrophysics. Aside from college, I am also an athlete. I play volleyball, tennis, and swimming. I wish to be able to prove to people that it is possible to have a professional athlete that is the most scholarly scholar. On the way to this goal, I have achieved an incredible fetish to mathematics. I love the challenge and the puzzle it brings, where there is always a solution. My favorite activity to within math is to do proofs of all the rules and formulas we've ever been taught; from simple area formulas, to the binomial theorem and derivatives.

Michael's Explanation:

This question is about how well you can map it in my opinion. You begin with 2 separate triangles, ABC and XYZ. You are told AC is congruent to XZ and they are later conjoined into a single segment, AZ. Now imagine putting the two triangles together but connecting XZ and AC together. In the end you get something at first that looks like a parallelogram, AYZB. However since no extra information was given about the triangles, you can't assume anything regarding the polygon being a parallelogram gram. Now here's the algebra. You are given two pairs of ratios. Beginning with angle YAZ and AZB, their ratio is 2 to 3. Si, YAZ = 2x and AZB = 3x. Now we do the same with YZA and BAZ, with 4 to 2. Now we can't use x as the common ratio because it doesn't necessarily have to be the same multiplying factor used as the previous one. For example, if the ratio is 2X/3x you get 2/3. Any value you put in x they will cancel, except zero of course. Then if you do 4Y/2Y you will get 4/2. Now Y can be the same value as X, or it can be different. After all, it will cancel out. Now with our ratios, we are given further information. Angles ZBA and ZYA are right angles. This makes our polygon look more rectangular, but not quite because consecutive angles are not right angles. This is useful because we have 2 right triangles. In a right triangle, the sum of 2 angles, excluding the right angle, is 90. They are complimentary. This means, for the upper triangle, 4Y+2X=90. For the lower, 3y+3x=90. Since they both equal 90, they equal each other. Thus we have 4y+2x = 3y + 3x. y = x. This is extremely useful because since we know y = x, we can make all the variables we used equal. So let's makes them all x. Using our equations again, we get: 3x + 3x = 90. X=15. With 15, we can find all the angles. Luckily, we get special right triangles when we plug 15 in for x. For triangle ZAB, we have a 45 45 90 triangle. Thus ZB and AB are congruent, and in the ratio of x:x:x (radical 2). In the triangle YZA we get a 30, 60,90 triangle, in the ratio of x,x (radical3),2x. We are given YZ is 1, which is opposite the 30 degree angle. This means in the ratio, x is one. We can find YA= radical 3 and ZA= 2. Since ZA is also part of the isosceles right triangle, it is following the special ratio. Since X (radical2) =2, the common ratio is 2. Thus AB is 2 and ZB is 2. To find the area, add the sum of the areas of the 2 triangles. A = 1/2 (1)(radical3) + 1/2 ( radical 2)(radical2) A= 1 +(Radical 3)/2 A= (2 + radical3)/2

Timothy White

Sir Wilifrid Laurier Secondary School

Ottawa, Ontario

Hey! My name is Timothy White. In June I graduated from Sir Wilfrid Laurier Secondary School, in Ottawa, Ontario. I'm currently a student attending Carleton University, studying for an undergraduate degree in Mathematics and hope to soon go on for a PhD in Mathematics with a concentration in Pure Math or Number Theory. I've loved math as far back as I can remember, and have wanted to be a mathematician since I first entered grade school. My goal in life is to solve the mystery of prime patterning, specifically, the Goldbach Theorem, but in the meantime, I settle for playing with numbers and making programs.

I am a fanatic of many things – I enjoy music, games, the performance arts and teaching. I play music in my spare time, as well as minor composition. I also enjoy a large variety of games, including wildly popular online games to smaller 1-on-1 fighting platformers to top-down roguelike shooters. I’m also a stage actor, having starred in my high school’s most recent award-winning musical and am a part of multiple acting clubs at my Uni, including musical and improv clubs. Finally, I work professionally as a tutor for math of all grade school levels. One day I wish I can use my teaching abilities to make math easier for everybody, and so they can enjoy the beauty of our number system.

I am very passionate about mathematics, to the point of having sought out the Waterloo math contests in high school, and placing incredibly high and earning many awards for my math abilities. This year I hope to continue that record through the Putnam contests.

I’m only just beginning my academic record with writing proofs and explanatory solutions to various problems, but I hope that my future’s work proves as good as my past’s.

Timothy's Explanation: http://i.imgur.com/Stz9a11.png