1550.42 – Pythagoras Party

The Biologist/Mathematician Collective is holding a vote to determine which dessert should be served at this year’s Pythagoras Party. The three nominees are: pi pie (a classic), cake designed like Wacław Sierpiński’s famous fractal triangles, and mitosis cupcakes.

In the election, a total of 15,424 votes are cast. The Wacław cake receives 1006 more than the pi pie and 1213 more than the mitosis cupcakes. How many votes does each dessert receive?

Solution

Pi pie: x votesMitosis cupcakes: y votesWacław cake: x+1006 and y+1213\begin{align*}\text{Pi pie: }& x \text{ votes} \\\text{Mitosis cupcakes: }& y \text{ votes} \\\text{Wacław cake: }& x + 1006 \text{ and } y + 1213 \\\end{align*}

x+1006+x+y=154242x+y=14418x+1006=y+1213x−y=2073x=14625x=4875 (pi pie)x+1006=5881 (Wacław cake)15424−(4875+5881)=4668 (mitosis cupcakes)\begin{align*}x + 1006 + x + y &= 15424 \\2x + y &= 14418 \\\\x + 1006 &= y + 1213 \\x - y &= 207 \\\\3x &= 14625 \\x &= 4875 \text{ (pi pie)} \\x + 1006 &= 5881 \text{ (Wacław cake)} \\15424 - (4875 + 5881) &= 4668 \text{ (mitosis cupcakes)}\end{align*}

A tight race – yet everyone is pleased with the result.

Can this problem be done without using formal algebra?

PS: For the inquiring (student) mind: With more than 15,000 people at the party, why would all the desserts have to be the same? Why not make all three kinds, basically in the ratio of how people voted? And – bonus points – how many of each might that be? And, finally, how might this vast number of desserts be served?

Further, how about seeing whether a student or team of students could re-cast the problem in a way that makes sense? Possibly some other kind of election?