olymmath

AMC 12 2018a Problem 22

  • Paper

The solutions to the equations z2=4+415iz^2=4+4\sqrt{15}i and z2=2+23i,z^2=2+2\sqrt 3i, where i=−1,i=\sqrt{-1}, form the vertices of a parallelogram in the complex plane. The area of this parallelogram can be written in the form pq−rs,p\sqrt q-r\sqrt s, where p,p, q,q, r,r, and ss are positive integers and neither qq nor ss is divisible by the square of any prime number. What is p+q+r+s?p+q+r+s?

(A) 20(B) 21(C) 22(D) 23(E) 24\textbf{(A) } 20 \textbf{(B) } 21 \textbf{(C) } 22 \textbf{(D) } 23 \textbf{(E) } 24

Dataset text AMC 12 2000 to 2025. Quoted with source.

Show the answer

A

Solutions

No solution is on olymmath yet. Official and human solutions come first when they arrive. AI solutions come after them, with a label.

Add a solution

Similar problems

These problems have the most words in common with this statement.

Locator amc12/2018a/p22 · Tag CX1H6SM4TH · Edit this page on GitHub · Keys: j next, k previous