1999 AMC 8 Problem 23

Attempt Problem 23 of the 1999 AMC 8 below, then check your answer against the professionally curated solution from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 1999 AMC 8 solutions, or check the answer key.

All problems are used with official legal permission of the Mathematical Association of America (MAA).

23.

Square ABCDABCD has sides of length 3.3. Segments CMCM and CNCN divide the square’s area into three equal parts. How long is segment CM?CM?
10\sqrt{10}
12\sqrt{12}
13\sqrt{13}
14\sqrt{14}
15\sqrt{15}
Answer: C
Concepts:triangle areaPythagorean Theorem
Small hint:
Each of the three regions has area one-third of the square
Big hint:
Use the area of △BMC\triangle BMC to find BMBM
Solution:
The area of the square is 32=9,3^2 = 9, which means that the area of one region is 9÷3=3.9 \div 3 = 3.
This means the area of △BMC\triangle BMC is 3,3, which means that 12⋅3⋅BM=3 \dfrac{1}{2} \cdot 3 \cdot BM = 3 BM=2. BM = 2.
Since △BMC\triangle BMC is right, we have CM=22+32=13. CM = \sqrt{2^2 + 3^2} = \sqrt{13}.
Thus, C is the correct answer.
LIVE by Po-Shen Loh
Programs

LIVE Classes

Self-Paced Classes

Stars

Scholarship

About Us

Who Is Po-Shen Loh?

Email Contact

Newsletter

Jobs

Become a Star

For a more Thought Full world.

© 2026 Expii, Inc.