Question

In $\triangle ABC,$ $\angle ABC=90^\circ$ and $BA=BC=\sqrt2.$ Points $P_1,P_2,\ldots,P_{2024}$ lie on hypotenuse $AC$ so that $AP_1=P_1P_2$ $=P_2P_3$ $=\cdots$ $=P_{2023}P_{2024}$ $=P_{2024}C.$

What is the length of the vector sum $\begin{aligned} &\vec{BP_1}+\vec{BP_2}+\vec{BP_3} \\ &\quad {}+\cdots+\vec{BP_{2024}}? \end{aligned}$

$1011$

$1012$

$2023$

$2024$

$2025$

Answer: D

Concepts:vector median (geometry)symmetry

Difficulty rating: 1430

Solution:

The points $P_k$ are symmetric about the midpoint $M$ of $AC,$ so pairing $P_k$ with its mirror gives $\vec{BP_k}+\vec{BP_{2025-k}}=2\,\vec{BM}.$ Hence the whole sum is $2024\,\vec{BM}.$ In a right triangle the median to the hypotenuse has length half the hypotenuse; here $AC=2,$ so $BM=1.$ The length of the sum is $2024\cdot1=2024.$ Thus, the correct answer is D.

2024 AMC 12A Problem 7

Answer: D

Difficulty rating: 1430

Solution:

Problem 7 in Other Years