2024 AMC 12A Problem 7

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

The points are symmetric about the midpoint

M

of

AC,

so the vectors pair up

The sum equals

2024\,\vec{BM};

the median to the hypotenuse of a right triangle is half the hypotenuse

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.

Problem 7 in other years