2025 AIME I Problems/Problem 14

Problem

Let $ABCDE$ be a convex pentagon with $AB=14,$ $BC=7,$ $CD=24,$ $DE=13,$ $EA=26,$ and $\angle B=\angle E=60^{\circ}.$ For each point $X$ in the plane, define $f(X)=AX+BX+CX+DX+EX.$ The least possible value of $f(X)$ can be expressed as $m+n\sqrt{p},$ where $m$ and $n$ are positive integers and $p$ is not divisible by the square of any prime. Find $m+n+p.$

Solution 1

Assume $AX=a, BX=b, CX=c$ , by Ptolemy inequality we have $a+2b\geq \sqrt{3}XE; a+2c\geq \sqrt{3}BX$ , while the inequality is reached when both $CXAB$ and $AXDE$ are concyclic. Since $\angle{BCA}=\angle{BXA}=\angle{EXA}=\angle{ADE}=90^{\circ}$ , so $B,X,E$ lie on the same line. Thus, the desired value is then $(1+\frac{\sqrt{3}}{2})BE$ .

Note $\cos(\angle{DAC})=\frac{1}{7}, \cos (\angle{EAB})=-\frac{11}{14}, BE=38$ by LOC, the answer is then $38+19\sqrt{3}\implies \boxed{060}$

~ Bluesoul

2025 AIME I (Problems • Answer Key • Resources)
Preceded by Problem 13	Followed by Problem 15
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2025 AIME I Problems/Problem 14

Problem

Solution 1

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