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Difference between revisions of "2025 AIME I Problems/Problem 14"

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Let <math>ABCDE</math> be a convex pentagon with <math>AB=14,</math> <math>BC=7,</math> <math>CD=24,</math> <math>DE=13,</math> <math>EA=26,</math> and <math>\angle B=\angle E=60^{\circ}.</math> For each point <math>X</math> in the plane, define <math>f(X)=AX+BX+CX+DX+EX.</math> The least possible value of <math>f(X)</math> can be expressed as <math>m+n\sqrt{p},</math> where <math>m</math> and <math>n</math> are positive integers and <math>p</math> is not divisible by the square of any prime. Find <math>m+n+p.</math>
 
Let <math>ABCDE</math> be a convex pentagon with <math>AB=14,</math> <math>BC=7,</math> <math>CD=24,</math> <math>DE=13,</math> <math>EA=26,</math> and <math>\angle B=\angle E=60^{\circ}.</math> For each point <math>X</math> in the plane, define <math>f(X)=AX+BX+CX+DX+EX.</math> The least possible value of <math>f(X)</math> can be expressed as <math>m+n\sqrt{p},</math> where <math>m</math> and <math>n</math> are positive integers and <math>p</math> is not divisible by the square of any prime. Find <math>m+n+p.</math>
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==Solution 1==
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Assume <math>AX=a, BX=b, CX=c</math>, by Ptolemy inequality we have <math>a+2b\geq \sqrt{3}XE; a+2c\geq \sqrt{3}BX</math>, while the inequality is reached when both <math>CXAB</math> and <math>AXDE</math> are concyclic. Since <math>\angle{BCA}=\angle{BXA}=\angle{EXA}=\angle{ADE}=90^{\circ}</math>, so <math>B,X,E</math> lie on the same line. Thus, the desired value is then <math>(1+\frac{\sqrt{3}}{2})BE</math>.
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Note <math>\cos(\angle{DAC})=\frac{1}{7}, \cos (\angle{EAB})=-\frac{11}{14}, BE=38</math> by LOC< the answer is then <math>38+19\sqrt{3}\implies \boxed{060}</math>
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~ Bluesoul
  
 
==See also==
 
==See also==

Revision as of 23:57, 13 February 2025

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

See also

2025 AIME I (Problems • Answer Key • Resources)
Preceded by
Problem 13 		Followed by
Problem 15
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15
All AIME Problems and Solutions

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