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

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==Problem==
  
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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>
  
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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{BXA}=\angle{BCA}=\angle{EDA}=\angle{EXA}=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
  
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==Solution 2==
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<asy>
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size(10cm);
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import math; import geometry; import olympiad;
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point A,B,C,D,F,P,X; A=(0,-7sqrt(3)); B=(-7,0); C=(0,0); D=(156/7,-36sqrt(3)/7); F=(169/7,-88sqrt(3)/7); P=(132/7,60sqrt(3)/7); X=(8580/2527,-10604sqrt(3)/2527);
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draw(A--B--C--P--D--F--A--C--D--A--P); draw(B--F); draw(circumcircle(A,B,C)); draw(circumcircle(A,D,F)); draw(circumcircle(C,P,D)); draw(C--X--D);
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label("A",A,SE); label("B",B,W); label("C",C,NW); label("P",P,N); label("D",D,E); label("E",F,SE); label("X",X,E);
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</asy>
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Firstly, note that <math>\triangle ABC</math> and <math>\triangle ADE</math> are just 30-60-90 triangles. Let <math>X</math> be the Fermat point of <math>\triangle ACD</math>, with motivation stemming from considering the pentagon as <math>\triangle ACD</math> with the two 30-60-90 extensions. Note that <math>AX+CX+DX</math> is minimized at this point when <math>\angle AXC=\angle CXD=\angle AXD=120^{\circ}</math>. Because we have <math>\angle ABC=\angle AED=60^{\circ}</math>, then <math>ABCX</math> and <math>AXDE</math> are both cyclic. Then we have <math>\angle AXE=\angle ADE=90^{\circ}</math> and <math>\angle BXA=\angle BCA=90^{\circ}</math>. Then it turns out that we actually have <math>\angle BXE=90^{\circ}+90^{\circ}=180^{\circ}</math>, implying that <math>B</math>, <math>X</math> and <math>E</math> are collinear. Now, by the triangle inequality, we must have <math>BX+XE\geq BE</math>, with equality occurring when <math>X</math> is on <math>BE</math>. Thus <math>AX+CX+DX</math> and <math>BX+EX</math> are minimized, so this point <math>X</math> is our desired point.
  
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yet
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Firstly, we will find <math>BX+EX=BE</math>. We have that <math>AC=7\sqrt{3}</math> and <math>AD=13\sqrt{3}</math>, so applying the Law of Cosines in <math>\triangle ACD</math>, we get <cmath>147+507-2(7\sqrt{3})(13\sqrt{3})\cos (\angle CAD)=576\implies \cos(\angle CAD)=\frac{1}{7}.</cmath> It follows as a result that <math>\sin (\angle CAD)=\frac{4\sqrt{3}}{7}</math>. Then we want to find <math>\cos (\angle BAE)</math>. We can do this by seeing <cmath>\cos (\angle BAE)=\cos (\angle CAD+60^{\circ})=\cos (\angle CAD)\cos 60^{\circ}-\sin (\angle CAD)\sin 60^{\circ}=\frac{1}{7}\cdot \frac{1}{2}-\frac{4\sqrt{3}}{7}\cdot \frac{\sqrt{3}}{2}=-\frac{11}{14}.</cmath> Applying the Law of Cosines again in <math>\triangle BAE</math>, then because <math>AB=14</math> and <math>AE=26</math>, we have <cmath>14^2+26^2-2(14)(26)\left (-\frac{11}{14}\right )=196+676-2\cdot 26\cdot (-11)=872+572=1444=BE^2,</cmath> so it follows that <math>BE=38=BX+EX</math>.
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Now, we will find the value of <math>AX+CX+DX</math>. Construct a point <math>P</math> outside such that <math>\triangle CPD</math> is equilateral, as shown. By property of fermat point, then <math>A</math>, <math>X</math>, and <math>P</math> are collinear. Additionally, <math>\angle CXD=120^{\circ}</math>, so <math>CPDX</math> is cyclic. Applying Ptolemy's Theorem, we have that <math>(CX)(PD)+(CP)(XD)=(XP)(CD)</math>. But since <math>\triangle CPD</math> is equilateral, it follows that <math>CX+DX=PX</math>. Then <math>AX+CX+DX=AX+PX=AP</math>, so we wish to find <math>AP</math>. Applying the Law of Cosines in <math>\triangle ACD</math>, we have that <cmath>(13\sqrt{3})^2+24^2-2(13\sqrt{3})(24)\cos (\angle ADC)=(7\sqrt{3})^2\implies \cos (\angle ADC)=\frac{\sqrt{3}}{2}\implies \angle ADC=30^{\circ}.</cmath> Then because <math>\angle CDP=60^{\circ}</math>, then <math>\angle ADP=90^{\circ}</math>, so we can find <math>AP</math> simply with the Pythagorean Theorem. We know <math>AD=13\sqrt{3}</math> and <math>DP=CD=24</math>, so <math>AP=\sqrt{(13\sqrt{3})^2+24^2}=19\sqrt{3}</math>.
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We then have <math>f(X)=AX+BX+CX+DX+EX=(BX+EX)+(AX+CX+DX)=BE+AP=38+19\sqrt{3}</math>, which is our minimum value. Therefore, the answer to the problem is <math>38+19+3=\boxed{060}</math>.
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~ethanzhang1001
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==See also==
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{{AIME box|year=2025|num-b=13|num-a=15|n=I}}
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{{MAA Notice}}

Latest revision as of 12:36, 14 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 $\angle{BXA}=\angle{BCA}=\angle{EDA}=\angle{EXA}=90^{\circ}$, 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

Solution 2

[asy] size(10cm); import math; import geometry; import olympiad; point A,B,C,D,F,P,X; A=(0,-7sqrt(3)); B=(-7,0); C=(0,0); D=(156/7,-36sqrt(3)/7); F=(169/7,-88sqrt(3)/7); P=(132/7,60sqrt(3)/7); X=(8580/2527,-10604sqrt(3)/2527); draw(A--B--C--P--D--F--A--C--D--A--P); draw(B--F); draw(circumcircle(A,B,C)); draw(circumcircle(A,D,F)); draw(circumcircle(C,P,D)); draw(C--X--D); label("A",A,SE); label("B",B,W); label("C",C,NW); label("P",P,N); label("D",D,E); label("E",F,SE); label("X",X,E); [/asy] Firstly, note that $\triangle ABC$ and $\triangle ADE$ are just 30-60-90 triangles. Let $X$ be the Fermat point of $\triangle ACD$, with motivation stemming from considering the pentagon as $\triangle ACD$ with the two 30-60-90 extensions. Note that $AX+CX+DX$ is minimized at this point when $\angle AXC=\angle CXD=\angle AXD=120^{\circ}$. Because we have $\angle ABC=\angle AED=60^{\circ}$, then $ABCX$ and $AXDE$ are both cyclic. Then we have $\angle AXE=\angle ADE=90^{\circ}$ and $\angle BXA=\angle BCA=90^{\circ}$. Then it turns out that we actually have $\angle BXE=90^{\circ}+90^{\circ}=180^{\circ}$, implying that $B$, $X$ and $E$ are collinear. Now, by the triangle inequality, we must have $BX+XE\geq BE$, with equality occurring when $X$ is on $BE$. Thus $AX+CX+DX$ and $BX+EX$ are minimized, so this point $X$ is our desired point.

Firstly, we will find $BX+EX=BE$. We have that $AC=7\sqrt{3}$ and $AD=13\sqrt{3}$, so applying the Law of Cosines in $\triangle ACD$, we get \[147+507-2(7\sqrt{3})(13\sqrt{3})\cos (\angle CAD)=576\implies \cos(\angle CAD)=\frac{1}{7}.\] It follows as a result that $\sin (\angle CAD)=\frac{4\sqrt{3}}{7}$. Then we want to find $\cos (\angle BAE)$. We can do this by seeing \[\cos (\angle BAE)=\cos (\angle CAD+60^{\circ})=\cos (\angle CAD)\cos 60^{\circ}-\sin (\angle CAD)\sin 60^{\circ}=\frac{1}{7}\cdot \frac{1}{2}-\frac{4\sqrt{3}}{7}\cdot \frac{\sqrt{3}}{2}=-\frac{11}{14}.\] Applying the Law of Cosines again in $\triangle BAE$, then because $AB=14$ and $AE=26$, we have \[14^2+26^2-2(14)(26)\left (-\frac{11}{14}\right )=196+676-2\cdot 26\cdot (-11)=872+572=1444=BE^2,\] so it follows that $BE=38=BX+EX$.

Now, we will find the value of $AX+CX+DX$. Construct a point $P$ outside such that $\triangle CPD$ is equilateral, as shown. By property of fermat point, then $A$, $X$, and $P$ are collinear. Additionally, $\angle CXD=120^{\circ}$, so $CPDX$ is cyclic. Applying Ptolemy's Theorem, we have that $(CX)(PD)+(CP)(XD)=(XP)(CD)$. But since $\triangle CPD$ is equilateral, it follows that $CX+DX=PX$. Then $AX+CX+DX=AX+PX=AP$, so we wish to find $AP$. Applying the Law of Cosines in $\triangle ACD$, we have that \[(13\sqrt{3})^2+24^2-2(13\sqrt{3})(24)\cos (\angle ADC)=(7\sqrt{3})^2\implies \cos (\angle ADC)=\frac{\sqrt{3}}{2}\implies \angle ADC=30^{\circ}.\] Then because $\angle CDP=60^{\circ}$, then $\angle ADP=90^{\circ}$, so we can find $AP$ simply with the Pythagorean Theorem. We know $AD=13\sqrt{3}$ and $DP=CD=24$, so $AP=\sqrt{(13\sqrt{3})^2+24^2}=19\sqrt{3}$.

We then have $f(X)=AX+BX+CX+DX+EX=(BX+EX)+(AX+CX+DX)=BE+AP=38+19\sqrt{3}$, which is our minimum value. Therefore, the answer to the problem is $38+19+3=\boxed{060}$.

~ethanzhang1001

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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