AIME 2025 I · 第 14 题

Solution 1

Assume $AX=a$, $BX=b$, $CX=c$, $DX=d$, $EX=e$, by Ptolemy inequality we have $a+2d\geq \sqrt{3}XE; a+2c\geq \sqrt{3}BX$, and as we are minimizing the value of the LHS, we want the inequality reached; this happens when both quadrilaterals $CXAB$ and $AXDE$ are concyclic. We notice that we have right angles $\angle{BXA}=\angle{BCA}$ and $\angle{EDA}=\angle{EXA}$ as they both subtend a diameter of circles by noticing that we have formed Pythagorean triples (along with being told that $\angle{B}=\angle{E}=60^{\circ}$. ）And so, since we have supplementary angles, $B$, $X$, and $E$ are collinear. Thus, we add up the two equalities to get $2a+2c+2d=\sqrt{3}(XE+BX)$, or $a+c+d=\frac{\sqrt{3}}{2}(BE)$. Thus we want to find $a+b+c+d+e=(1+\frac{\sqrt{3}}{2})BE$.

By Law of Cosines, we know that $\cos(\angle{DAC})=\frac{1}{7}$, and since $\angle{CAB}=\angle{DAE}=30^{\circ}$, we know $\cos(\angle{EAB})=cos(\angle{CAB}+\angle{DAE}+\arccos(\frac{1}{7}))=cos(60^{\circ}+\arccos(\frac{1}{7}))$, which we find with the cosine angle addition formula and the fact that $\sin(\arccos(\frac{1}{7}))=\sqrt{1-(\frac{1}{7})^2}$. Then, once again with Law of Cosines, we find that $BE=38$, and thus the minimum of $a+b+c+d+e$, or $(1+\frac{\sqrt{3}}{2})BE$, is $38+19\sqrt{3}\implies\boxed{060}$.

~Bluesoul ~hashbrown2009

Rewritten and edited by Juwushu.

Solution 2

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

Solution 3(Fermat Point)

$AE = 26$, $DE = 13$, and $\angle{E}={60}^{\circ}$, which means that $\triangle{AED}$ is a 30-60-90 triangle, so $AD = 13\sqrt3$, $\angle{EAD}={30}^{\circ}$, and $\angle{ADE}={90}^{\circ}$. Similar with $\triangle{ABC}$, $AD = 7\sqrt3$, $\angle{BAC}={30}^{\circ}$, and $\angle{ACB}={90}^{\circ}$

To solve the question, we would have to locate point $X$ first. We first consider the points $B$ and $E$. For the distance of $X$ to $B$ and $E$ to become the shortest, $X$ should lay on $BE$. For $X$ to be closest to point $A$, it should be on the foot of perpendicular from $A$ to line $BE$.

Consider about $C$ and $D$. $\angle{ADE}=\angle{AXE}={90}^{\circ}$, so $AXDE$ is cyclic. Therefore, $\angle{EXD}=\angle{EAD}=\angle{BXC}={30}^{\circ}$. $\angle{DXC}=\angle{AXD}=\angle{AXD}={120}^{\circ}$, so $X$ is coincidently the Fermat Point of $ADC$.

To calculate the $f(X)$, we divide it into 2 parts: the sum of distance to $A$, $C$, and $D$ and the sum of distance to $B$ and $E$. Applying Ptolemy's Theorem in $AXDE$ and $AXCB$,

$$AX+2DX=\sqrt3EX$$ $$and$$ $$AX+2CX=\sqrt3BX$$ We get that $AX+DX+CX = \frac{\sqrt3}{2}\cdot(EX+BX) = \frac{\sqrt3}{2}BE$

Construct equilateral triangle $\triangle{ADE}$ outside of $\triangle{ADC}$ on side $AD$. Because $X$ is the Fermat Point, $FC=AX+DX+CX$. To calculate $FC$, we needed to utilize $\angle{FDC}$

$$\angle{FDC}=\angle{FDA}+\angle{ADC}={60}^{\circ}+\angle{ADC}$$ .

From $\triangle{ADC}$, we know: \begin{align*} \cos{\angle{ADC}} & = \frac{(13\sqrt3)^2+{24}^2-(7\sqrt3)^2}{2\cdot13\sqrt3\cdot24}\\ &= \frac{936}{2\cdot13\sqrt3\cdot24}\\ &= \frac{\sqrt3}{2}\\ \end{align*}

This shows that $\angle{ADC} = {30}^{\circ}$, which means that $\angle{FDC} = {60}^{\circ}+{30}^{\circ}={90}^{\circ}$

Using the Pythagorean Theorem,

$$FC = \sqrt{(13\sqrt3)^2+24^2)}=19\sqrt3=AX+DX+CD$$ $$f(X) = 19\sqrt3(1+\frac{2}{\sqrt3})=38+19\sqrt3$$ The answer is $38+19+3=\boxed{060}$

~cassphe

Video Solution 1 by StressedPineapple

https://youtube.com/watch?v=ttQrUg6A4PA&t=334s

Video Solution 2

2025 AIME I #14

MathProblemSolvingSkills.com

Great solution!