AIME 2026 I · 第 14 题

Solution 1

Let the vertices of the pentagon be $A, B, C, D,$ and $E$. Since the pentagon is equiangular, each interior angle is $108^\circ$. Let $a=AB, b=BC, c=CD, d=DE, e=EA$.We define the following:$\phi = \frac{1+\sqrt{5}}{2}$ is the golden ratio, satisfying $\phi^2 = \phi + 1$ and $\phi^3 = 2\phi + 1$.$J = ab+bc+cd+de+ea$ is the sum of the products of adjacent sides. $K = ac+bd+ce+da+eb$ is the sum of the products of non-adjacent sides (diagonals). The idea will be to use the law of cosines and take cyclic sums. We want $(a+b+c+d+e)^2 = a^2 + b^2 + c^2 + d^2 + e^2 + 2J + 2K = 308 + 2J + 2K$.

Finding the value of J

By the law of cosines on triangle $ABC$, we have: $AC^2 = a^2 + b^2 - 2ab \cos(108^\circ)$. Taking the cyclic sum over all five triangles ($ABC, BCD, CDE, DEA, EAB$) gives:\begin{align*}\sum AC^2 &= \sum (a^2 + b^2) - 2\cos(108^\circ) \sum ab \end{align*} \begin{align*} 800 &= 308 + 308 - 2\cos(108^\circ)J \end{align*} Using $\cos(108^\circ) = \frac{1-\sqrt{5}}{4} = \frac{1-\phi}{2}$:\begin{align*}800 &= 616 - 2\left(\frac{1-\phi}{2}\right)J \end{align*} \begin{align*} 184 &= (\phi - 1)J\end{align*}Since $\phi - 1 = \frac{1}{\phi}$, we have $J = 184\phi$.

Finding the value of K

Extend line $EA$ through $A$ and $CB$ through $B$ to intersect at point $F$. Since the interior angles are $108^\circ$, the exterior angles $\angle FAB$ and $\angle FBA$ are both $72^\circ$. Consequently, in $\triangle FAB$, we find $\angle AFB = 180^\circ - 72^\circ - 72^\circ = 36^\circ$. Applying the Law of Sines in $\triangle FAB$ with side $AB=a$ and $AF=FB$, we determine the relationship to the golden ratio: $AF = a\phi = FB$. We obtain $FE = e + a\phi, FC = b + a\phi$. Using the Law of Cosines on $\triangle CFE$ for diagonal $CE$:

$$CE^2 = (e+a\phi)^2 + (b+a\phi)^2 - 2(e+a\phi)(b+a\phi)\cos(36^\circ).$$ Expanding the expression:

$$CE^2 = e^2 + b^2 + 2ea\phi + 2ab\phi + 2a^2\phi^2 - 2eb\cos(36^\circ) - 2ea\phi\cos(36^\circ) - 2ab\phi\cos(36^\circ) - 2a^2\phi^2\cos(36^\circ).$$ Taking the cyclic sum across all five diagonals ($CE, DA, EB, AC, BD$) gives:

$$\sum CE^2 = 2\sum a^2 + 4\phi J + 2\phi^2 \sum a^2 - 2\cos(36^\circ) K - 4\phi\cos(36^\circ) J - 2\phi^2\cos(36^\circ) \sum a^2$$ Substitute $\sum a^2 = 308$, $\cos(36^\circ) = \frac{\phi}{2}$, and by the definitions of $J$ and $K$ we find \begin{align*} 800 &= 2(308) + 4\phi J + 2\phi^2(308) - 2\left(\frac{\phi}{2}\right)K - 4\phi\left(\frac{\phi}{2}\right)J - 2\phi^2\left(\frac{\phi}{2}\right)(308) \end{align*} \begin{align*} 800 &= 616 + 4\phi J + 616\phi^2 - \phi K - 2\phi^2 J - 308\phi^3\end{align*} Simplify using $\phi^2 = \phi + 1$ and $\phi^3 = 2\phi + 1$:\begin{align*}800 &= 616 + 4\phi J + 616(\phi + 1) - \phi K - 2(\phi + 1)J - 308(2\phi + 1) \end{align*} \begin{align*} 800 &= 616 + 4\phi J + 616\phi + 616 - \phi K - 2\phi J - 2J - 616\phi - 308 \end{align*} \begin{align*} 800 &= 924 + 2\phi J - 2J - \phi K \end{align*} \begin{align*} 800 &= 924 + 2J(\phi - 1) - \phi K\end{align*}Substitute $J(\phi - 1) = 184$:\begin{align*}800 &= 924 + 2(184) - \phi K \end{align*} \begin{align*} 800 &= 1292 - \phi K \end{align*} \begin{align*} \phi K &= 492 \implies K = \frac{492}{\phi}\end{align*}Using $1/\phi = \phi - 1$, we obtain $K = 492(\phi - 1) = 492\phi - 492$.

Square of the Perimeter

\begin{align*} P^2 &= 308 + 2(184\phi) + 2(492\phi - 492) \end{align*} \begin{align*} P^2 &= 308 + 368\phi + 984\phi - 984 \end{align*} \begin{align*} P^2 &= 1352\phi - 676\end{align*} Substitute $\phi = \frac{1+\sqrt{5}}{2}$:\begin{align*}P^2 &= 1352\left(\frac{1+\sqrt{5}}{2}\right) - 676 \end{align*} \begin{align*} P^2 &= 676(1+\sqrt{5}) - 676 = 676\sqrt{5}\end{align*} With $m = 676$ and $n = 5$, we have $m + n = 676 + 5 = \boxed{681}$. I WILL PERSONALLY MARCH YOU ACROSS THE MICHIGAN AVENUE BRIDGE and drop you right into the Grand River! I WILL PERSONALLY MARCH YOU ACROSS THE MICHIGAN AVENUE BRIDGE and drop you right into the Grand River! I WILL PERSONALLY MARCH YOU ACROSS THE MICHIGAN AVENUE BRIDGE and drop you right into the Grand River! I WILL PERSONALLY MARCH YOU ACROSS THE MICHIGAN AVENUE BRIDGE and drop you right into the Grand River! I WILL PERSONALLY MARCH YOU ACROSS THE MICHIGAN AVENUE BRIDGE and drop you right into the Grand River! I WILL PERSONALLY MARCH YOU ACROSS THE MICHIGAN AVENUE BRIDGE and drop you right into the Grand River! I WILL PERSONALLY MARCH YOU ACROSS THE MICHIGAN AVENUE BRIDGE and drop you right into the Grand River! I WILL PERSONALLY MARCH YOU ACROSS THE MICHIGAN AVENUE BRIDGE and drop you right into the Grand River! I WILL PERSONALLY MARCH YOU ACROSS THE MICHIGAN AVENUE BRIDGE and drop you right into the Grand River!

~Hyperbolic_c0sine

Solution 2

idk what the above guy is doing but We can use complex numbers. Let $\omega=e^{2\pi i/5}$. Let $a,b,c,d,e$ be the side lengths in order. The first condition gives

$$a^2+b^2+c^2+d^2+e^2=308,$$ and as $|x|^2=x\overline x$, the second condition gives

$$\sum_\text{cyc}(a+b\omega)(a+b\omega^{-1})=2(a^2+b^2+c^2+d^2+e^2)+(\omega+\omega^{-1})(ab+bc+cd+de+ea)=800.$$ This is as, for example, the vector $AC$ is the vector $AB$ plus the vector $BC$. Alternatively, the vector $AC$ is also the vector $AE$ plus the vector $ED$ plus the vector $DC$, so the second condition also gives

$$\sum_\text{cyc}(a+b\omega+c\omega^2)(a+b\omega^{-1}+c\omega^{-2})=3(a^2+b^2+c^2+d^2+e^2)+2(\omega+\omega^{-1})(ab+bc+cd+de+ea)+(\omega^2+\omega^{-2})(ac+bd+ce+da+eb)$$ We can just solve for $a^2+b^2+c^2+d^2+e^2,ab+bc+cd+de+ea,ac+bd+ce+da+eb$ as this is a system of three variables and three equations. Then we want to find

$$(a+b+c+d+e)^2=a^2+b^2+c^2+d^2+e^2+2(ab+bc+cd+de+ea)+2(ac+bd+ce+da+eb).$$ Notice that $\omega+\omega^{-1}=2\cos72^\circ,\omega^2+\omega^{-2}=2\cos144^\circ$. I highly recommend memorizing these useful values if a similar problem pops up in the future.

Solution 3 (very smart cheese)

From the previous two solutions (reading those two before this one is highly suggested), it is easy to see that the square of the perimeter should equal $308+184(1+\sqrt{5})+X$. The problem is to find this value of $X$.

This is definitely the hardest part of the problem, so what we do instead is try to guess. Simplify the answer expression to $492+184\sqrt{5}+X$. We know the final answer must be of the form $m\sqrt{n}$, and in this case, we can easily see that $n$ should be $5$. Also, it is intuitive that $X=Y(\sqrt{5}-1)$ for some value of $Y$. If one can't see why, it's still quite straightforward to observe that this makes the most sense. Then, the answer expression should simplify to $(492-Y)+(184+Y)\sqrt{5}$. In order to get the answer as the product of an integer and $\sqrt{5}$, the rational part, $492-Y$, must equal to $0$. Thus, $Y=492$, and the answer expression should be $(184+492)\sqrt{5}=676\sqrt{5}$, making the final answer $676+5=\boxed{681}$.

~KevinChen_Yay

Solution 4

Let the five sides have length $a,b,c,d,e$. Then, we have $a^2+b^2+c^2+d^2+e^2=308$. Let $x=\sin(18^{\circ})=-\cos(108^{\circ})=\frac{\sqrt5-1}{4}$, so the diagonal connecting the endpoints of $a,b$ has length $a^2+b^2-2ab\cos(108^{\circ})=a^2+b^2+2abx$. Thus, the second condition is

$$2(a^2+b^2+c^2+d^2+e^2)+2x(ab+bc+cd+de+ea)=800$$ so $ab+bc+cd+de+ea=\frac{92}{x}=92\sqrt5+92$.

We want to find the value of $(a+b+c+d+e)^2=308+2(ab+bc+cd+de+ea)+2(ac+bd+ce+da+be)$, so we want to find $ac+bd+ce+da+be$.

Let the extensions of $a$ and $c$ intersect at point $P$ and the extensions of $a$ and $d$ intersect at point $Q$. Let $c$ and $d$ intersect at $A$. Then, clearly $\angle PQA=36^{\circ}$, $\angle QPA=36^{\circ}$ by the fact that the pentagon is equiangular. So, $PAQ$ is isosceles, and $PA=QA$. Notice that by the definition of the sine, $QA=\frac{e}{2x}+d$ and $PA=\frac{b}{2x}+c$. Thus,

$$\frac{e}{2x}+d=\frac{b}{2x}+c$$ To form $ac$, we want to multiply the equation by another thing that has a term of $\frac{b}{2x}+a$. Using the same method for another two sides, we get

$$\frac{b}{2x}+a=\frac{d}{2x}+e$$ multiplying these two give

$$(\frac{b}{2x}+c)(\frac{b}{2x}+a)=(\frac{e}{2x}+d)(\frac{d}{2x}+e)$$ $$\frac{b^2}{4x^2}+\frac{bc}{2x}+\frac{ab}{2x}+ac=\frac{de}{4x^2}+de+\frac{e^2}{2x}+\frac{d^2}{2x}$$ $$ac=\frac{de}{4x^2}+de+\frac{e^2}{2x}+\frac{d^2}{2x}-\frac{b^2}{4x^2}-\frac{bc}{2x}-\frac{ab}{2x}$$ we want to find $ac+bd+ce+da+be$, so we can add all equations of this type cyclically, obtaining

$$\sum_{cyc}ac=(\frac{1}{4x^2}-\frac{1}{x}+1)\sum_{cyc}ab+(\frac{1}{x}-\frac{1}{4x^2})\sum_{cyc}a^2$$ $$\sum_{cyc}ac=(92\sqrt5+92)(\frac{1}{4x^2}-\frac{1}{x}+1)+308(\frac{1}{x}-\frac{1}{4x^2})$$ Now, $x=\frac{\sqrt5-1}{4}$, so $\frac{1}{x}=\sqrt5+1$ and $\frac{1}{4x^2}=\frac{3+\sqrt5}{2}$. So,

$$\sum_{cyc}ac=(92\sqrt5+92)(\frac{3+\sqrt5}{2}-\sqrt5-1+1)+308(\sqrt5+1-\frac{3+\sqrt5}{2})$$ $$\sum_{cyc}ac=(92\sqrt5+92)(\frac{3-\sqrt5}{2})+308(\frac{\sqrt5-1}{2})$$ $$\sum_{cyc}ac=246\sqrt5-246$$ Plugging this back into $(a+b+c+d+e)^2=308+2(ab+bc+cd+de+ea)+2(ac+bd+ce+da+be)$ we have

$$(a+b+c+d+e)^2=308+184\sqrt5+184+492\sqrt5-492=676\sqrt5$$ answer extraction gives $676+5=\boxed{681}$.

~steve4916

Solution 5 ~ Jesse Zhang (FUNKCCP)

Consider an equiangular pentagon in the complex plane with vertices $V_1, V_2, V_3, V_4, V_5$ in counterclockwise order. Let the side lengths be $a_k = |V_{k+1} - V_k|$ for $k=1, \dots, 5$ (indices modulo 5), and let the diagonal lengths be $d_k = |V_{k+2} - V_k|$.

Since the pentagon is equiangular, each interior angle is given by $\dfrac{(5-2) \times 180^\circ}{5} = 108^\circ$.

We first establish a relationship between the side lengths and the diagonal lengths using the Law of Cosines on triangle $\triangle V_k V_{k+1} V_{k+2}$. The sides of this triangle are $a_k$, $a_{k+1}$, and the diagonal $d_k$. The included angle at vertex $V_{k+1}$ is $108^\circ$.

$$d_k^2 = a_k^2 + a_{k+1}^2 - 2 a_k a_{k+1} \cos(108^\circ).$$ We define the golden ratio $\phi = \frac{1+\sqrt{5}}{2}$. Recall that $\cos(108^\circ) = -\sin(18^\circ) = -\frac{\sqrt{5}-1}{4} = -\frac{1}{2\phi}$. Substituting this value, the equation becomes:

$$d_k^2 = a_k^2 + a_{k+1}^2 + \frac{1}{\phi} a_k a_{k+1}.$$ Let $S_1 = \sum_{k=1}^5 a_k^2$ and $S_2 = \sum_{k=1}^5 a_k a_{k+1}$. Summing the expression for $d_k^2$ over $k=1, \dots, 5$, we obtain the sum of the squares of the diagonals, denoted as $D^2$:

$$\sum_{k=1}^5 d_k^2 = \sum_{k=1}^5 a_k^2 + \sum_{k=1}^5 a_{k+1}^2 + \frac{1}{\phi} \sum_{k=1}^5 a_k a_{k+1}$$ $$D^2 = 2 S_1 + \frac{1}{\phi} S_2.$$ Given $S_1 = 308$ and $D^2 = 800$, we substitute these values:

$$800 = 2(308) + \frac{1}{\phi} S_2 \implies 800 = 616 + \frac{S_2}{\phi} \implies 184 = \frac{S_2}{\phi}.$$ Thus, $S_2 = 184\phi$.

Next, we utilize the geometric closure of the pentagon. Viewing the sides as vectors in the complex plane, let $\omega = e^{i 72^\circ}$. The vector corresponding to side $a_k$ points in the direction of $\omega^{k-1}$ (up to a global rotation). The condition $\sum_{k=1}^5 a_k \omega^{k-1} = 0$ implies:

$$\left| \sum_{k=1}^5 a_k \omega^{k-1} \right|^2 = 0.$$ Expanding this squared modulus:

$$\sum_{j=1}^5 \sum_{k=1}^5 a_j a_k \omega^{j-k} = 0.$$ We group the terms by the difference in indices $m \equiv j-k \pmod 5$. Let $S_3 = \sum_{k=1}^5 a_k a_{k+2}$. The coefficients are powers of $\omega$:

$\bullet$ For $m=0$, the coefficient is $\omega^0 = 1$. The sum is $S_1$.
$\bullet$ For $m=\pm 1$, the coefficient is $\omega + \omega^{-1} = 2\cos(72^\circ) = \frac{1}{\phi}$. The sum is $S_2$.
$\bullet$ For $m=\pm 2$, the coefficient is $\omega^2 + \omega^{-2} = 2\cos(144^\circ) = -\phi$. The sum is $S_3$.

The closure equation becomes:

$$S_1 + \frac{1}{\phi}S_2 - \phi S_3 = 0.$$ Substituting the known values $S_1 = 308$ and $\frac{S_2}{\phi} = 184$:

$$308 + 184 - \phi S_3 = 0 \implies 492 = \phi S_3 \implies S_3 = \frac{492}{\phi}.$$ We are asked to find the square of the perimeter $P = \sum_{k=1}^5 a_k$.

$$P^2 = \left( \sum_{k=1}^5 a_k \right)^2 = \sum_{k=1}^5 a_k^2 + 2 \sum_{k=1}^5 a_k a_{k+1} + 2 \sum_{k=1}^5 a_k a_{k+2}.$$ $$P^2 = S_1 + 2 S_2 + 2 S_3.$$ Substituting the expressions in terms of $\phi$:

$$P^2 = 308 + 2(184\phi) + 2\left(\frac{492}{\phi}\right).$$ Using the identity $\frac{1}{\phi} = \phi - 1$:

$$P^2 = 308 + 368\phi + 984(\phi - 1) = 308 + 368\phi + 984\phi - 984.$$ $$P^2 = 1352\phi - 676.$$ Finally, substitute $\phi = \frac{1+\sqrt{5}}{2}$:

$$P^2 = 1352\left(\frac{1+\sqrt{5}}{2}\right) - 676 = 676(1+\sqrt{5}) - 676 = 676 + 676\sqrt{5} - 676.$$ $$P^2 = 676\sqrt{5}.$$ This is in the form $m\sqrt{n}$ with $m = 676$ and $n = 5$. Since $5$ is a prime number, it is square-free. The value $m+n$ is:

Video Solution

https://youtu.be/9y_ixU2AYfk