AIME 2026 II · 第 8 题

Solution 1

Since $\triangle ABC$ is isosceles with $AB = BC$, the angle bisector, median, and altitude from $B$ to $AC$ coincide, which is the perpendicular bisector of $AC$. The incenter $I$ lies on this common line. Let $D$ be the intersection of this line with $AC$, so $D$ is the midpoint of $AC$ and $ID \perp AC$.

We set $AC = 2$, so $AD = DC = 1$. Let $ID = x$.

The sides $AI$ and $CI$ are equal, by Pythagorean Theorem,

$$AI = CI = \sqrt{AD^2 + ID^2} = \sqrt{1 + x^2}.$$ Thus the perimeter of $\triangle AIC$ is:

$$\text{Perimeter}(\triangle AIC) = AC + AI + CI = 2 + 2\sqrt{1 + x^2}.$$ Let $\angle IAD = \theta$. Then

$$\cos\theta = \frac{AD}{AI} = \frac{1}{\sqrt{1 + x^2}}.$$ Using the double-angle formula for cosine:

$$\cos 2\theta = 2\cos^2\theta - 1 = 2\left(\frac{1}{1 + x^2}\right) - 1 = \frac{2 - (1 + x^2)}{1 + x^2} = \frac{1 - x^2}{1 + x^2}.$$ In $\triangle ABC$, $\angle BAC = 2\theta$, and $AD = 1$ is half of $AC$. Thus:

$$AB = \frac{AD}{\cos 2\theta} = \frac{1}{\frac{1 - x^2}{1 + x^2}} = \frac{1 + x^2}{1 - x^2}.$$ Since $AB = BC$, the perimeter of $\triangle ABC$ is:

$$\text{Perimeter}(\triangle ABC) = AB + BC + AC = 2AB + 2 = 2\left(\frac{1 + x^2}{1 - x^2}\right) + 2.$$ The ratio of perimeters is given as $125:6$:

$$\frac{2\left(\frac{1 + x^2}{1 - x^2}\right) + 2}{2 + 2\sqrt{1 + x^2}} = \frac{125}{6}.$$ We simplify the left-hand side:

$$\frac{2\left(\frac{1 + x^2 + 1 - x^2}{1 - x^2}\right)}{2(1 + \sqrt{1 + x^2})} = \frac{\frac{2}{1 - x^2}}{1 + \sqrt{1 + x^2}} = \frac{2}{(1 - x^2)(1 + \sqrt{1 + x^2})}.$$ Thus:

$$\frac{2}{(1 - x^2)(1 + \sqrt{1 + x^2})} = \frac{125}{6}.$$ Cross-multiplying gives:

$$(1 - x^2)(1 + \sqrt{1 + x^2}) = \frac{12}{125}.$$ We make a substitution, let $\sqrt{1 + x^2} = \dfrac{a}{5}$ for some positive integer $a$. Then:

$$1 + x^2 = \frac{a^2}{25} \implies x^2 = \frac{a^2 - 25}{25},$$ $$1 - x^2 = 1 - \frac{a^2 - 25}{25} = \frac{25 - a^2 + 25}{25} = \frac{50 - a^2}{25},$$ $$1 + \sqrt{1 + x^2} = 1 + \frac{a}{5} = \frac{5 + a}{5}.$$ We substitute into the equation:

$$\frac{50 - a^2}{25} \cdot \frac{5 + a}{5} = \frac{12}{125}.$$ Simplifying gives:

$$\frac{(50 - a^2)(5 + a)}{125} = \frac{12}{125} \implies (50 - a^2)(5 + a) = 12.$$ We guess and check small positive integer values for $a$, and find $a = 7$ satisfies the equation:

$$(50 - 7^2)(5 + 7) = (50 - 49)(12) = 1 \cdot 12 = 12.$$ Thus $a = 7$, so:

$$\sqrt{1 + x^2} = \frac{7}{5} \implies 1 + x^2 = \frac{49}{25} \implies x^2 = \frac{24}{25}.$$ From $AB = \dfrac{1 + x^2}{1 - x^2}$:

$$1 - x^2 = 1 - \frac{24}{25} = \frac{1}{25}, \quad AB = \frac{\frac{49}{25}}{\frac{1}{25}} = 49.$$ In our initial setup, $AC = 2$ and $AI = \dfrac{7}{5}$. To make all sides integers, we scale by a factor of $5$, so $AC = 10$, $AI = 7$, and:

$$AB_{\text{min}} = 49 \times 5 = \boxed{245}.$$ ~Steven Zheng

Solution 2

Since $\triangle ABC$ is isosceles with $AB = BC$, the angle bisector and altitude from $B$ to $AC$ coincide, so $BI \perp AC$. Letting $BC=b$, it follows that $AB = \frac{b}{2\cos A}$ and $AI=\frac{b}{2\cos \frac{A}{2}}$. Substituting into perimeter, and letting $\cos \frac{A}{2}=x$,

$$\dfrac{2AB+b}{2AI+b} = \dfrac{\frac{b}{\cos A}+b}{\frac{b}{\cos \frac{A}{2}}+b}=\dfrac{\frac{1}{2x^2-1}+1}{\frac{1}{x}+1}=\dfrac{125}{6}$$ Expanding out that last in equality, we find that

$$238x^3+250x^2-125x-125=0$$ Fortunately, we can make the substitutions $x = 5y$, then $y = \frac{1}{z}$, which simplifies the equations into

$$238x^3+250x^2-125x-125=0 \implies 238y^3+50y^2-5y-1=0 \implies z^3+5z^2-50z-238=0$$ Much nicer. Checking via rational root theorem, we find that $z=7$ is one such solution, and there are no other rational solutions. Therefore,

$$\cos \frac{A}{2}= x = 5y = \frac{5}{z} = \frac{5}{7}\text{, and } \cos A = 2x^2-1 = \frac{1}{49}$$ Thus, $2AB = 49b$ and $10AI = 7b$, so $10 \mid b$. The smallest value for $b$ is 10. Thus, the minimum value of $AB$ is $\frac{49\cdot 10}{2}=\boxed{245}$.

~sillybone

Side Note (Making Things a Little Easier)

Consider the equation $238x^3+250x^2-125x-125=0$ It is cubic and has no apparent real root.

According to Vieta's Theorem, the numerator of a real root of $x$ should be a factor of $125$, and the denominator should be a factor of $238$.

Because $0<\frac{A}{2}<\frac{\pi}{2}$, $\frac{\sqrt{2}}{2}<\cos{\frac{A}{2}}<1$, there are only $2$ possible values: $\frac{5}{7}$ or $\frac{25}{34}$. Plug in $\frac{5}{7}$ to make $\cos{A} = 2\cos^{2}{\frac{A}{2}} - 1 = \frac{1}{49}$. We get lengths of the sides of the $2$ iscoceles, letting half of the base to be $1$ —— $49$ and $\frac{7}{5}$, which fits the $125:6$ ratio.

We don't need a rigorous proof for why $\frac{25}{34}$ is unsuitable in this situation: Even if it is an actual root of the equation, $\cos{A} = \frac{2\cdot25^2-34^2}{34^2}$, a value with a large, complicated numerator. It would result in a much larger minimum $AB$ than $\frac{5}{7}$

~cassphe

Note that we must have some rational root, since we need $\frac{AB}{BC}$ to be rational. Also, the motivation for substitution is to simplify the cubic, so Rational Root theorem is simpler. Motivation for me was that I can calculate cubics faster for integers inputs, rather than fractions.

~sillybone

Solution 3 (Guess & Check)

Clearly, $AB$ and $AC$ must make up most of triangle $\triangle ABC$'s perimeter. We can set the perimeter of triangle $\triangle ABC$ to $125x,$ and the perimeter of triangle $\triangle AIC$ to $6x.$ We do not need to check many values of $x,$ because if $x$ was too large then side $AB$ would be greater than $1000.$ After checking values of $x,$ we can find that the value $x=4$ works if side $AB$ is $245,$ side $AC$ is $10,$ and side $IA=7.$ We can double check this by finding the inradius of triangle $\triangle ABC$ using the formula $\frac{A}{s},$ where $A$ is the area and $s$ is the semiperimeter. Using Heron's formula for the area, we can find that the inradius is equal to $\sqrt{24},$ or $2 \sqrt 6.$ If we draw the lines connecting the incircle to the sides of the triangle, we know that they form right angles with the sides from the tangent circle theorem. Using the Pythagorean Theorem on these sides, we can see that the side lengths do in fact work, giving us $AB=\boxed {245}$ as our answer.

~Nodskin Goble

Video Solution 1 by TheBeautyofMath

With Discussion about various problem solving techniques. Least mentally taxing solution process: https://youtu.be/ivK4vL-ovCY

~IceMatrix