AIME 2026 II · 第 10 题

Solution 1

Let $BC=x$. By Angle Bisector Theorem, we have

$$BD = \frac{200}{200+225}x=\frac{8x}{17},\ CD = \frac{9x}{17}$$ Since both of these positive integers, we write $x=17y$, for some $y\in \mathbb{N}$. Then, $BD=8y$ and $CD=9y$.

Since $BD$ is tangent to $\omega$, it follows that $\angle BDE = \angle EAD$. Similarly, $\angle CDF = \angle FAC$. Thus, $\triangle BDE \sim \triangle BAD$ and $\triangle CDF \sim \triangle CAD$. We can use this to solve for $BE$ and $CF$:

$$\frac{BD}{AB}= \frac{BE}{BD}\implies BE = \frac{8y^2}{25}, CF = \frac{9y^2}{25}$$ Thus, $25\mid y^2\implies 5\mid y$. Hence, we can let $y=5z, z\in \mathbb{Z}^+$, so $BC=85z$.

Now, since $E$ is on $AB$ and $E\neq A$, we have $EB < AB\implies 8z^2< 200 \implies z < 5$. Using $F$ and $AC$ gives $z< 5$ as well. Thus, the maximum value of $BC$ is attained when $z=4$, or $BC = 4\cdot 85 = \boxed{340}$.

~sillybone

Solution 2

1. Power of a point at $B$ and $C$

Since $\omega$ is tangent to $BC$ at $D$, we have

$$\operatorname{Pow}\omega(B)=BD^2,\qquad \operatorname{Pow}\omega(C)=CD^2.$$ Because line $BA$ intersects $\omega$ at $A$ and $E$, we also have

$$\operatorname{Pow}_\omega(B)=BA\cdot BE=200(200-AE).$$ Hence

$$BD^2=200(200-AE). \tag{1}$$ Similarly, line $CA$ intersects $\omega$ at $A$ and $F$, so

$$\operatorname{Pow}_\omega(C)=CA\cdot CF=225(225-AF),$$ and therefore

$$CD^2=225(225-AF). \tag{2}$$ 2. Convert integrality into square conditions

Let $BE=200-AE$. Since $AE$ is a positive integer, $BE\in{1,2,\dots,199}$.

From (1),

$$BD^2=200\cdot BE = (2^3\cdot 5^2)\cdot BE.$$ For this to be a perfect square, we must have

$$BE=2s^2$$ for some integer $s\ge 1$. Then

$$BD^2=200\cdot 2s^2=400s^2 \implies BD=20s. \tag{3}$$ Also $BE<200$ implies $2s^2<200$, so $s\le 9$.

Next let $CF=225-AF$. From (2),

$$CD^2=225\cdot CF=15^2\cdot CF,$$ so $CD$ is an integer if and only if $CF$ is a perfect square:

$$CF=t^2 \implies CD=15t. \tag{4}$$ Since $t^2<225$, we have $t\le 14$.

Thus

$$BD=20s,\ 1\le s\le 9;\qquad CD=15t,\ 1\le t\le 14.$$ 3. Angle Bisector Theorem

Because $AD$ bisects $\angle BAC$,

$$\frac{BD}{DC}=\frac{AB}{AC}=\frac{200}{225}=\frac{8}{9}.$$ Substitute (3) and (4):

$$\frac{20s}{15t}=\frac{8}{9} \implies \frac{4s}{3t}=\frac{8}{9} \implies 36s=24t \implies 3s=2t.$$ So $s$ is even. Let $s=2k$, then $t=3k$.

Bounds give

$$2k\le 9 \Rightarrow k\le 4,\qquad 3k\le 14 \Rightarrow k\le 4,$$ so $k\in{1,2,3,4}$.

4. Maximize $BC$

We have

$$BC=BD+DC=20s+15t=20(2k)+15(3k)=85k.$$ Hence the maximum value of $k$ is $4$

$$85 \times 4=340$$ Therefore,

$$\boxed{340}.$$ (by a 16-year-old student, if see any mistake, feel free to edit)

Edit for International version: (Sum of all possible values asked instead of maximum)

Same as Solution 1 until getting 85k. As k can be equal to 1, 2, 3, 4

Answer = 85(1+2+3+4) = 85(10) = 850

Solution 3

Let $AB=c=200$, $AC=b=225$, and $BC=a$. By the Angle Bisector Theorem, we have $BD:CD = c:b = 200:225 = 8:9$. Let $BD = 8k$ and $CD = 9k$ for some real number $k$. Then $BC = 17k$.

Since the circle $\omega$ is tangent to $BC$ at $D$ and passes through $E$ and $F$ on $AB$ and $AC$ respectively, by the Power of a Point Theorem (specifically the tangent-secant theorem), we have $BD^2 = BE \cdot BA$ and $CD^2 = CF \cdot CA$. We can express the lengths of $AE$ and $AF$ as:

$$AE = AB - BE = AB - \frac{BD^2}{AB} = AB \left( 1 - \left(\frac{BD}{AB}\right)^2 \right)$$ $$AF = AC - CF = AC - \frac{CD^2}{AC} = AC \left( 1 - \left(\frac{CD}{AC}\right)^2 \right)$$ Substituting the known values:

$$AE = 200 \left( 1 - \left(\frac{8k}{200}\right)^2 \right) = 200 - \frac{8k^2}{25}$$ $$AF = 225 \left( 1 - \left(\frac{9k}{225}\right)^2 \right) = 225 - \frac{9k^2}{25}$$ The problem states that $AE$ and $AF$ are positive integers. For $AE$ and $AF$ to be integers, the terms $\frac{8k^2}{25}$ and $\frac{9k^2}{25}$ must be integers. Since $\gcd(8, 25)=1$ and $\gcd(9, 25)=1$, it follows that $k^2$ must be a multiple of $25$. Thus, $k$ must be a multiple of $5$.

Next, we establish the bounds for $k$. Since $AE$ and $AF$ must be positive:

$$200 - \frac{8k^2}{25} > 0 \implies k^2 < 625 \implies k < 25$$ Additionally, by the Triangle Inequality on $\triangle ABC$:

$$AB + AC > BC \implies 200 + 225 > 17k \implies 425 > 17k \implies k < 25$$ $$AB + BC > AC \implies 200 + 17k > 225 \implies 17k > 25 \implies k > \frac{25}{17} \approx 1.47$$ Given $k$ is a multiple of $5$ and $1.47 < k < 25$, the possible values for $k$ are $5, 10, 15, 20$. The problem asks for the greatest possible value of $BC$:

$$17 \times 20 = \boxed{340}$$ ~YFX

Solution 4

Let $BD = a, CD = b, AE = c, AF = d$ so that $BE = 200 - c, CF = 225 - d$. By PoP, we have

$a^2 = (200 - c)(200) = (400 - 2c)(100) \implies (\frac{a}{10})^2 = 400 - 2c$ so we know $\frac{a}{10}$ is even and an integer so $a = 20t$ By PoP we have

$b^2 = (225 - d)(225) \implies 225 - d = (\frac{b}{15})^2$ so $b = 15k$. Now Angle Bisector Theorem tells

$\frac{200}{225} = \frac{8}{9} = \frac{a}{b}$ so $a = 8m, b = 9m$. We know $lcm(20, 8) = 40$ which implies $a = 40p$ and we also know $lcm(15, 9) = 45$ so $b = 45q$. But since $a = 8m, b = 9m$, we need $p = q$. Thus, $a = 40p, b = 45p$. So $BC = a + b = 85p$. Now note that $400 - 2c = (4p)^2 = 16p^2 \implies 200 - c = 8p^2 \implies c = 200 - 8p^2 \implies 8p^2 < 200$ and $225 - d = (3p)^2 = 9p^2 \implies d = 225 - 9p^2 \implies 9p^2 < 225$. Elegantly, solving $8p^2 < 200$ gives $p^2 < 25$ and solving $9p^2 < 225$ also gives $p^2 < 25$. Obviously $p < 5$ and note $p$ also has to be a positive integer. So all possible values are $p = 1,2,3,4$ and the maximum possible value is obviously $85(4) = \boxed{340}$.

~ilikemath247365

Solution 5

Let the side lengths of $\triangle ABC$ be denoted by $AB = c$, $AC = b$, and $BC = a$. We are given $c = 200$ and $b = 225$. Let $D$ be a point on $BC$ such that $AD$ is the internal angle bisector of $\angle BAC$. By the Angle Bisector Theorem, the ratio of the segments of the opposite side is proportional to the adjacent sides:

$$\frac{BD}{CD} = \frac{AB}{AC} = \frac{200}{225} = \frac{8}{9}.$$ Since $BD$ and $CD$ are positive integers and $\gcd(8, 9) = 1$, there must exist a positive integer $m$ such that

$$BD = 8m \quad \text{and} \quad CD = 9m.$$ Consequently, the total length of side $BC$ is given by $a = BD + CD = 17m$.

Let $\omega$ be the circle passing through $A$ and tangent to the segment $BC$ at $D$. We apply the Power of a Point Theorem. For point $B$, lying on the extension of the tangent segment $BD$ and the secant line $BA$, the power of the point with respect to $\omega$ is given by:

$$\mathcal{P}(B) = BD^2 = BE \cdot BA.$$ Substituting the known expressions into this equation:

$$(8m)^2 = BE \cdot 200 \implies 64m^2 = 200 \cdot BE \implies BE = \frac{64m^2}{200} = \frac{8m^2}{25}.$$ Since $E$ lies on the segment $AB$, the length $AE$ is given by $AE = AB - BE = 200 - \frac{8m^2}{25}$. We are given that $AE$ is a positive integer. For $AE$ to be an integer, $BE$ must be an integer (since $AB$ is an integer). This implies:

$$200 \mid 64m^2 \iff 25 \mid 8m^2.$$ Since $\gcd(25, 8) = 1$, it must be that $25 \mid m^2$, which implies $5 \mid m$. Thus, let $m = 5k$ for some positive integer $k$.

Similarly, we consider the power of point $C$ with respect to $\omega$. The line $BC$ is tangent at $D$, and $CA$ is a secant intersecting $\omega$ at $F$ and $A$. Thus:

$$\mathcal{P}(C) = CD^2 = CF \cdot CA.$$ Substituting the values:

$$(9m)^2 = CF \cdot 225 \implies 81m^2 = 225 \cdot CF \implies CF = \frac{81m^2}{225} = \frac{9m^2}{25}.$$ With $m = 5k$, we substitute into the expressions for $AE$ and $AF$:

$$BE = \frac{8(5k)^2}{25} = 8k^2 \implies AE = 200 - 8k^2,$$ $$CF = \frac{9(5k)^2}{25} = 9k^2 \implies AF = 225 - 9k^2.$$ The problem states that $AE$ and $AF$ are positive integers. This imposes bounds on $k$:

$$200 - 8k^2 > 0 \implies 8k^2 < 200 \implies k^2 < 25 \implies k < 5,$$ $$225 - 9k^2 > 0 \implies 9k^2 < 225 \implies k^2 < 25 \implies k < 5.$$ Additionally, the Triangle Inequality must hold for $\triangle ABC$. The sides are $AB=200$, $AC=225$, and $BC = 17m = 17(5k) = 85k$.

$$AB + AC > BC \implies 200 + 225 > 85k \implies 425 > 85k \implies k < 5.$$ The other triangle inequalities ($AB+BC > AC$ and $AC+BC > AB$) are satisfied for all $k \ge 1$. Thus, the possible integer values for $k$ are $k \in \{1, 2, 3, 4\}$.

To find the greatest possible value of $BC$, we maximize $k$. The largest valid integer is $k = 4$. Substituting $k=4$:

$$BC = 85(4) = 340.$$ We verify the other segments are integers:

$$AE = 200 - 8(4)^2 = 200 - 128 = 72 \quad (\text{Integer } > 0),$$ $$AF = 225 - 9(4)^2 = 225 - 144 = 81 \quad (\text{Integer } > 0).$$ The lengths $BD = 8(20) = 160$ and $CD = 9(20) = 180$ are also integers. The triangle exists since $200+225 > 340$.

The greatest possible value of $BC$ is $340$.

~ Jesse Zhang (FUNKCCP)