AIME 2026 I · 第 10 题

Solution 1

Since a $13$-$14$-$15$ triangle is composed of a $5$-$12$-$13$ triangle connected to a $9$-$12$-$15$ triangle, we set up coordinates

$$A = (5,12),\quad B = (0,0),\quad C = (14,0).$$ From this we have

$$[ABC] = \frac{1}{2} \cdot 14 \cdot 12 = 84,$$ and thus

$$R = \frac{abc}{4[ABC]} = \frac{65}{8}.$$ We compute the circumcenter as the intersection of two perpendicular bisectors. The perpendicular bisector of $BC$ is clearly

$$x = 7.$$ The perpendicular bisector of $AB$ passes through $\left(\frac{5}{2}, 6\right)$ and has slope $-\frac{5}{12}$, so its equation is

$$y - 6 = -\frac{5}{12}\left(x - \frac{5}{2}\right).$$ The intersection is

$$O = \left(7,\frac{33}{8}\right).$$ If the triangle is rotated by $\theta$, then the rotation maps $(9,-12)$ to

$$\left(9\cos\theta + 12\sin\theta,\; 9\sin\theta - 12\cos\theta\right).$$ Since $BC$ is horizontal, $A'C'$ must be vertical, so the $x$-component is $0$:

$$9\cos\theta + 12\sin\theta = 0.$$ Thus

$$3\cos\theta + 4\sin\theta = 0,$$ giving

$$\tan\theta = -\frac{3}{4}.$$ We take

$$\cos\theta = \frac{4}{5}, \quad \sin\theta = -\frac{3}{5}.$$ To rotate about $O = \left(7,\frac{33}{8}\right)$, for any point $P$ we use

$$P' = O + R_\theta(P - O),$$ where

$$R_\theta(x,y) = \left( \frac{4}{5}x + \frac{3}{5}y,\; -\frac{3}{5}x + \frac{4}{5}y \right).$$ Compute vectors from $O$:

$$A - O = \left(-2,\frac{63}{8}\right),$$ $$B - O = \left(-7,-\frac{33}{8}\right),$$ $$C - O = \left(7,-\frac{33}{8}\right).$$ Rotate:

$$R_\theta(A - O) = \left( \frac{4}{5}(-2) + \frac{3}{5}\cdot\frac{63}{8},\; -\frac{3}{5}(-2) + \frac{4}{5}\cdot\frac{63}{8} \right) = \left(\frac{25}{8},\frac{15}{2}\right),$$ $$R_\theta(C - O) = \left(\frac{25}{8},-\frac{15}{2}\right),$$ $$R_\theta(B - O) = \left(-\frac{323}{40},\frac{9}{10}\right).$$ Hence

$$A' = \left(\frac{81}{8},\frac{93}{8}\right),$$ $$C' = \left(\frac{81}{8},-\frac{27}{8}\right),$$ $$B' = \left(-\frac{43}{40},\frac{201}{40}\right).$$ Using the shoelace formula on hexagon $A,A',C,C',B,B'$ gives area

$$\frac{1557}{10}.$$ The closest integer is

$$\boxed{156}.$$ - spectraldragon8

Solution 2 (No words)

$$s=\frac{13+14+15}{2}=21$$ $$[ABC]=\sqrt{21\cdot8\cdot7\cdot6}=84$$ $$[ABC]=\frac{13\cdot14\cdot15}{4R}$$ $$R=\frac{65}{8}$$ $$\angle AOB=\gamma,\quad \angle BOC=\beta,\quad \angle COA=\alpha,\quad \alpha+\beta+\gamma=2\pi$$ $$13^2=2R^2(1-\cos\gamma),\quad 14^2=2R^2(1-\cos\beta),\quad 15^2=2R^2(1-\cos\alpha)$$ $$\cos\gamma=-\frac{7}{25},\quad \cos\beta=-\frac{2047}{4225},\quad \cos\alpha=-\frac{119}{169}$$ $$\cos\frac{\gamma}{2}=\sqrt{\frac{1+\cos\gamma}{2}}=\frac{3}{5},\quad \sin\frac{\gamma}{2}=\sqrt{\frac{1-\cos\gamma}{2}}=\frac{4}{5}$$ $$\cos\theta=\frac{4}{5},\quad \sin\theta=-\frac{3}{5}$$ $$[POP']=\frac12R^2\sin\theta$$ $$[AA'CC'BB'] =\frac12R^2\left( 3\sin\theta-\sin(\alpha+\theta)-\sin(\beta+\theta)-\sin(\gamma+\theta) \right)$$ $$[AA'CC'BB'] =\frac12\left(\frac{65}{8}\right)^2\cdot\frac{99648}{21125}$$ $$[AA'CC'BB']=\frac{1557}{10}$$ $$\boxed{156}$$ ~Gray_Wolf (and Silver Rush for pointing out flaws.)

Solution 3 (minimal trig)

Let $A’C’$ and $BC$ intersect at $P$, and connect $A’B$.

First, we will find the angle of rotation. Since the angle between $AC$ and $BC$ and the angle between $A’C’$ and $BC$ differ by $90^{\circ}-\angle ACB$, that is the angle of rotation, which we will denote by $\theta$. Thus, we have $\angle AOA’ = \angle COC’ = \angle BOB’ = \theta$. Also, by the inscribed angle theorem, $\angle CA’C’=\angle BCC’=\theta/2$.

By drawing the altitude from $A$ to $BC$ in $\triangle ABC$, we derive that $\sin{\theta}=3/5$. We can use the half angle formula or draw an angle bisector in a $3$-$4$-$5$ triangle to find $\sin({\theta/2})$, which is $1/\sqrt{10}$, so $\tan{\theta/2}=1/3$. Now, let $CP=x$ and $C’P=y$. We know that $A’C’=15$ and $BC=14$, so $3x+y=15$ and $3y+x=14$. Solving, $x=\frac{31}{8}$ and $y=\frac{27}{8}$. Using Pythagorean Theorem on $\triangle CPC’$ gives $CC’=\frac{13\sqrt{10}}{8}$, and because $AA’$, $BB’$, and $CC’$ are all on the arc of measure $\theta$, $AA’=BB’=\frac{13\sqrt{10}}{8}$. Also, $\triangle ABP$ is similar to $\triangle CPC’$ with ratio $3$, so $A’B= \frac{39\sqrt{10}}{8}$.

By the inscribed angle theorem, $\angle AA’B=\angle ACB=\angle A’C’B’=\angle A’BB’$. As opposite angles in a cyclic quadrilateral are supplementary, $\angle A’BB’+\angle AB’B=\angle AA’B+\angle AB’B=180^{\circ}$. Thus, $AB’ \parallel A’B$, which makes $AA’BB’$ a trapezoid with equal base angles, or an isosceles trapezoid. Draw the altitudes of the trapezoid from $A$ and $B’$, and denote the feet as $M$ and $N$, respectively. We also have $\sin {\angle AA’B}=4/5$, so $AM=B’N= \frac{13\sqrt{10}}{10}$, and $A’M=BN= \frac{39\sqrt{10}}{40}$. Subtracting twice the latter value from $A’B$ yields $AB’= \frac{117\sqrt{10}}{40}$. Finally, we use the area formula for a trapezoid, which gives us $[AA’BB’]=50.7$.

The diagonals of $A’CC’B$ are perpendicular and have lengths $15$ and $14$, so it has area $\frac{1}{2}\cdot{15}\cdot{14}=105$. The area of hexagon $AA’CC’BB’$ is the sum of the areas of $A’CC’B$ and $AA’BB’$, which is $155.7$, and that rounds to $\boxed{156}$.

~joiceeliu

If sin theta = 3/5, wouldn't sin theta/2 = 1/sqrt(10)? Sorry I meant tan, fixed it

Solution 4 (lots of trig)

We first have a claim.

Claim. The area of an isosceles triangle with vertex $\alpha$ and leg length $R$ is $\frac{R^2\sin\alpha}{2}$.

Proof. Notice that the height is $\cos\frac{\alpha}{2}*R$ and the base is $2*\sin\frac{\alpha}{2}*R$. Thus, the area of the triangle is

$$\frac{1}{2}*\cos\frac{\alpha}{2}*R*2*\sin\frac{\alpha}{2}*R=\cos\frac{\alpha}{2}*\sin\frac{\alpha}{2}*R^2=\frac{R^2\sin\alpha}{2}$$ ---

Let the new triangle be rotated by $\theta$ clockwise.

Notice that now the area of $AA'BB'CC'$ can be represented as the sum of the areas of six isosceles triangles. So, we will find the sines of the vertices of the six isosceles triangles to calculate their areas.

Let $A'C'$ and $AC$ intersect at $E$ and $A'C'$ and $BC$ intersect at $D$. Then, since $A'B'C'$ is $ABC$ rotated by $\theta$, the angle between $A'C'$ and $AC$ is $\theta$. Also, because $\angle A'DC=90^{\circ}$ by the problem statement, $\theta=90^{\circ}-C$. Notice that by the law of cosines $\sin C=\frac{4}{5}$ and $\cos C=\frac{3}{5}$, so $\sin\theta=\sin(90-C)=\cos C=\frac{3}{5}$ and $\cos \theta=\sin C=\frac{4}{5}$.

Notice that $\angle BOC'=\angle BOC-\angle COC'=2A-\theta$. Similarly, $\angle AOB'=2C-\theta$ and $\angle A'OC=2B-\theta$. So,

$$\sin\angle BOC'=\sin(2A)\cos \theta-\cos(2A)\sin\theta$$ $$=2\cos A\sin A\cos\theta-(2\cos^2A-1)\sin\theta$$ by the formulas for $\sin(x+y)$ and $\cos(x+y)$. By the law of cosines, $\cos A=\frac{33}{65}$, so $\sin A=\frac{\sqrt{65^2-33^2}}{65}=\frac{56}{65}$. So,

$$\sin\angle BOC'=2*\frac{33}{65}*\frac{56}{65}*\frac{4}{5}-(2*\frac{33^2}{65^2}-1)*\frac{3}{5}$$ $$=\frac{2*33*56*4-(2*33^2-65^2)*3}{65^2*5}$$ $$=\frac{20925}{65^2*5}$$ $$=\frac{837}{845}$$ similarly,

$$\sin\angle A'OC=2\cos B\sin B\cos\theta-(2\cos^2B-1)\sin\theta$$ $$=2*\frac{5}{13}*\frac{12}{13}*\frac{4}{5}-(2*\frac{5^2}{13^2}-1)*\frac{3}{5}$$ $$=\frac{2*5*12*4-(2*5^2-13^2)*3}{13^2*5}$$ $$=\frac{837}{845}$$ $$\sin\angle AOB'=2\cos C\sin C\cos\theta-(2\cos^2C-1)\sin\theta$$ $$=2*\frac{3}{5}*\frac{4}{5}*\frac{4}{5}-(2*\frac{9}{25}-1)*\frac{3}{5}$$ $$=\frac{96-(18-25)*3}{125}$$ $$=\frac{117}{125}$$ Now, we will use the claim to calculate the areas of the six isosceles triangles.

$$[AA'BB'CC'] =[AOA']+[A'OB]+[BOB']+[B'OC]+[COC']+[C'OA]$$ $$=\frac12R^2\bigl(3\sin\theta+\sin\angle BOC'+\sin\angle A'OC+\sin\angle AOB'\bigr)$$ $$=\frac12R^2\left(3\cdot\frac35+\frac{837}{845}+\frac{837}{845}+\frac{117}{125}\right)$$ $$=\frac12R^2\cdot\frac{99648}{21125} =R^2\cdot\frac{49824}{21125}.$$ By the circumradius formula, $R=\frac{65}{8}$, so

$$[AA'BB'CC']=\frac{4225}{64}*\frac{49824}{21125}=\frac{1557}{10}=155.7,$$ answer extraction gives $\boxed{156}.$

Solution 5 (Almost No Trigonometry)

Connect $A'B$. This divides the hexagon $AA'CC'BB'$ into two quadrilaterals: $A'BC'C$ and $AA'BB'$.

First, compute the area of quadrilateral $A'BC'C$. We have $[A'BC'C] = \frac{1}{2} \cdot BC \cdot A'C' = \frac{1}{2} \cdot 14 \cdot 15 = 105.$

Let $M = AB \cap A'B'$ and $Q = BC \cap B'C'$. Then the area of quadrilateral $AA'BB'$ can be expressed as $[AA'BB'] = \frac{1}{2} \cdot AB \cdot A'B' \cdot \sin \angle BMB'.$

Observe that $\angle BMB' = \angle BQB'=\angle CQC'$. Therefore, $\sin \angle BMB' = \sin \angle BQB' = \sin \angle CQC'=\cos \angle A'C'B'.$

In $\triangle A'C'B'$, we know $A'C' = 15$, $B'C' = 14$, and $A'B' = 13$. By the Law of Cosines, $\cos \angle A'C'B' = \frac{14^2 + 15^2 - 13^2}{2 \cdot 14 \cdot 15} = \frac{196 + 225 - 169}{420} = \frac{252}{420} = \frac{3}{5}.$ Thus $\sin \angle BMB' = \frac{3}{5}$.

Now compute the area: $[AA'BB'] = \frac{1}{2} \cdot 13 \cdot 13 \cdot \frac{3}{5} = \frac{507}{10} = 50.7.$

Finally, the area of hexagon $AA'CC'BB'$ is $[AA'CC'BB'] = [A'BC'C] + [AA'BB'] = 105 + 50.7 = 155.7.$

Rounding to the nearest integer gives $\boxed{156}$.

- WUGOUYU

Video Solution (Fast and Easy 🔥🚀)

2026 AIME I #10

By piacademyus.org