AIME 2020 I · 第 15 题

Solution 1

The following is a power of a point solution to this menace of a problem:

Let points be what they appear as in the diagram below. Note that $3HX = HY$ is not insignificant; from here, we set $XH = HE = \frac{1}{2} EY = HL = 2$ by PoP and trivial construction. Now, $D$ is the reflection of $A$ over $H$. Note $AO \perp XY$, and therefore by Pythagorean theorem we have $AE = XD = \sqrt{5}$. Consider $HD = 3$. We have that $\triangle HXD \cong HLK$, and therefore we are ready to PoP with respect to $(BHC)$. Setting $BL = x, LC = y$, we obtain $xy = 10$ by PoP on $(ABC)$, and furthermore, we have $KH^2 = 9 = (KL - x)(KL + y) = (\sqrt{5} - x)(\sqrt{5} + y)$. Now, we get $4 = \sqrt{5}(y - x) - xy$, and from $xy = 10$ we take

$$\frac{14}{\sqrt{5}} = y - x.$$ However, squaring and manipulating with $xy = 10$ yields that $(x + y)^2 = \frac{396}{5}$ and from here, since $AL = 5$ we get the area to be $3\sqrt{55} \implies \boxed{058}$. ~awang11's sol

Solution 1a

As in the diagram, let ray $AH$ extended hits BC at L and the circumcircle at say $P$. By power of the point at H, we have $HX \cdot HY = AH \cdot HP$. The three values we are given tells us that $HP=\frac{2\cdot 6}{3}=4$. L is the midpoint of $HP$(see here: https://www.cut-the-knot.org/Curriculum/Geometry/AltitudeAndCircumcircle.shtml ), so $HL=LP=2$.

As in the diagram provided, let K be the intersection of $BC$ and $XY$. By power of a point on the circumcircle of triangle $HBC$, $KH^{2}=KB \cdot KC$. By power of a point on the circumcircle of triangle $ABC$, $KB \cdot KC=KX \cdot KY$, thus $KH^{2}=(KH-2)(KH+6)$. Solving gives $4KH=12$ or $KH=3$.

By the Pythagorean Theorem on triangle $HKL$, $KL=\sqrt{5}$. Now continue with solution 1.

Solution 2

Diagram not to scale.

We first observe that $H'$, the image of the reflection of $H$ over line $BC$, lies on circle $O$. This is because $\angle HBC = 90 - \angle C = \angle H'AC = \angle H'BC$. This is a well known lemma. The result of this observation is that circle $O'$, the circumcircle of $\triangle BHC$ is the image of circle $O$ over line $BC$, which in turn implies that $\overline{AH} = \overline{OO'}$ and thus $AHO'O$ is a parallelogram. That $AHO'O$ is a parallelogram implies that the radius $AO$ is perpendicular to $\overline{XY}$, and thus divides segment $\overline{XY}$ in two equal pieces, $\overline{XD}$ and $\overline{DY}$, of length $4$.

Using Power of a Point,

$$\overline{AH} \cdot \overline{HH'} = \overline{XH} \cdot \overline{HY} \Longrightarrow 3 \cdot \overline{HH'} = 2 \cdot 6 \Longrightarrow \overline{HH'} = 4$$ This means that $\overline{HL} = \frac12 \cdot 4 = 2$ and $\overline{AL} = 2 + 3 = 5$, where $L$ is the foot of the altitude from $A$ onto $BC$. All that remains to be found is the length of segment $\overline{BC}$.

Looking at right triangle $\triangle AHD$, we find that

$$\overline{AD} = \sqrt{\overline{AH}^2 - \overline{HD}^2} = \sqrt{3^2 - 2^2} = \sqrt{5}$$ Looking at right triangle $\triangle ODY$, we get the equation

$$\overline{OY}^2 - \overline{DY}^2 = \overline{OD}^2 = \left(\overline{AO} - \overline{AD}\right)^2$$ Plugging in known values, and letting $R$ be the radius of the circle, we find that

$$R^2 - 16 = (R - \sqrt{5})^2 = R^2 - 2\sqrt5 R + 5 \Longrightarrow R = \frac{21\sqrt5}{10}$$ Recall that $AHO'O$ is a parallelogram, so $\overline{AH} = \overline{OO'} = 3$. So, $\overline{OM} = \frac32$, where $M$ is the midpoint of $\overline{BC}$. This means that

$$\overline{BC} = 2\overline{BM} = 2\sqrt{R^2 - \left(\frac32\right)^2} = 2\sqrt{\frac{441}{20} - \frac{9}{4}} = \frac{6\sqrt{55}}{5}$$ Thus, the area of triangle $\triangle ABC$ is

$$\frac{\overline{AL} \cdot \overline{BC}}{2} = \frac{5 \cdot \frac{6\sqrt{55}}{5}}{2} = {3\sqrt{55}}$$ The answer is $3 + 55 = \boxed{058}$.

Solution 3 (Official MAA 1)

Extend $\overline{AH}$ to intersect $\omega$ again at $P$. The Power of a Point Theorem yields $HP = \tfrac{HX \cdot HY}{HA} = 4$. Because $\angle CAP=\angle CBP$, and $\angle CAP$ and $\angle CBH$ are both complements to $\angle C$, it follows that $\angle CBP = \angle CBH$, implying that $\overline{BC}$ bisects $\overline{HP}$, so the length of the altitude from $A$ to $\overline{BC}$ is $h_a = AH + \tfrac12 HP = 5$.

Let the circumcircle of $\triangle BCH$ be $\omega'$. Because $\triangle BCH \cong \triangle BCP$, the two triangles must have the same circumradius. Because the circumcircle of $\triangle BCP$ is $\omega$, the circles $\omega$ and $\omega'$ have the same radius $R$. Denote the centers of $\omega$ and $\omega'$ by $O$ and $O'$, respectively, and let $M$ be the midpoint of $\overline{XY}$. Note that trapezoid $HMOO'$ has $\angle H = \angle M = 90^\circ$. Also $HM = XM - XH = \frac12\cdot XY - HX = 2$ and $HO' = R$. Because $\omega$ is a translation of $\omega'$ in the direction of $\overline{AH}$, it follows that $OO' = AH = 3$. Finally, the Pythagorean Theorem applied to $\triangle XMO$ yields $MO = \sqrt{R^2-16}$. Let $T$ be the projection of $O$ onto $\overline{HO'}$. Then $TO' = R-MO$, so the Pythagorean Theorem applied to $\triangle TOO'$ yields

$$R - \sqrt{R^2-16} = \sqrt{3^2 - 2^2} = \sqrt{5}.$$ Solving for $R$ gives $R = \tfrac{21}{2\sqrt5}$. It follows from properties of the orthocenter $H$ that

$$\cos\angle A = \dfrac{AH}{2R} = \dfrac{\sqrt5}{7},$$ so

$$\sin\angle A = \sqrt{1 - \cos^2\angle A} = \dfrac{2\sqrt{11}}{7}.$$ Therefore by the Extended Law of Sines

$$a = BC = 2R \sin\angle A = \dfrac{6\sqrt{11}}{\sqrt5},$$ so

$$[\triangle ABC] = \frac12 a h_a = \frac12 \cdot \frac{6\sqrt{11}}{\sqrt{5}} \cdot 5 = 3\sqrt{55}.$$ The requested sum is $3+55 = 58$.

Solution 3a (slightly modified Solution 3)

Note that $\overline{BC}$ bisects $\overline{OO'}$. Using the same method from Solution 3 we find $R = \frac{21}{2\sqrt{5}}$. Let the midpoint of $\overline{BC}$ be $N$, then by the Pythagorean Theorem we have $BN^2 = OB^2 - ON^2 = R^2 - \frac{3}{2}^2$, so $BN = \frac{3\sqrt{11}}{\sqrt{5}}$. Since $h_a = 5$ we have that the area of ABC is $3\sqrt{55}$ so the answer is $3 + 55 = \boxed{58}$

~bobjoebilly

Solution 4 (Official MAA 2)

Let $D$ be the intersection point of line $AH$ and $\overline{BC}$, noting that $\overline{AD}\perp\overline{BC}$. Because the area of $\triangle ABC$ is $\tfrac12\cdot AD\cdot BC$, it suffices to compute $AD$ and $BC$ separately. As in the previous solution, $AD = 5$. The value of $BC$ can be found using the following lemma.

Lemma: Triangle $AXY$ is isosceles with base $\overline{XY}$.

Proof: Because the circumcircle of $\triangle BCH$, $\omega'$, and $\omega$ have the same radius, there exists a translation $\Phi$ sending the former to the latter. Because $\overline{AH}$ is parallel to the line connecting the centers of the two circles, $\Phi$ must send $H$ to $A$, meaning $\Phi$ also sends $\overline{XY}$ to the tangent to $\omega$ at $A$. But this means that this tangent is parallel to $\overline{XY}$, which implies the conclusion.

Applying Stewart's Theorem to $\triangle AXY$ yields

$$AX^2 = AH^2 + HX\cdot HY = 3^2 + 2\cdot 6 = 21,$$ implying $AX = AY= \sqrt{21}.$

By the Law of Cosines

$$\cos \angle XAY = \frac{21 + 21 - 64}{2 \cdot 21} = -\frac{11}{21},$$ so

$$\sin \angle XAY = \dfrac{4\sqrt{20}}{21}.$$ Let $R$ be the radius of $\omega$. By the Extended Law of Sines

$$R = \frac{XY}{2 \cdot \sin \angle XAY} = \dfrac{21}{\sqrt{20}}.$$ Then the solution proceeds as in Solution 3.

Solution 5 (Official MAA 3)

Define points $D$ and $P$ as above, and note that $AD=5$ and $DH=PD=AD - AH = 2$. Let the circumcircle of $\triangle BCH$ be $\omega'$.

Extend $\overline{XY}$ past $X$ until it intersects line $BC$ at point $Z$. Because line $BC$ is a radical axis of $\omega$ and $\omega'$, it follows from the Power of a Point Theorem that

$$ZX \cdot ZY = ZX \cdot(ZX + 8) = ZH^2 = (ZX+2)^2,$$ from which $ZX=1$. By Pythagorean Theorem $ZD=\sqrt{ZH^2 - DH^2} = \sqrt{5}$.

Let $m=CD$ and $n=BD$. By the Power of a Point Theorem

$$mn= AD\cdot PD = 10.$$ On the other hand,

$$ZH^2 = 9 = ZB \cdot ZC = (\sqrt{5} - n)(\sqrt{5} + m) = 5 + \sqrt{5}(m-n) - mn,$$ from which $m-n = \frac{14}{\sqrt{5}}$. Therefore

$$(m+n)^2 = (m-n)^2 + 4mn = \frac{196}{5} + 40 = \frac{396}{5}.$$ Thus $BC = m+n=\sqrt{\frac{396}{5}} = 6\sqrt{\frac{11}{5}}$. Therefore $[\triangle ABC] = \frac{1}{2}BC \cdot AD = 3\sqrt{55},$ as above.

Solution 6

Let $O$ be circumcenter of $ABC,$ let $R$ be circumradius of $ABC,$ let $\omega'$ be the image of circle $\omega$ over line $BC$ (the circumcircle of $HBC$).

Let $P$ be the image of the reflection of $H$ over line $BC, P$ lies on circle $\omega.$ Let $M$ be the midpoint of $XY.$ Then $P$ lies on $\omega, OA = O'H, OA || O'H.$

(see here: https://brilliant.org/wiki/triangles-orthocenter/ or https://en.wikipedia.org/wiki/Altitude_(triangle) Russian)

$P$ lies on $\omega \implies OA = OP = R,$

$OA = O'H, OA || O'H \implies M$ lies on $OA.$

We use properties of crossing chords and get

$$AH \cdot HP = XH \cdot HY = 2 \cdot 6 \implies HP = 4, AP = AH + HP = 7.$$ We use properties of radius perpendicular chord and get

$$MH = \frac{XH + HY}{2} – HY = 2.$$ We find

$$\sin OAH =\frac{MH}{AH} = \frac{2}{3} \implies \cos OAH = \frac{\sqrt{5}}{3}.$$ We use properties of isosceles $\triangle OAP$ and find $\hspace{5mm}R = \frac{AP}{2\cos OAP} = \frac{7}{2\frac {\sqrt{5}}{3}} = \frac{21}{2\sqrt{5}}.$

We use $OM' = \frac{AH}{2} = \frac {3}{2}$ and find $\hspace{25mm} \frac{BC}{2} = \sqrt{R^2 – OM'^2} = 3 \sqrt {\frac {11}{5}}.$

The area of $ABC$

$$[ABC]=\frac{BC}{2} \cdot (AH + HD) = 3\cdot \sqrt{55} \implies 3+55 = \boldsymbol{\boxed{058}}.$$ vladimir.shelomovskii@gmail.com, vvsss

Two Video Solutions by MOP 2024

• https://youtube.com/watch?v=qgM-GbAWXCc (Power of a Point and Symmetry)
• https://youtube.com/watch?v=XOUssYh14aw (Geometric Transformations, Angle Chasing, Trig)

~r00tsOfUnity

Video Solution by On the Spot STEM

https://www.youtube.com/watch?v=L7B20E95s4M