AIME 2020 II · 第 15 题

Solution 1

Let $O$ be the circumcenter of $\triangle ABC$; say $OT$ intersects $BC$ at $M$; draw segments $XM$, and $YM$. We have $MT=3\sqrt{15}$.

Since $\angle A=\angle CBT=\angle BCT$, we have $\cos A=\tfrac{11}{16}$. Notice that $AXTY$ is cyclic, so $\angle XTY=180^{\circ}-A$, so $\cos XTY=-\cos A$, and the cosine law in $\triangle TXY$ gives

$$1143-2XY^2=-\frac{11}{8}\cdot XT\cdot YT.$$ Since $\triangle BMT \cong \triangle CMT$, we have $TM\perp BC$, and therefore quadrilaterals $BXTM$ and $CYTM$ are cyclic. Let $P$ (resp. $Q$) be the midpoint of $BT$ (resp. $CT$). So $P$ (resp. $Q$) is the center of $(BXTM)$ (resp. $CYTM$). Then $\theta=\angle ABC=\angle MTX$ and $\phi=\angle ACB=\angle YTM$. So $\angle XPM=2\theta$, so

$$\frac{\frac{XM}{2}}{XP}=\sin \theta,$$ which yields $XM=2XP\sin \theta=BT(=CT)\sin \theta=TY$. Similarly we have $YM=XT$.

Ptolemy's theorem in $BXTM$ gives

$$16TY=11TX+3\sqrt{15}BX,$$ while Pythagoras' theorem gives $BX^2+XT^2=16^2$. Similarly, Ptolemy's theorem in $YTMC$ gives

$$16TX=11TY+3\sqrt{15}CY$$ while Pythagoras' theorem in $\triangle CYT$ gives $CY^2+YT^2=16^2$. Solve this for $XT$ and $TY$ and substitute into the equation about $\cos XTY$ to obtain the result $XY^2=\boxed{717}$.

(Notice that $MXTY$ is a parallelogram, which is an important theorem in Olympiad, and there are some other ways of computation under this observation.)

-Fanyuchen20020715

Solution 2 (Official MAA)

Let $M$ denote the midpoint of $\overline{BC}$. The critical claim is that $M$ is the orthocenter of $\triangle AXY$, which has the circle with diameter $\overline{AT}$ as its circumcircle. To see this, note that because $\angle BXT = \angle BMT = 90^\circ$, the quadrilateral $MBXT$ is cyclic, it follows that

$$\angle MXA = \angle MXB = \angle MTB = 90^\circ - \angle TBM = 90^\circ - \angle A,$$ implying that $\overline{MX} \perp \overline{AC}$. Similarly, $\overline{MY} \perp \overline{AB}$. In particular, $MXTY$ is a parallelogram.

Hence, by the Parallelogram Law,

$$TM^2 + XY^2 = 2(TX^2 + TY^2) = 2(1143-XY^2).$$ But $TM^2 = TB^2 - BM^2 = 16^2 - 11^2 = 135$. Therefore

$$XY^2 = \frac13(2 \cdot 1143-135) = \boxed{717}.$$

Solution 3 (Law of Cosines)

Let $H$ be the orthocenter of $\triangle AXY$.

Lemma 1: $H$ is the midpoint of $BC$.

Proof: Let $H'$ be the midpoint of $BC$, and observe that $XBH'T$ and $TH'CY$ are cyclical. Define $H'Y \cap BA=E$ and $H'X \cap AC=F$, then note that:

$$\angle H'BT=\angle H'CT=\angle H'XT=\angle H'YT=\angle A.$$ That implies that $\angle H'XB=\angle H'YC=90^\circ-\angle A$, $\angle CH'Y=\angle EH'B=90^\circ-\angle B$, and $\angle BH'Y=\angle FH'C=90^\circ-\angle C$. Thus $YH'\perp AX$ and $XH' \perp AY$; $H'$ is indeed the same as $H$, and we have proved lemma 1.

Since $AXTY$ is cyclical, $\angle XTY=\angle XHY$ and this implies that $XHYT$ is a paralelogram. By the Law of Cosines:

$$XY^2=XT^2+TY^2+2(XT)(TY)\cdot \cos(\angle A)$$ $$XY^2=XH^2+HY^2+2(XH)(HY) \cdot \cos(\angle A)$$ $$HT^2=HX^2+XT^2-2(HX)(XT) \cdot \cos(\angle A)$$ $$HT^2=HY^2+YT^2-2(HY)(YT) \cdot \cos(\angle A).$$ We add all these equations to get:

$$HT^2+XY^2=2(XT^2+TY^2) \qquad (1).$$ We have that $BH=HC=11$ and $BT=TC=16$ using our midpoints. Note that $HT \perp BC$, so by the Pythagorean Theorem, it follows that $HT^2=135$. We were also given that $XT^2+TY^2=1143-XY^2$, which we multiply by $2$ to use equation $(1)$.

$$2(XT^2+TY^2)=2286-2 \cdot XY^2$$ Since $2(XT^2+TY^2)=2(HT^2+TY^2)=HT^2+XY^2$, we have

$$135+XY^2=2286-2 \cdot XY^2$$ $$3 \cdot XY^2=2151.$$ Therefore, $XY^2=\boxed{717}$. ~ MathLuis

Solution 4 (Similarity and median)

Using the Claim (below) we get $\triangle ABC \sim \triangle XTM \sim \triangle YMT.$

Corresponding sides of similar $\triangle XTM \sim \triangle YMT$ is $MT,$ so

$\triangle XTM = \triangle YMT \implies MY = XT, MX = TY \implies XMYT$ – parallelogram.

$$4 TD^2 = MT^2 = \sqrt{BT^2 - BM^2} =\sqrt{153}.$$ The formula for median $DT$ of triangle $XYT$ is

$$2 DT^2 = XT^2 + TY^2 – \frac{XY^2}{2},$$ $$3 \cdot XY^2 = 2XT^2 + 2TY^2 + 2XY^2 – 4 DT^2,$$ $$3 \cdot XY^2 = 2 \cdot 1143-153 = 2151 \implies XY^2 = \boxed{717}.$$

Claim

Let $\triangle ABC$ be an acute scalene triangle with circumcircle $\omega$. The tangents to $\omega$ at $B$ and $C$ intersect at $T$. Let $X$ be the projections of $T$ onto line $AB$. Let M be midpoint BC. Then triangle ABC is similar to triangle XTM.

Proof

$\angle BXT = \angle BMT = 90^o \implies$ the quadrilateral $MBXT$ is cyclic.

$BM \perp MT, TX \perp AB \implies \angle MTX = \angle MBA.$ $\angle CBT = \angle BAC = \frac {\overset{\Large\frown} {BC}}{ 2} \implies \triangle ABC \sim \triangle XTM.$

vladimir.shelomovskii@gmail.com, vvsss

Solution 5 (Symmedian and Stewarts)

Let $M$ be the midpoint of $BC$. Note that $\angle XYT = \angle XAT = \angle MAC$ because $AT$ is a symmedian. Similarly, $\angle TXY = \angle MAB$. $XT = 16\sin{C}$ and $YT = 16\sin{C}$. By law of sines on $XYT$, $\frac{16\sin{C}}{\sin{XYT}} = \frac{XY}{\sin{A}}$. However by law of sines on $MAC$, $\frac{\sin{C}}{\sin{MAC}} = \frac{AM}{11}$. Combining these two yields, $\frac{16}{11} AM = \frac{XY}{\sin{A}}$. Since $\sin{A} = \frac{\sqrt{135}}{16}$, we have $XY^2 = \frac{135}{121} AM^2$.

Letting $AB = x$, $AC = y$, and $AM = d$, we have $x^2 + y^2 = 2d^2 + 242$ by Stewarts. Since $x = 2R\sin{C}$, and $y = 2R\sin{B}$ by extended law of sines, we can write $\sin^2{B} + \sin^2{C} = \frac{2d^2 + 242}{4R^2}$. By law of cosines on $ABC$, $x^2 + y^2 - 2xy(\frac{11}{16}) = 484$, $\frac{11}{8} xy + 484 = 2d^2 + 242$, $xy = \frac{8}{11} (2d^2 - 242)$. Then similarly as before we can write $\sin{B}\sin{C} = \frac{\frac{8}{11} (2d^2 - 242)}{4R^2}$.

By law of cosines on $XYT$ and using $XT^2 + XY^2 + YT^2 = 1143$, we have $512(\sin^2{B} + \sin^2{C}) - 1143 = -512\sin{B}\sin{C}(\frac{11}{16})$. Substituting our previous values here and using $R = \frac{176}{\sqrt{135}}$ yields a value for $d^2$, and multiplying by $\frac{135}{121}$ gives $\boxed{717}$.

~sdfgfjh