Difference between revisions of "2020 AIME II Problems/Problem 15"
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==Solution== | ==Solution== | ||
− | Assume <math>O</math> to be the center of triangle <math>ABC</math>, <math>OT</math> cross <math>BC</math> at <math>M</math>, link <math>XM</math>, <math>YM</math>. Let <math>P</math> be the middle point of <math>BT</math> and <math>Q</math> be the middle point of <math>CT</math>, so we have <math>MT=3\sqrt{15}</math>. Since <math>\angle A=\angle CBT=\angle BCT</math>, we have <math>\cos A=\frac{11}{16}</math>. Notice that | + | Assume <math>O</math> to be the center of triangle <math>ABC</math>, <math>OT</math> cross <math>BC</math> at <math>M</math>, link <math>XM</math>, <math>YM</math>. Let <math>P</math> be the middle point of <math>BT</math> and <math>Q</math> be the middle point of <math>CT</math>, so we have <math>MT=3\sqrt{15}</math>. Since <math>\angle A=\angle CBT=\angle BCT</math>, we have <math>\cos A=\frac{11}{16}</math>. Notice that <math>\angle XTY=180^{\circ}-A</math>, so <math>\cos XYT=-\cos A</math>, and this gives us <math>1143-2XY^2=\frac{-11}{8}XT\cdot YT</math>. Since <math>TM</math> is perpendicular to <math>BC</math>, <math>BXTM</math> and <math>CYTM</math> cocycle (respectively), so <math>\theta_1=\angle ABC=\angle MTX</math> and <math>\theta_2=\angle ACB=\angle YTM</math>. So <math>\angle XPM=2\theta_1</math>, so <cmath>\frac{\frac{XM}{2}}{XP}=\sin \theta_1</cmath>, which yields <math>XM=2XP\sin \theta_1=BT(=CT)\sin \theta_1=TY.</math> So same we have <math>YM=XT</math>. Apply Ptolemy theorem in <math>BXTM</math> we have <math>16TY=11TX+3\sqrt{15}BX</math>, and use Pythagoras theorem we have <math>BX^2+XT^2=16^2</math>. Same in <math>YTMC</math> and triangle <math>CYT</math> we have <math>16TX=11TY+3\sqrt{15}CY</math> and <math>CY^2+YT^2=16^2</math>. Solve this for <math>XT</math> and <math>TY</math> and submit into the equation about <math>\cos XYT</math>, we can obtain the result <math>XY^2=\boxed{717}</math>. |
+ | |||
+ | (Notice that <math>MXTY</math> is a parallelogram, which is an important theorem in Olympiad, and there are some other ways of computation under this observation.) | ||
-Fanyuchen20020715 | -Fanyuchen20020715 | ||
− | == | + | ==Solution 2 (Official MAA)== |
− | + | Let <math>M</math> denote the midpoint of <math>\overline{BC}</math>. The critical claim is that <math>M</math> is the orthocenter of <math>\triangle AXY</math>, which has the circle with diameter <math>\overline{AT}</math> as its circumcircle. To see this, note that because <math>\angle BXT = \angle BMT = 90^\circ</math>, the quadrilateral <math>MBXT</math> is cyclic, it follows that | |
+ | <cmath>\angle MXA = \angle MXB = \angle MTB = 90^\circ - \angle TBM = 90^\circ - \angle A,</cmath> implying that <math>\overline{MX} \perp \overline{AC}</math>. Similarly, <math>\overline{MY} \perp \overline{AB}</math>. In particular, <math>MXTY</math> is a parallelogram. | ||
+ | <asy> | ||
+ | defaultpen(fontsize(8pt)); | ||
+ | unitsize(0.8cm); | ||
+ | |||
+ | pair A = (0,0); | ||
+ | pair B = (-1.26,-4.43); | ||
+ | pair C = (-1.26+3.89, -4.43); | ||
+ | pair M = (B+C)/2; | ||
+ | pair O = circumcenter(A,B,C); | ||
+ | pair T = (0.68, -6.49); | ||
+ | pair X = foot(T,A,B); | ||
+ | pair Y = foot(T,A,C); | ||
+ | path omega = circumcircle(A,B,C); | ||
+ | real rad = circumradius(A,B,C); | ||
+ | |||
+ | |||
+ | |||
+ | filldraw(A--B--C--cycle, rgb(0.98,0.81,0.69)); | ||
+ | label("$\omega$", O + rad*dir(45), SW); | ||
+ | filldraw(T--Y--M--X--cycle, rgb(173/255,216/255,230/255)); | ||
+ | draw(M--T); | ||
+ | draw(X--Y); | ||
+ | draw(B--T--C); | ||
+ | draw(A--X--Y--cycle); | ||
+ | draw(omega); | ||
+ | dot("$X$", X, W); | ||
+ | dot("$Y$", Y, E); | ||
+ | dot("$O$", O, W); | ||
+ | dot("$T$", T, S); | ||
+ | dot("$A$", A, N); | ||
+ | dot("$B$", B, W); | ||
+ | dot("$C$", C, E); | ||
+ | dot("$M$", M, N); | ||
+ | |||
+ | |||
+ | </asy> | ||
+ | Hence, by the Parallelogram Law, | ||
+ | <cmath> TM^2 + XY^2 = 2(TX^2 + TY^2) = 2(1143-XY^2).</cmath> But <math>TM^2 = TB^2 - BM^2 = 16^2 - 11^2 = 135</math>. Therefore <cmath>XY^2 = \frac13(2 \cdot 1143-135) = 717.</cmath> | ||
+ | |||
+ | ==Solution 3 (Law of Cosines)== | ||
+ | Let <math>H</math> be the orthocenter of <math>\triangle AXY</math>. | ||
+ | |||
+ | <b>Lemma 1:</b> <math>H</math> is the midpoint of <math>BC</math>. | ||
+ | |||
+ | <b>Proof:</b> Let <math>H'</math> be the midpoint of <math>BC</math>, and observe that <math>XBH'T</math> and <math>TH'CY</math> are cyclical. Define <math>H'Y \cap BA=E</math> and <math>H'X \cap AC=F</math>, then note that: | ||
+ | <cmath>\angle H'BT=\angle H'CT=\angle H'XT=\angle H'YT=\angle A.</cmath> | ||
+ | That implies that <math>\angle H'XB=\angle H'YC=90^\circ-\angle A</math>, <math>\angle CH'Y=\angle EH'B=90^\circ-\angle B</math>, and <math>\angle BH'Y=\angle FH'C=90^\circ-\angle C</math>. Thus <math>YH'\perp AX</math> and <math>XH' \perp AY</math>; <math>H'</math> is indeed the same as <math>H</math>, and we have proved lemma 1. | ||
+ | |||
+ | Since <math>AXTY</math> is cyclical, <math>\angle XTY=\angle XHY</math> and this implies that <math>XHYT</math> is a paralelogram. | ||
+ | By the Law of Cosines: | ||
+ | <cmath>XY^2=XT^2+TY^2+2(XT)(TY)\cdot \cos(\angle A)</cmath> | ||
+ | <cmath>XY^2=XH^2+HY^2+2(XH)(HY) \cdot \cos(\angle A)</cmath> | ||
+ | <cmath>HT^2=HX^2+XT^2-2(HX)(XT) \cdot \cos(\angle A)</cmath> | ||
+ | <cmath>HT^2=HY^2+YT^2-2(HY)(YT) \cdot \cos(\angle A).</cmath> | ||
+ | We add all these equations to get: | ||
+ | <cmath>HT^2+XY^2=2(XT^2+TY^2) \qquad (1).</cmath> | ||
+ | We have that <math>BH=HC=11</math> and <math>BT=TC=16</math> using our midpoints. Note that <math>HT \perp BC</math>, so by the Pythagorean Theorem, it follows that <math>HT^2=135</math>. We were also given that <math>XT^2+TY^2=1143-XY^2</math>, which we multiply by <math>2</math> to use equation <math>(1)</math>. <cmath>2(XT^2+TY^2)=2286-2 \cdot XY^2</cmath> Since <math>2(XT^2+TY^2)=2(HT^2+TY^2)=HT^2+XY^2</math>, we have | ||
+ | <cmath>135+XY^2=2286-2 \cdot XY^2</cmath> <cmath>3 \cdot XY^2=2151.</cmath> | ||
+ | Therefore, <math>XY^2=\boxed{717}</math>. ~ MathLuis | ||
==See Also== | ==See Also== | ||
{{AIME box|year=2020|n=II|num-b=14|after=Last Problem}} | {{AIME box|year=2020|n=II|num-b=14|after=Last Problem}} | ||
+ | [[Category:Intermediate Geometry Problems]] | ||
{{MAA Notice}} | {{MAA Notice}} |
Revision as of 12:47, 15 April 2021
Problem
Let be an acute scalene triangle with circumcircle . The tangents to at and intersect at . Let and be the projections of onto lines and , respectively. Suppose , , and . Find .
Solution
Assume to be the center of triangle , cross at , link , . Let be the middle point of and be the middle point of , so we have . Since , we have . Notice that , so , and this gives us . Since is perpendicular to , and cocycle (respectively), so and . So , so , which yields So same we have . Apply Ptolemy theorem in we have , and use Pythagoras theorem we have . Same in and triangle we have and . Solve this for and and submit into the equation about , we can obtain the result .
(Notice that 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 denote the midpoint of . The critical claim is that is the orthocenter of , which has the circle with diameter as its circumcircle. To see this, note that because , the quadrilateral is cyclic, it follows that implying that . Similarly, . In particular, is a parallelogram. Hence, by the Parallelogram Law, But . Therefore
Solution 3 (Law of Cosines)
Let be the orthocenter of .
Lemma 1: is the midpoint of .
Proof: Let be the midpoint of , and observe that and are cyclical. Define and , then note that: That implies that , , and . Thus and ; is indeed the same as , and we have proved lemma 1.
Since is cyclical, and this implies that is a paralelogram. By the Law of Cosines: We add all these equations to get: We have that and using our midpoints. Note that , so by the Pythagorean Theorem, it follows that . We were also given that , which we multiply by to use equation . Since , we have Therefore, . ~ MathLuis
See Also
2020 AIME II (Problems • Answer Key • Resources) | ||
Preceded by Problem 14 |
Followed by Last Problem | |
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