Difference between revisions of "2020 AIME I Problems/Problem 15"
(→Solution) |
R00tsofunity (talk | contribs) |
||
(30 intermediate revisions by 14 users not shown) | |||
Line 1: | Line 1: | ||
− | |||
== Problem == | == Problem == | ||
+ | Let <math>\triangle ABC</math> be an acute triangle with circumcircle <math>\omega,</math> and let <math>H</math> be the intersection of the altitudes of <math>\triangle ABC.</math> Suppose the tangent to the circumcircle of <math>\triangle HBC</math> at <math>H</math> intersects <math>\omega</math> at points <math>X</math> and <math>Y</math> with <math>HA=3,HX=2,</math> and <math>HY=6.</math> The area of <math>\triangle ABC</math> can be written in the form <math>m\sqrt{n},</math> where <math>m</math> and <math>n</math> are positive integers, and <math>n</math> is not divisible by the square of any prime. Find <math>m+n.</math> | ||
+ | |||
+ | == Solution 1== | ||
+ | The following is a power of a point solution to this menace of a problem: | ||
− | |||
− | |||
<asy> | <asy> | ||
− | + | defaultpen(fontsize(12)+0.6); size(250); | |
− | + | pen p=fontsize(10)+gray+0.4; | |
− | + | ||
− | + | var phi=75.5, theta=130, r=4.8; | |
− | + | pair A=r*dir(270-phi-57), C=r*dir(270+phi-57), B=r*dir(theta-57), H=extension(A,foot(A,B,C),B,foot(B,C,A)); | |
− | + | path omega=circumcircle(A,B,C), c=circumcircle(H,B,C); | |
− | + | pair O=circumcenter(H,B,C), Op=2*H-O, Z=bisectorpoint(O,Op), X=IP(omega,L(H,Z,0,50)), Y=IP(omega,L(H,Z,50,0)), D=2*H-A, K=extension(B,C,X,Y), E=extension(A,origin,X,Y), L=foot(H,B,C); | |
− | + | ||
− | draw(( | + | draw(K--C^^omega^^c^^L(A,D,0,1)); draw(Y--K^^L(A,origin,0,1.5)^^L(X,D,0.7,3.5),fuchsia+0.4); |
− | draw( | + | draw(CR(H,length(H-L)),royalblue); |
− | + | draw(C--K,p); | |
− | + | dot("$A$",A,dir(120)); dot("$B$",B,down); dot("$C$",C,down); dot("$H$",H,down); dot("$X$",X,up); dot("$Y$",Y,down); | |
− | draw( | + | dot("$O$",origin,down); dot("$O'$",O,down); dot("$K$",K,up); dot("$L$",L,dir(H-X)); |
− | draw( | + | dot("$D$",D,down); dot("$E$",E,down); |
− | draw(( | + | |
− | + | //draw(O--origin,p); | |
− | + | //draw(origin--4*dir(57),fuchsia); | |
− | + | //draw(Arc(H,3,160,200),royalblue); //draw(Arc(H,2,30,70),heavygreen); //draw(Arc(H,6,220,240),fuchsia); | |
− | + | </asy> | |
− | + | Let points be what they appear as in the diagram below. Note that <math>3HX = HY</math> is not insignificant; from here, we set <math>XH = HE = \frac{1}{2} EY = HL = 2</math> by PoP and trivial construction. Now, <math>D</math> is the reflection of <math>A</math> over <math>H</math>. Note <math>AO \perp XY</math>, and therefore by Pythagorean theorem we have <math>AE = XD = \sqrt{5}</math>. Consider <math>HD = 3</math>. We have that <math>\triangle HXD \cong HLK</math>, and therefore we are ready to PoP with respect to <math>(BHC)</math>. Setting <math>BL = x, LC = y</math>, we obtain <math>xy = 10</math> by PoP on <math>(ABC)</math>, and furthermore, we have <math>KH^2 = 9 = (KL - x)(KL + y) = (\sqrt{5} - x)(\sqrt{5} + y)</math>. Now, we get <math>4 = \sqrt{5}(y - x) - xy</math>, and from <math>xy = 10</math> we take <cmath>\frac{14}{\sqrt{5}} = y - x.</cmath> However, squaring and manipulating with <math>xy = 10</math> yields that <math>(x + y)^2 = \frac{396}{5}</math> and from here, since <math>AL = 5</math> we get the area to be <math>3\sqrt{55} \implies \boxed{058}</math>. ~awang11's sol | |
− | + | ||
− | + | ==Solution 1a== | |
− | + | As in the diagram, let ray <math>AH</math> extended hits BC at L and the circumcircle at say <math>P</math>. By power of the point at H, we have <math>HX \cdot HY = AH \cdot HP</math>. The three values we are given tells us that <math>HP=\frac{2\cdot 6}{3}=4</math>. L is the midpoint of <math>HP</math>(see here: https://www.cut-the-knot.org/Curriculum/Geometry/AltitudeAndCircumcircle.shtml ), so <math>HL=LP=2</math>. | |
− | + | ||
− | + | As in the diagram provided, let K be the intersection of <math>BC</math> and <math>XY</math>. By power of a point on the circumcircle of triangle <math>HBC</math>, <math>KH^{2}=KB \cdot KC</math>. By power of a point on the circumcircle of triangle <math>ABC</math>, <math>KB \cdot KC=KX \cdot KY</math>, thus <math>KH^{2}=(KH-2)(KH+6)</math>. Solving gives <math>4KH=12</math> or <math>KH=3</math>. | |
− | + | ||
− | label("$ | + | By the Pythagorean Theorem on triangle <math>HKL</math>, <math>KL=\sqrt{5}</math>. Now continue with solution 1. |
− | + | ||
− | label("$ | + | == Solution 2 == |
− | + | <asy> | |
− | label("$ | + | size(10cm); |
− | + | pair A, B, C, D, H, K, O, P, L, M, X, Y; | |
− | label("$ | + | A = (-15, 27); |
− | + | B = (-24, 0); | |
− | label("$ | + | C = (24, 0); |
− | + | D = (-8.28, 18.04); | |
− | label("$ | + | O = (0, 7); |
− | + | P = (0, -7); | |
− | label("$ | + | H = (-15, 13); |
− | + | K = (-15, -13); | |
− | label("$ | + | M = (0, 0); |
− | + | L = (-15, 0); | |
− | + | X = (-24.9569, 5.53234); | |
+ | Y = (8.39688, 30.5477); | ||
+ | draw(circle(O, 25)); | ||
+ | draw(circle(P, 25)); | ||
+ | draw(A--B--C--cycle); | ||
+ | draw(H -- K); | ||
+ | draw(A -- O -- P -- H -- cycle); | ||
+ | draw(X -- Y); | ||
+ | draw(O -- X, dashed); | ||
+ | draw(O -- Y, dashed); | ||
+ | draw(O -- B, dashed); | ||
+ | draw(O -- C, dashed); | ||
+ | |||
+ | label("$O$", O, ENE); | ||
+ | label("$A$", A, NW); | ||
+ | label("$B$", B, W); | ||
+ | label("$C$", C, E); | ||
+ | label("$H$", H, E); | ||
+ | label("$H'$", K, NE); | ||
+ | label("$X$", X, W); | ||
+ | label("$Y$", Y, NE); | ||
+ | label("$O'$", P, E); | ||
+ | label("$M$", M, NE); | ||
+ | label("$L$", L, NE); | ||
+ | label("$D$", D, NNE); | ||
+ | |||
+ | label("$2$", X -- H, NW); | ||
+ | label("$3$", H -- A, SW); | ||
+ | label("$6$", H -- Y, NW); | ||
+ | label("$R$", O -- Y, E); | ||
+ | |||
+ | dot(O); | ||
+ | dot(P); | ||
+ | dot(D); | ||
+ | dot(H); | ||
+ | |||
</asy> | </asy> | ||
+ | Diagram not to scale. | ||
+ | |||
+ | |||
+ | We first observe that <math>H'</math>, the image of the reflection of <math>H</math> over line <math>BC</math>, lies on circle <math>O</math>. This is because <math>\angle HBC = 90 - \angle C = \angle H'AC = \angle H'BC</math>. This is a well known lemma. The result of this observation is that circle <math>O'</math>, the circumcircle of <math>\triangle BHC</math> is the image of circle <math>O</math> over line <math>BC</math>, which in turn implies that <math>\overline{AH} = \overline{OO'}</math> and thus <math>AHO'O</math> is a parallelogram. That <math>AHO'O</math> is a parallelogram implies that <math>AO</math> is perpendicular to <math>\overline{XY}</math>, and thus divides segment <math>\overline{XY}</math> in two equal pieces, <math>\overline{XD}</math> and <math>\overline{DY}</math>, of length <math>4</math>. | ||
+ | |||
+ | |||
+ | Using Power of a Point, | ||
+ | <cmath>\overline{AH} \cdot \overline{HH'} = \overline{XH} \cdot \overline{HY} \Longrightarrow 3 \cdot \overline{HH'} = 2 \cdot 6 \Longrightarrow \overline{HH'} = 4</cmath> | ||
+ | This means that <math>\overline{HL} = \frac12 \cdot 4 = 2</math> and <math>\overline{AL} = 2 + 3 = 5</math>, where <math>L</math> is the foot of the altitude from <math>A</math> onto <math>BC</math>. All that remains to be found is the length of segment <math>\overline{BC}</math>. | ||
+ | |||
+ | Looking at right triangle <math>\triangle AHD</math>, we find that | ||
+ | <cmath>\overline{AD} = \sqrt{\overline{AH}^2 - \overline{HD}^2} = \sqrt{3^2 - 2^2} = \sqrt{5}</cmath> | ||
+ | Looking at right triangle <math>\triangle ODY</math>, we get the equation | ||
+ | <cmath>\overline{OY}^2 - \overline{DY}^2 = \overline{OD}^2 = \left(\overline{AO} - \overline{AD}\right)^2</cmath> | ||
+ | Plugging in known values, and letting <math>R</math> be the radius of the circle, we find that | ||
+ | <cmath>R^2 - 16 = (R - \sqrt{5})^2 = R^2 - 2\sqrt5 R + 5 \Longrightarrow R = \frac{21\sqrt5}{10}</cmath> | ||
+ | |||
+ | Recall that <math>AHO'O</math> is a parallelogram, so <math>\overline{AH} = \overline{OO'} = 3</math>. So, <math>\overline{OM} = \frac32</math>, where <math>M</math> is the midpoint of <math>\overline{BC}</math>. This means that | ||
+ | <cmath>\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}</cmath> | ||
+ | |||
+ | Thus, the area of triangle <math>\triangle ABC</math> is | ||
+ | <cmath>\frac{\overline{AL} \cdot \overline{BC}}{2} = \frac{5 \cdot \frac{6\sqrt{55}}{5}}{2} = {3\sqrt{55}}</cmath> | ||
+ | The answer is <math>3 + 55 = \boxed{058}</math>. | ||
− | Let | + | ==Solution 3 (Official MAA 1)== |
+ | Extend <math>\overline{AH}</math> to intersect <math>\omega</math> again at <math>P</math>. The Power of a Point Theorem yields <math>HP = \tfrac{HX \cdot HY}{HA} = 4</math>. Because <math>\angle CAP=\angle CBP</math>, and <math>\angle CAP</math> and <math>\angle CBH</math> are both complements to <math>\angle C</math>, it follows that <math>\angle CBP = \angle CBH</math>, implying that <math>\overline{BC}</math> bisects <math>\overline{HP}</math>, so the length of the altitude from <math>A</math> to <math>\overline{BC}</math> is <math>h_a = AH + \tfrac12 HP = 5</math>. | ||
+ | |||
+ | Let the circumcircle of <math>\triangle BCH</math> be <math>\omega'</math>. Because <math>\triangle BCH \cong \triangle BCP</math>, the two triangles must have the same circumradius. Because the circumcircle of <math>\triangle BCP</math> is <math>\omega</math>, the circles <math>\omega</math> and <math>\omega'</math> have the same radius <math>R</math>. Denote the centers of <math>\omega</math> and <math>\omega'</math> by <math>O</math> and <math>O'</math>, respectively, and let <math>M</math> be the midpoint of <math>\overline{XY}</math>. Note that trapezoid <math>HMOO'</math> has <math>\angle H = \angle M = 90^\circ</math>. Also <math>HM = XM - XH = \frac12\cdot XY - HX = 2</math> and <math>HO' = R</math>. Because <math>\omega</math> is a translation of <math>\omega'</math> in the direction of <math>\overline{AH}</math>, it follows that <math>OO' = AH = 3</math>. Finally, the Pythagorean Theorem applied to <math>\triangle XMO</math> yields <math>MO = \sqrt{R^2-16}</math>. Let <math>T</math> be the projection of <math>O</math> onto <math>\overline{HO'}</math>. Then <math>TO' = R-MO</math>, so the Pythagorean Theorem applied to <math>\triangle TOO'</math> yields | ||
+ | <cmath>R - \sqrt{R^2-16} = \sqrt{3^2 - 2^2} = \sqrt{5}.</cmath>Solving for <math>R</math> gives <math>R = \tfrac{21}{2\sqrt5}</math>. It follows from properties of the orthocenter <math>H</math> that<cmath>\cos\angle A = \dfrac{AH}{2R} = \dfrac{\sqrt5}{7},</cmath>so<cmath>\sin\angle A = \sqrt{1 - \cos^2\angle A} = \dfrac{2\sqrt{11}}{7}.</cmath>Therefore by the Extended Law of Sines<cmath>a = BC = 2R \sin\angle A = \dfrac{6\sqrt{11}}{\sqrt5},</cmath>so | ||
+ | <cmath>[\triangle ABC] = \frac12 a h_a = \frac12 \cdot \frac{6\sqrt{11}}{\sqrt{5}} \cdot 5 = 3\sqrt{55}.</cmath>The requested sum is <math>3+55 = 58</math>. | ||
+ | |||
+ | <asy> | ||
+ | unitsize(0.6 cm); | ||
+ | |||
+ | pair A, B, C, D, H, M, O, Op, P, T, X, Y, Z; | ||
+ | real R = 21/(2*sqrt(5)); | ||
+ | |||
+ | A = (7/sqrt(5),7/2); | ||
+ | O = (0,0); | ||
+ | B = intersectionpoint((0,-3/2)--(R,-3/2),Circle(O,R)); | ||
+ | C = intersectionpoint((0,-3/2)--(-R,-3/2),Circle(O,R)); | ||
+ | H = A + B + C; | ||
+ | P = reflect(B,C)*(H); | ||
+ | D = (H + P)/2; | ||
+ | Op = reflect(B,C)*(O); | ||
+ | X = intersectionpoint(H--(H + scale(2)*rotate(90)*(Op - H)), Circle(O,R)); | ||
+ | Y = intersectionpoint(H--(H + scale(2)*rotate(90)*(H - Op)), Circle(O,R)); | ||
+ | Z = extension(X, Y, B, C); | ||
+ | M = (X + Y)/2; | ||
+ | T = H + O - M; | ||
+ | |||
+ | draw(Circle(O,R)); | ||
+ | draw(Circle(Op,R)); | ||
+ | draw(A--B--C--cycle); | ||
+ | draw(A--P); | ||
+ | draw(B--Z--Y); | ||
+ | draw(H--Op--O--M); | ||
+ | draw(O--T); | ||
+ | draw(O--X); | ||
+ | |||
+ | dot("$A$", A, NE); | ||
+ | dot("$B$", B, SW); | ||
+ | dot("$C$", C, W); | ||
+ | dot("$D$", D, SW); | ||
+ | dot("$H$", H, NE); | ||
+ | dot("$M$", M, NE); | ||
+ | dot("$O$", O, W); | ||
+ | dot("$O'$", Op, W); | ||
+ | dot("$P$", P, SE); | ||
+ | dot("$T$", T, N); | ||
+ | dot("$X$", X, E); | ||
+ | dot("$Y$", Y, NW); | ||
+ | dot("$Z$", Z, E); | ||
+ | |||
+ | label("$\omega$", R*dir(140), dir(140)); | ||
+ | label("$\omega'$", Op + R*dir(220), dir(220)); | ||
+ | </asy> | ||
+ | |||
+ | ==Solution 3a (slightly modified Solution 3)== | ||
+ | |||
+ | Note that <math>\overline{BC}</math> bisects <math>\overline{OO'}</math>. Using the same method from Solution 3 we find <math>R = \frac{21}{2\sqrt{5}}</math>. Let the midpoint of <math>\overline{BC}</math> be <math>N</math>, then by the Pythagorean Theorem we have <math>BN^2 = OB^2 - ON^2 = R^2 - \frac{3}{2}^2</math>, so <math>BN = \frac{3\sqrt{11}}{\sqrt{5}}</math>. Since <math>h_a = 5</math> we have that the area of ABC is <math>3\sqrt{55}</math> so the answer is <math>3 + 55 = \boxed{58}</math> | ||
+ | |||
+ | ~bobjoebilly | ||
+ | |||
+ | ==Solution 4 (Official MAA 2)== | ||
+ | Let <math>D</math> be the intersection point of line <math>AH</math> and <math>\overline{BC}</math>, noting that <math>\overline{AD}\perp\overline{BC}</math>. Because the area of <math>\triangle ABC</math> is <math>\tfrac12\cdot AD\cdot BC</math>, it suffices to compute <math>AD</math> and <math>BC</math> separately. As in the previous solution, <math>AD = 5</math>. The value of <math>BC</math> can be found using the following lemma. | ||
+ | |||
+ | Lemma: Triangle <math>AXY</math> is isosceles with base <math>\overline{XY}</math>. | ||
+ | |||
+ | Proof: Because the circumcircle of <math>\triangle BCH</math>, <math>\omega'</math>, and <math>\omega</math> have the same radius, there exists a translation <math>\Phi</math> sending the former to the latter. Because <math>\overline{AH}</math> is parallel to the line connecting the centers of the two circles, <math>\Phi</math> must send <math>H</math> to <math>A</math>, meaning <math>\Phi</math> also sends <math>\overline{XY}</math> to the tangent to <math>\omega</math> at <math>A</math>. But this means that this tangent is parallel to <math>\overline{XY}</math>, which implies the conclusion. | ||
+ | |||
+ | Applying Stewart's Theorem to <math>\triangle AXY</math> yields<cmath>AX^2 = AH^2 + HX\cdot HY = 3^2 + 2\cdot 6 = 21,</cmath>implying <math>AX = AY= \sqrt{21}.</math> | ||
+ | |||
+ | By the Law of Cosines | ||
+ | <cmath>\cos \angle XAY = \frac{21 + 21 - 64}{2 \cdot 21} = -\frac{11}{21},</cmath> | ||
+ | so<cmath>\sin \angle XAY = \dfrac{4\sqrt{20}}{21}.</cmath> | ||
+ | Let <math>R</math> be the radius of <math>\omega</math>. By the Extended Law of Sines<cmath>R = \frac{XY}{2 \cdot \sin \angle XAY} = \dfrac{21}{\sqrt{20}}.</cmath> | ||
+ | Then the solution proceeds as in Solution 3. | ||
+ | |||
+ | |||
+ | ==Solution 5 (Official MAA 3)== | ||
+ | Define points <math>D</math> and <math>P</math> as above, and note that <math>AD=5</math> and <math>DH=PD=AD - AH = 2</math>. Let the circumcircle of <math>\triangle BCH</math> be <math>\omega'</math>. | ||
+ | |||
+ | Extend <math>\overline{XY}</math> past <math>X</math> until it intersects line <math>BC</math> at point <math>Z</math>. Because line <math>BC</math> is a radical axis of <math>\omega</math> and <math>\omega'</math>, it follows from the Power of a Point Theorem that | ||
+ | <cmath> | ||
+ | ZX \cdot ZY = ZX \cdot(ZX + 8) = ZH^2 = (ZX+2)^2, | ||
+ | </cmath>from which <math>ZX=1</math>. By Pythagorean Theorem <math>ZD=\sqrt{ZH^2 - DH^2} = \sqrt{5}</math>. | ||
+ | |||
+ | |||
+ | |||
+ | Let <math>m=CD</math> and <math>n=BD</math>. By the Power of a Point Theorem | ||
+ | <cmath> | ||
+ | mn= AD\cdot PD = 10. | ||
+ | </cmath>On the other hand, | ||
+ | <cmath> | ||
+ | ZH^2 = 9 = ZB \cdot ZC = (\sqrt{5} - n)(\sqrt{5} + m) = 5 + \sqrt{5}(m-n) - mn, | ||
+ | </cmath>from which <math>m-n = \frac{14}{\sqrt{5}}</math>. Therefore | ||
+ | |||
+ | <cmath> | ||
+ | (m+n)^2 = (m-n)^2 + 4mn = \frac{196}{5} + 40 = \frac{396}{5}. | ||
+ | </cmath> | ||
+ | Thus <math>BC = m+n=\sqrt{\frac{396}{5}} = 6\sqrt{\frac{11}{5}}</math>. Therefore | ||
+ | <math>[\triangle ABC] = \frac{1}{2}BC \cdot AD = 3\sqrt{55},</math> as above. | ||
+ | |||
+ | ==Solution 6== | ||
+ | [[File:AIME-I-2020-15.png|400px|right]] | ||
+ | Let <math>O</math> be circumcenter of <math>ABC,</math> let <math>R</math> be circumradius of <math>ABC,</math> let <math>\omega'</math> be the image of circle <math>\omega</math> over line <math>BC</math> (the circumcircle of <math>HBC</math>). | ||
+ | |||
+ | Let <math>P</math> be the image of the reflection of <math>H</math> over line <math>BC, P</math> lies on circle <math>\omega.</math> Let <math>M</math> be the midpoint of <math>XY.</math> | ||
+ | Then <math>P</math> lies on <math>\omega, OA = O'H, OA || O'H.</math> | ||
+ | |||
+ | (see here: https://brilliant.org/wiki/triangles-orthocenter/ | ||
+ | or https://en.wikipedia.org/wiki/Altitude_(triangle) Russian) | ||
+ | |||
+ | <math>P</math> lies on <math>\omega \implies OA = OP = R,</math> | ||
+ | |||
+ | <math>OA = O'H, OA || O'H \implies M</math> lies on <math>OA.</math> | ||
+ | |||
+ | We use properties of crossing chords and get | ||
+ | <cmath>AH \cdot HP = XH \cdot HY = 2 \cdot 6 \implies HP = 4, AP = AH + HP = 7.</cmath> | ||
+ | We use properties of radius perpendicular chord and get | ||
+ | <cmath>MH = \frac{XH + HY}{2} – HY = 2.</cmath> | ||
+ | We find | ||
+ | <cmath>\sin OAH =\frac{MH}{AH} = \frac{2}{3} \implies \cos OAH = \frac{\sqrt{5}}{3}.</cmath> | ||
+ | We use properties of isosceles <math>\triangle OAP</math> and find <math>\hspace{5mm}R = \frac{AP}{2\cos OAP} = \frac{7}{2\frac {\sqrt{5}}{3}} = \frac{21}{2\sqrt{5}}.</math> | ||
+ | |||
+ | We use <math>OM' = \frac{AH}{2} = \frac {3}{2}</math> and find <math>\hspace{25mm} \frac{BC}{2} = \sqrt{R^2 – OM'^2} = 3 \sqrt {\frac {11}{5}}.</math> | ||
+ | |||
+ | The area of <math>ABC</math> | ||
+ | <cmath>[ABC]=\frac{BC}{2} \cdot (AH + HD) = 3\cdot \sqrt{55} \implies 3+55 = \boldsymbol{\boxed{058}}.</cmath> | ||
+ | '''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 | ||
==See Also== | ==See Also== | ||
{{AIME box|year=2020|n=I|num-b=14|after=Last Problem}} | {{AIME box|year=2020|n=I|num-b=14|after=Last Problem}} | ||
+ | |||
+ | [[Category:Intermediate Geometry Problems]] | ||
{{MAA Notice}} | {{MAA Notice}} |
Latest revision as of 22:45, 27 November 2023
Contents
Problem
Let be an acute triangle with circumcircle
and let
be the intersection of the altitudes of
Suppose the tangent to the circumcircle of
at
intersects
at points
and
with
and
The area of
can be written in the form
where
and
are positive integers, and
is not divisible by the square of any prime. Find
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
is not insignificant; from here, we set
by PoP and trivial construction. Now,
is the reflection of
over
. Note
, and therefore by Pythagorean theorem we have
. Consider
. We have that
, and therefore we are ready to PoP with respect to
. Setting
, we obtain
by PoP on
, and furthermore, we have
. Now, we get
, and from
we take
However, squaring and manipulating with
yields that
and from here, since
we get the area to be
. ~awang11's sol
Solution 1a
As in the diagram, let ray extended hits BC at L and the circumcircle at say
. By power of the point at H, we have
. The three values we are given tells us that
. L is the midpoint of
(see here: https://www.cut-the-knot.org/Curriculum/Geometry/AltitudeAndCircumcircle.shtml ), so
.
As in the diagram provided, let K be the intersection of and
. By power of a point on the circumcircle of triangle
,
. By power of a point on the circumcircle of triangle
,
, thus
. Solving gives
or
.
By the Pythagorean Theorem on triangle ,
. Now continue with solution 1.
Solution 2
Diagram not to scale.
We first observe that , the image of the reflection of
over line
, lies on circle
. This is because
. This is a well known lemma. The result of this observation is that circle
, the circumcircle of
is the image of circle
over line
, which in turn implies that
and thus
is a parallelogram. That
is a parallelogram implies that
is perpendicular to
, and thus divides segment
in two equal pieces,
and
, of length
.
Using Power of a Point,
This means that
and
, where
is the foot of the altitude from
onto
. All that remains to be found is the length of segment
.
Looking at right triangle , we find that
Looking at right triangle
, we get the equation
Plugging in known values, and letting
be the radius of the circle, we find that
Recall that is a parallelogram, so
. So,
, where
is the midpoint of
. This means that
Thus, the area of triangle is
The answer is
.
Solution 3 (Official MAA 1)
Extend to intersect
again at
. The Power of a Point Theorem yields
. Because
, and
and
are both complements to
, it follows that
, implying that
bisects
, so the length of the altitude from
to
is
.
Let the circumcircle of be
. Because
, the two triangles must have the same circumradius. Because the circumcircle of
is
, the circles
and
have the same radius
. Denote the centers of
and
by
and
, respectively, and let
be the midpoint of
. Note that trapezoid
has
. Also
and
. Because
is a translation of
in the direction of
, it follows that
. Finally, the Pythagorean Theorem applied to
yields
. Let
be the projection of
onto
. Then
, so the Pythagorean Theorem applied to
yields
Solving for
gives
. It follows from properties of the orthocenter
that
so
Therefore by the Extended Law of Sines
so
The requested sum is
.
Solution 3a (slightly modified Solution 3)
Note that bisects
. Using the same method from Solution 3 we find
. Let the midpoint of
be
, then by the Pythagorean Theorem we have
, so
. Since
we have that the area of ABC is
so the answer is
~bobjoebilly
Solution 4 (Official MAA 2)
Let be the intersection point of line
and
, noting that
. Because the area of
is
, it suffices to compute
and
separately. As in the previous solution,
. The value of
can be found using the following lemma.
Lemma: Triangle is isosceles with base
.
Proof: Because the circumcircle of ,
, and
have the same radius, there exists a translation
sending the former to the latter. Because
is parallel to the line connecting the centers of the two circles,
must send
to
, meaning
also sends
to the tangent to
at
. But this means that this tangent is parallel to
, which implies the conclusion.
Applying Stewart's Theorem to yields
implying
By the Law of Cosines
so
Let
be the radius of
. By the Extended Law of Sines
Then the solution proceeds as in Solution 3.
Solution 5 (Official MAA 3)
Define points and
as above, and note that
and
. Let the circumcircle of
be
.
Extend past
until it intersects line
at point
. Because line
is a radical axis of
and
, it follows from the Power of a Point Theorem that
from which
. By Pythagorean Theorem
.
Let and
. By the Power of a Point Theorem
On the other hand,
from which
. Therefore
Thus
. Therefore
as above.
Solution 6
Let be circumcenter of
let
be circumradius of
let
be the image of circle
over line
(the circumcircle of
).
Let be the image of the reflection of
over line
lies on circle
Let
be the midpoint of
Then
lies on
(see here: https://brilliant.org/wiki/triangles-orthocenter/ or https://en.wikipedia.org/wiki/Altitude_(triangle) Russian)
lies on
lies on
We use properties of crossing chords and get
We use properties of radius perpendicular chord and get
We find
We use properties of isosceles
and find
We use and find
The area of
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
See Also
2020 AIME I (Problems • Answer Key • Resources) | ||
Preceded by Problem 14 |
Followed by Last Problem | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
All AIME Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.