Difference between revisions of "2020 AIME I Problems/Problem 15"
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== 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 | + | 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== | == Solution 1== | ||
The following is a power of a point solution to this menace of a problem: | 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; | |
− | + | ||
− | pen | + | 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(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(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); | |
− | + | 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(( | + | //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); | |
− | |||
− | dot( | ||
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− | dot | ||
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− | dot | ||
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− | dot | ||
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− | dot | ||
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− | dot | ||
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− | dot( | ||
− | |||
− | dot | ||
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</asy> | </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 | 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== | ==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, 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>. | 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>. | ||
− | By the Pythagorean Theorem, <math>KL=\sqrt{5}</math>. Now continue with solution 1. | + | |
+ | By the Pythagorean Theorem on triangle <math>HKL</math>, <math>KL=\sqrt{5}</math>. Now continue with solution 1. | ||
== Solution 2 == | == Solution 2 == | ||
Line 122: | Line 99: | ||
<cmath>\overline{AD} = \sqrt{\overline{AH}^2 - \overline{HD}^2} = \sqrt{3^2 - 2^2} = \sqrt{5}</cmath> | <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 | Looking at right triangle <math>\triangle ODY</math>, we get the equation | ||
− | <cmath>\overline{OY}^2 - \overline{ | + | <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 | 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> | <cmath>R^2 - 16 = (R - \sqrt{5})^2 = R^2 - 2\sqrt5 R + 5 \Longrightarrow R = \frac{21\sqrt5}{10}</cmath> | ||
Line 130: | Line 107: | ||
Thus, the area of triangle <math>\triangle ABC</math> is | 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} = | + | <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>. | The answer is <math>3 + 55 = \boxed{058}</math>. | ||
− | == Video Solution == | + | ==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 | 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 21:45, 27 November 2023
Contents
[hide]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 thatsoTherefore by the Extended Law of Sinesso 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 yieldsimplying
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
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