Difference between revisions of "2024 AIME II Problems/Problem 10"
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Let <math>\triangle ABC</math> have circumcenter <math>O</math> and incenter <math>I</math> with <math>\overline{IA}\perp\overline{OI}</math>, circumradius <math>13</math>, and inradius <math>6</math>. Find <math>AB\cdot AC</math>. | Let <math>\triangle ABC</math> have circumcenter <math>O</math> and incenter <math>I</math> with <math>\overline{IA}\perp\overline{OI}</math>, circumradius <math>13</math>, and inradius <math>6</math>. Find <math>AB\cdot AC</math>. | ||
− | ==Solution 1 (Similar Triangles)== | + | ==Solution 1 (Similar Triangles and PoP)== |
+ | |||
+ | Start off by (of course) drawing a diagram! Let <math>I</math> and <math>O</math> be the incenter and circumcenters of triangle <math>ABC</math>, respectively. Furthermore, extend <math>AI</math> to meet <math>BC</math> at <math>L</math> and the circumcircle of triangle <math>ABC</math> at <math>D</math>. | ||
+ | |||
+ | <asy> | ||
+ | size(300); | ||
+ | import olympiad; | ||
+ | real c=8.1,a=5*(c+sqrt(c^2-64))/6,b=5*(c-sqrt(c^2-64))/6; | ||
+ | pair B=(0,0),C=(c,0), D = (c/2-0.01, -2.26); | ||
+ | pair A = (c/3,8.65*c/10); | ||
+ | draw(circumcircle(A,B,C)); | ||
+ | pair I=incenter(A,B,C); | ||
+ | pair O=circumcenter(A,B,C); | ||
+ | pair L=extension(A,I,C,B); | ||
+ | dot(I^^O^^A^^B^^C^^D^^L); | ||
+ | draw(A--L); | ||
+ | draw(A--D); | ||
+ | path midangle(pair d,pair e,pair f) {return e--e+((f-e)/length(f-e)+(d-e)/length(d-e))/2;} | ||
+ | draw(C--B--D--cycle); | ||
+ | draw(A--C--B); | ||
+ | draw(A--B); | ||
+ | draw(B--I--C^^A--I); | ||
+ | draw(incircle(A,B,C)); | ||
+ | label("$B$",B,SW); | ||
+ | label("$C$",C,SE); | ||
+ | label("$A$",A,N); | ||
+ | label("$D$",D,S); | ||
+ | label("$I$",I,NW); | ||
+ | label("$L$",L,SW); | ||
+ | label("$O$",O,E); | ||
+ | label("$\alpha$",B,5*dir(midangle(A,B,I)),fontsize(8)); | ||
+ | label("$\alpha$",B,5*dir(midangle(I,B,C)),fontsize(8)); | ||
+ | label("$\beta$",C,12*dir(midangle(B,C,I)),fontsize(8)); | ||
+ | label("$\beta$",C,12*dir(midangle(I,C,A)),fontsize(8)); | ||
+ | label("$\gamma$",A,5*dir(midangle(B,A,I)),fontsize(8)); | ||
+ | label("$\gamma$",A,5*dir(midangle(I,A,C)),fontsize(8)); | ||
+ | |||
+ | draw(I--O); | ||
+ | draw(A--O); | ||
+ | draw(rightanglemark(A,I,O)); | ||
+ | </asy> | ||
+ | |||
+ | |||
+ | We'll tackle the initial steps of the problem in two different manners, both leading us to the same final calculations. | ||
+ | ==Solution 1.1== | ||
+ | Since <math>I</math> is the incenter, <math>\angle BAL \cong \angle DAC</math>. Furthermore, <math>\angle ABC</math> and <math>\angle ADC</math> are both subtended by the same arc <math>AC</math>, so <math>\angle ABC \cong \angle ADC.</math> Therefore by AA similarity, <math>\triangle ABL \sim \triangle ADC</math>. | ||
+ | From this we can say that <cmath>\frac{AB}{AD} = \frac{AL}{AC} \implies AB \cdot AC = AL \cdot AD </cmath> | ||
+ | |||
+ | Since <math>AD</math> is a chord of the circle and <math>OI</math> is a perpendicular from the center to that chord, <math>OI</math> must bisect <math>AD</math>. This can be seen by drawing <math>OD</math> and recognizing that this creates two congruent right triangles. Therefore, <cmath>AD = 2 \cdot ID \implies AB \cdot AC = 2 \cdot AL \cdot ID</cmath> | ||
+ | |||
+ | We have successfully represented <math>AB \cdot AC</math> in terms of <math>AL</math> and <math>ID</math>. Solution 1.2 will explain an alternate method to get a similar relationship, and then we'll rejoin and finish off the solution. | ||
+ | |||
+ | ==Solution 1.2== | ||
+ | |||
+ | <math>\angle ALB \cong \angle DLC</math> by vertical angles and <math>\angle LBA \cong \angle CDA</math> because both are subtended by arc <math>AC</math>. Thus <math>\triangle ABL \sim \triangle CDL</math>. | ||
+ | |||
+ | Thus <cmath>\frac{AB}{CD} = \frac{AL}{CL} \implies AB = CD \cdot \frac{AL}{CL}</cmath> | ||
+ | |||
+ | Symmetrically, we get <math>\triangle ALC \sim \triangle BLD</math>, so | ||
+ | <cmath>\frac{AC}{BD} = \frac{AL}{BL} \implies AC = BD \cdot \frac{AL}{BL}</cmath> | ||
+ | |||
+ | Substituting, we get <cmath>AB \cdot AC = CD \cdot \frac{AL}{CL} \cdot BD \cdot \frac{AL}{BL}</cmath> | ||
+ | |||
+ | Lemma 1: BD = CD = ID | ||
+ | |||
+ | Proof: | ||
+ | |||
+ | We commence angle chasing: we know <math>\angle DBC \cong DAC = \gamma</math>. Therefore <cmath>\angle IBD = \alpha + \gamma</cmath>. | ||
+ | Looking at triangle <math>ABI</math>, we see that <math>\angle IBA = \alpha</math>, and <math>\angle BAI = \gamma</math>. Therefore because the sum of the angles must be <math>180</math>, <math>\angle BIA = 180-\alpha - \gamma</math>. Now <math>AD</math> is a straight line, so <cmath>\angle BID = 180-\angle BIA = \alpha+\gamma</cmath>. | ||
+ | Since <math>\angle IBD = \angle BID</math>, triangle <math>IBD</math> is isosceles and thus <math>ID = BD</math>. | ||
+ | |||
+ | A similar argument should suffice to show <math>CD = ID</math> by symmetry, so thus <math>ID = BD = CD</math>. | ||
+ | |||
+ | Now we regroup and get <cmath>CD \cdot \frac{AL}{CL} \cdot BD \cdot \frac{AL}{BL} = ID^2 \cdot \frac{AL^2}{BL \cdot CL}</cmath> | ||
+ | |||
+ | Now note that <math>BL</math> and <math>CL</math> are part of the same chord in the circle, so we can use Power of a point to express their product differently. <cmath>BL \cdot CL = AL \cdot LD \implies AB \cdot AC = ID^2 \cdot \frac{AL}{LD}</cmath> | ||
+ | |||
+ | ==Solution 1 (Continued)== | ||
+ | Now we have some sort of expression for <math>AB \cdot AC</math> in terms of <math>ID</math> and <math>AL</math>. Let's try to find <math>AL</math> first. | ||
+ | |||
+ | Drop an altitude from <math>D</math> to <math>BC</math>, <math>I</math> to <math>AC</math>, and <math>I</math> to <math>BC</math>: | ||
+ | |||
+ | <asy> | ||
+ | size(300); | ||
+ | import olympiad; | ||
+ | real c=8.1,a=5*(c+sqrt(c^2-64))/6,b=5*(c-sqrt(c^2-64))/6; | ||
+ | pair B=(0,0),C=(c,0), D = (c/2-0.01, -2.26), E = (c/2-0.01,0); | ||
+ | pair A = (c/3,8.65*c/10); | ||
+ | pair F = (2*c/3-0.14, 4-0.29); | ||
+ | pair G = (c/2-0.68,0); | ||
+ | draw(circumcircle(A,B,C)); | ||
+ | pair I=incenter(A,B,C); | ||
+ | pair O=circumcenter(A,B,C); | ||
+ | pair L=extension(A,I,C,B); | ||
+ | dot(I^^O^^A^^B^^C^^D^^L^^E^^F^^G); | ||
+ | draw(A--L); | ||
+ | draw(A--D); | ||
+ | draw(D--E); | ||
+ | draw(I--F); | ||
+ | draw(I--G); | ||
+ | path midangle(pair d,pair e,pair f) {return e--e+((f-e)/length(f-e)+(d-e)/length(d-e))/2;} | ||
+ | draw(C--B--D--cycle); | ||
+ | draw(A--C--B); | ||
+ | draw(A--B); | ||
+ | draw(B--I--C^^A--I); | ||
+ | draw(incircle(A,B,C)); | ||
+ | label("$B$",B,SW); | ||
+ | label("$C$",C,SE); | ||
+ | label("$A$",A,N); | ||
+ | label("$D$",D,S); | ||
+ | label("$I$",I,NW); | ||
+ | label("$L$",L,SW); | ||
+ | label("$O$",O,E); | ||
+ | label("$E$",E,N); | ||
+ | label("$F$",F,NE); | ||
+ | label("$G$",G,SW); | ||
+ | label("$\alpha$",B,5*dir(midangle(A,B,I)),fontsize(8)); | ||
+ | label("$\alpha$",B,5*dir(midangle(I,B,C)),fontsize(8)); | ||
+ | label("$\beta$",C,12*dir(midangle(B,C,I)),fontsize(8)); | ||
+ | label("$\beta$",C,12*dir(midangle(I,C,A)),fontsize(8)); | ||
+ | label("$\gamma$",A,5*dir(midangle(B,A,I)),fontsize(8)); | ||
+ | label("$\gamma$",A,5*dir(midangle(I,A,C)),fontsize(8)); | ||
+ | |||
+ | |||
+ | draw(I--O); | ||
+ | draw(A--O); | ||
+ | draw(rightanglemark(A,I,O)); | ||
+ | draw(rightanglemark(B,E,D)); | ||
+ | draw(rightanglemark(I,F,A)); | ||
+ | draw(rightanglemark(I,G,L)); | ||
+ | </asy> | ||
+ | |||
+ | Since <math>\angle DBE \cong \angle IAF</math> and <math>\angle BED \cong \angle IFA</math>, <math>\triangle BDE \sim \triangle AIF</math>. | ||
+ | |||
+ | Furthermore, we know <math>BD = ID</math> and <math>AI = ID</math>, so <math>BD = AI</math>. Since we have two right similar triangles and the corresponding sides are equal, these two triangles are actually congruent: this implies that <math>DE = IF = 6</math> since <math>IF</math> is the inradius. | ||
+ | |||
+ | Now notice that <math>\triangle IGL \sim \triangle DEL</math> because of equal vertical angles and right angles. Furthermore, <math>IG</math> is the inradius so it's length is <math>6</math>, which equals the length of <math>DE</math>. Therefore these two triangles are congruent, so <math>IL = DL</math>. | ||
+ | |||
+ | Since <math>IL+DL = ID</math>, <math>ID = 2 \cdot IL</math>. Furthermore, <math>AL = AI + IL = ID + IL = 3 \cdot IL</math>. | ||
+ | |||
+ | We can now plug back into our initial equations for <math>AB \cdot AC</math>: | ||
+ | |||
+ | From <math>1.1</math>, <math>AB \cdot AC = 2 \cdot AL \cdot ID = 2 \cdot 3 \cdot IL \cdot 2 \cdot IL</math> | ||
+ | |||
+ | <cmath>\implies AB \cdot AC = 3 \cdot (2 \cdot IL) \cdot (2 \cdot IL) = 3 \cdot ID^2</cmath> | ||
+ | |||
+ | Alternatively, from <math>1.2</math>, <math>AB \cdot AC = ID^2 \cdot \frac{AL}{DL}</math> | ||
+ | <cmath>\implies AB \cdot AC = ID^2 \cdot \frac{3 \cdot IL}{IL} = 3 \cdot ID^2</cmath> | ||
+ | |||
+ | Now all we need to do is find <math>ID</math>. | ||
+ | |||
+ | The problem now becomes very simple if one knows Euler's Formula for the distance between the incenter and the circumcenter of a triangle. This formula states that <math>OI^2 = R(R-2r)</math>, where <math>R</math> is the circumradius and <math>r</math> is the inradius. We will prove this formula first, but if you already know the proof, skip this part. | ||
+ | |||
+ | Theorem: in any triangle, let <math>d</math> be the distance from the circumcenter to the incenter of the triangle. Then <math>d^2 = R \cdot (R-2r)</math>, where <math>R</math> is the circumradius of the triangle and <math>r</math> is the inradius of the triangle. | ||
+ | |||
+ | Proof: | ||
+ | |||
+ | Construct the following diagram: | ||
+ | |||
<asy> | <asy> | ||
− | size( | + | size(300); |
import olympiad; | import olympiad; | ||
real c=8.1,a=5*(c+sqrt(c^2-64))/6,b=5*(c-sqrt(c^2-64))/6; | real c=8.1,a=5*(c+sqrt(c^2-64))/6,b=5*(c-sqrt(c^2-64))/6; | ||
− | pair | + | pair B=(0,0),C=(c,0), D = (c/2-0.01, -2.26), E = (c/2-0.01,0); |
− | pair | + | pair A = (c/3,8.65*c/10); |
+ | pair F = (2*c/3-0.14, 4-0.29); | ||
+ | pair G = (c/2-0.68,0); | ||
+ | draw(circumcircle(A,B,C)); | ||
pair I=incenter(A,B,C); | pair I=incenter(A,B,C); | ||
− | pair L=extension( | + | pair O=circumcenter(A,B,C); |
− | dot(I^^A^^B^^C^^D); | + | pair L=extension(A,I,C,B); |
− | draw( | + | dot(I^^O^^A^^B^^C^^D^^L^^F); |
+ | draw(A--L); | ||
+ | draw(A--D); | ||
+ | draw(I--F); | ||
path midangle(pair d,pair e,pair f) {return e--e+((f-e)/length(f-e)+(d-e)/length(d-e))/2;} | path midangle(pair d,pair e,pair f) {return e--e+((f-e)/length(f-e)+(d-e)/length(d-e))/2;} | ||
− | draw( | + | draw(C--B--D--cycle); |
− | |||
draw(A--C--B); | draw(A--C--B); | ||
− | draw(A | + | draw(A--B); |
+ | draw(A--I); | ||
draw(incircle(A,B,C)); | draw(incircle(A,B,C)); | ||
− | label("$ | + | label("$B$",B,SW); |
− | label("$ | + | label("$C$",C,SE); |
− | label("$ | + | label("$A$",A,N); |
− | label("$D$",D,S | + | label("$D$",D,S); |
− | label("$I$",I, | + | label("$I$",I,NW); |
− | label("$L$",L,SW, | + | label("$L$",L,SW); |
− | label("$\ | + | label("$O$",O,S); |
− | label("$\ | + | label("$F$",F,NE); |
− | + | label("$\gamma$",A,5*dir(midangle(B,A,I)),fontsize(8)); | |
− | label("$ | + | label("$\gamma$",A,5*dir(midangle(I,A,C)),fontsize(8)); |
− | label("$ | + | |
− | + | pair H = (10*c/8-1.46,2*c/3-1.85), J = (-0.55,1.4); | |
+ | dot(H^^J); | ||
+ | label("$H$", H, E); | ||
+ | label("$J$", J, W); | ||
+ | |||
+ | |||
+ | draw(I--O); | ||
+ | draw(I--H); | ||
+ | draw(I--J); | ||
+ | draw(rightanglemark(I,F,A)); | ||
</asy> | </asy> | ||
− | + | ||
+ | Let <math>OI = d</math>, <math>OH = R</math>, <math>IF = r</math>. By the Power of a Point, <math>IH \cdot IJ = AI \cdot ID</math>. | ||
+ | <math>IH = R+d</math> and <math>IJ = R-d</math>, so <cmath>(R+d) \cdot (R-d) = AI \cdot ID = AI \cdot CD</cmath> | ||
+ | |||
+ | Now consider <math>\triangle ACD</math>. Since all three points lie on the circumcircle of <math>\triangle ABC</math>, the two triangles have the same circumcircle. Thus we can apply law of sines and we get <math>\frac{CD}{\sin(\angle DAC)} = 2R</math>. This implies | ||
+ | |||
+ | <cmath>(R+d)\cdot (R-d) = AI \cdot 2R \cdot \sin(\angle DAC)</cmath> | ||
+ | |||
+ | Also, <math>\sin(\angle DAC)) = \sin(\angle IAF))</math>, and <math>\triangle IAF</math> is right. Therefore <cmath>\sin(\angle IAF) = \frac{IF}{AI} = \frac{r}{AI}</cmath> | ||
+ | |||
+ | Plugging in, we have | ||
+ | |||
+ | <cmath>(R+d)\cdot (R-d) = AI \cdot 2R \cdot \frac{r}{AI} = 2R \cdot r</cmath> | ||
+ | |||
+ | Thus <cmath>R^2-d^2 = 2R \cdot r \implies d^2 = R \cdot (R-2r)</cmath> | ||
+ | |||
+ | |||
+ | |||
+ | |||
+ | Now we can finish up our solution. We know that <math>AB \cdot AC = 3 \cdot ID^2</math>. Since <math>ID = AI</math>, <math>AB \cdot AC = 3 \cdot AI^2</math>. Since <math>\triangle AOI</math> is right, we can apply the pythagorean theorem: <math>AI^2 = AO^2-OI^2 = 13^2-OI^2</math>. | ||
+ | |||
+ | Plugging in from Euler's formula, <math>OI^2 = 13 \cdot (13 - 2 \cdot 6) = 13</math>. | ||
+ | |||
+ | Thus <math>AI^2 = 169-13 = 156</math>. | ||
+ | |||
+ | Finally <math>AB \cdot AC = 3 \cdot AI^2 = 3 \cdot 156 = \textbf{468}</math>. | ||
+ | |||
+ | |||
~KingRavi | ~KingRavi | ||
− | ==Solution== | + | ==Solution 2 (Excenters)== |
By Euler's formula <math>OI^{2}=R(R-2r)</math>, we have <math>OI^{2}=13(13-12)=13</math>. Thus, by the Pythagorean theorem, <math>AI^{2}=13^{2}-13=156</math>. Let <math>AI\cap(ABC)=M</math>; notice <math>\triangle AOM</math> is isosceles and <math>\overline{OI}\perp\overline{AM}</math> which is enough to imply that <math>I</math> is the midpoint of <math>\overline{AM}</math>, and <math>M</math> itself is the midpoint of <math>II_{a}</math> where <math>I_{a}</math> is the <math>A</math>-excenter of <math>\triangle ABC</math>. Therefore, <math>AI=IM=MI_{a}=\sqrt{156}</math> and <cmath>AB\cdot AC=AI\cdot AI_{a}=3\cdot AI^{2}=\boxed{468}.</cmath> | By Euler's formula <math>OI^{2}=R(R-2r)</math>, we have <math>OI^{2}=13(13-12)=13</math>. Thus, by the Pythagorean theorem, <math>AI^{2}=13^{2}-13=156</math>. Let <math>AI\cap(ABC)=M</math>; notice <math>\triangle AOM</math> is isosceles and <math>\overline{OI}\perp\overline{AM}</math> which is enough to imply that <math>I</math> is the midpoint of <math>\overline{AM}</math>, and <math>M</math> itself is the midpoint of <math>II_{a}</math> where <math>I_{a}</math> is the <math>A</math>-excenter of <math>\triangle ABC</math>. Therefore, <math>AI=IM=MI_{a}=\sqrt{156}</math> and <cmath>AB\cdot AC=AI\cdot AI_{a}=3\cdot AI^{2}=\boxed{468}.</cmath> | ||
Line 43: | Line 243: | ||
− | ==Solution | + | ==Solution 3== |
Denote <math>AB=a, AC=b, BC=c</math>. By the given condition, <math>\frac{abc}{4A}=13; \frac{2A}{a+b+c}=6</math>, where <math>A</math> is the area of <math>\triangle{ABC}</math>. | Denote <math>AB=a, AC=b, BC=c</math>. By the given condition, <math>\frac{abc}{4A}=13; \frac{2A}{a+b+c}=6</math>, where <math>A</math> is the area of <math>\triangle{ABC}</math>. | ||
− | Moreover, since <math>OI\bot AI</math>, the second intersection of the line <math>AI</math> and <math>(ABC)</math> is the reflection of <math>A</math> about <math>I</math>, denote that as <math>D</math>. By the incenter-excenter lemma, <math>DI=BD=CD=\frac{AD}{2}\implies BD(a+b)=2BD\cdot c\implies a+b=2c</math>. | + | Moreover, since <math>OI\bot AI</math>, the second intersection of the line <math>AI</math> and <math>(ABC)</math> is the reflection of <math>A</math> about <math>I</math>, denote that as <math>D</math>. By the incenter-excenter lemma with Ptolemy's Theorem, <math>DI=BD=CD=\frac{AD}{2}\implies BD(a+b)=2BD\cdot c\implies a+b=2c</math>. |
Thus, we have <math>\frac{2A}{a+b+c}=\frac{2A}{3c}=6, A=9c</math>. Now, we have <math>\frac{abc}{4A}=\frac{abc}{36c}=\frac{ab}{36}=13\implies ab=\boxed{468}</math> | Thus, we have <math>\frac{2A}{a+b+c}=\frac{2A}{3c}=6, A=9c</math>. Now, we have <math>\frac{abc}{4A}=\frac{abc}{36c}=\frac{ab}{36}=13\implies ab=\boxed{468}</math> | ||
Line 53: | Line 253: | ||
~Bluesoul | ~Bluesoul | ||
− | ==Solution | + | ==Solution 4 (Trig)== |
Denote by <math>R</math> and <math>r</math> the circumradius and inradius, respectively. | Denote by <math>R</math> and <math>r</math> the circumradius and inradius, respectively. | ||
First, we have | First, we have | ||
− | \[ | + | <cmath>\[ |
r = 4 R \sin \frac{A}{2} \sin \frac{B}{2} \sin \frac{C}{2} \hspace{1cm} (1) | r = 4 R \sin \frac{A}{2} \sin \frac{B}{2} \sin \frac{C}{2} \hspace{1cm} (1) | ||
− | \] | + | \]</cmath> |
Second, because <math>AI \perp IO</math>, | Second, because <math>AI \perp IO</math>, | ||
Line 119: | Line 319: | ||
~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com) | ~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com) | ||
+ | |||
+ | |||
+ | |||
+ | |||
+ | |||
+ | ==Solution 5 (Trig)== | ||
+ | |||
+ | [[File:2024AIMEIIProblem10.png|450px|center]] | ||
+ | |||
+ | |||
+ | Firstly, we can construct the triangle <math>\triangle ABC</math> by drawing the circumcirlce (centered at <math>O</math> with radius <math>R = OA = 13</math>) and incircle (centered at <math>I</math> with radius <math>r = 6</math>). Next, from <math>A</math>, construct tangent lines to the incircle meeting the circumcirlce at point <math>B</math> and <math>C</math>, say, as shown in the diagram. By Euler's theorem (relating the distance between <math>O</math> and <math>I</math> to the circumradius and inradius), we have | ||
+ | <cmath> | ||
+ | OI = \sqrt{R^2 - 2rR} = \sqrt{13}. | ||
+ | </cmath> | ||
+ | This leads to | ||
+ | <cmath> | ||
+ | AI = \sqrt{R^2 - OI^2} = \sqrt{156}. | ||
+ | </cmath> | ||
+ | Let <math>P</math> be the point of tangency where the incircle meets the side <math>\overline{AC}</math>. Now we denote | ||
+ | <cmath> | ||
+ | \theta \coloneqq \angle BAI = \angle IAP \qquad \text{and} \qquad \phi \coloneqq \angle OAI. | ||
+ | </cmath> | ||
+ | Notice that <math>\angle BAO = \angle BAI - \angle OAI = \theta - \phi</math>. Finally, the crux move is to recognize | ||
+ | <cmath> | ||
+ | AB = 2R \cos(\theta - \phi) \qquad \text{and} \qquad AC = 2R \cos(\theta + \phi) | ||
+ | </cmath> | ||
+ | since <math>O</math> is the circumcenter. Then multiply these two expressions and apply the compound-angle formula to get | ||
+ | \begin{aligned} | ||
+ | AB \cdot AC | ||
+ | &= 4R^2 \cos(\theta - \phi) \cos(\theta + \phi) \\[0.3em] | ||
+ | &= 4R^2\left( | ||
+ | \cos^2\theta \cos^2\phi - \sin^2\theta \sin^2\phi | ||
+ | \right) \\[0.3em] | ||
+ | &= 4\cos^2\theta(\underbrace{R\cos\phi}_{AI \, = \, \sqrt{156}})^2 - 4\sin^2\theta(\underbrace{R\sin\phi}_{OI \, = \, \sqrt{13}})^2 \\[0.3em] | ||
+ | &= 52 (12\cos^2\theta - \sin^2 \theta) \\[0.3em] | ||
+ | AB \cdot AC | ||
+ | &= 52 (12 - 13\sin^2\theta), | ||
+ | \end{aligned} | ||
+ | where in the last equality, we make use of the substitution <math>\cos^2\theta = 1 - \sin^2\theta</math>. Looking at <math>\triangle AIP</math>, we learn that <math>\sin \theta = \frac{r}{AI} = \frac{6}{\sqrt{156}}</math> which means <math>\sin^2 \theta = \frac{3}{13}</math>. Hence we have | ||
+ | <cmath> | ||
+ | AB \cdot AC = 52\left( 12 - 13 \cdot \tfrac{3}{13} \right) = 52 \cdot 9 = \boxed{468}. | ||
+ | </cmath> | ||
+ | This completes the solution | ||
+ | |||
+ | -- VensL. | ||
+ | |||
+ | ==Solution 6 (Close to Solution 3)== | ||
+ | [[File:2024 AIME II 10.png|230px|right]] | ||
+ | Denote <math>E = \odot ABC \cap AI, AB = c, AC = b, BC=a, r</math> is inradius. | ||
+ | <cmath>AO = EO = R \implies AI = EI.</cmath> | ||
+ | It is known that <math>\frac {AI}{EI} = \frac {b+c}{a} – 1 = 1 \implies b + c = 2a.</math> | ||
+ | *[[Barycentric_coordinates | Points on bisectors]] | ||
+ | <cmath>[ABC] =\frac{ (a+b+c) r}{2} = \frac {3ar}{2} = \frac {abc}{4R} \implies bc = 6Rr = \boxed{468}.</cmath> | ||
+ | '''vladimir.shelomovskii@gmail.com, vvsss''' | ||
+ | |||
+ | ==Solution 7== | ||
+ | |||
+ | Call side <math>BC = a</math>, and similarly label the other sides. Note that <math>{OI}^2 = R^2 - 2Rr</math>. Also note that <math>AO = R</math>, so by the right angle, <math>AI = \sqrt{2Rr}</math>. However, we can double Angle Bisector theorem. The length of the angle bisector from A is <math>\sqrt{(bc)(1 - \frac{a^2}{(b+c)^2})}</math>. As a direct result, the length AI simplifies down to <math>\frac{\sqrt{(bc)(b+c-a)}}{\sqrt{{a+b+c}}}</math>. | ||
+ | |||
+ | Draw the incircle and call the tangent to side AB F. Then, <math>AF = \frac{b+c-a}{2}</math>. But this length, by Pythagorean, is <math>\sqrt{120}</math>, so <math>b+c-a = 2\sqrt{120}</math>. | ||
+ | |||
+ | Also note that the area of the triangle is <math>[ABC] = \frac{abc}{52}</math>, by <math>\frac{abc}{4S} = R</math>. By the incircle, we know that <math>\sin{\frac{A}{2}} = \frac{6}{\sqrt{156}}</math>, and similarly, <math>\cos{\frac{A}{2}} = \frac{\sqrt{120}}{\sqrt{156}}</math>. By double-angle, <math>\sin A = \frac{\sqrt{120}}{13}</math>. But the area of the triangle <math>[ABC]</math> is simply <math>\frac{1}{2}bc \sin A</math>, which is also <math>2\sqrt{120}bc</math>. But we know this is <math>abc</math> from above, so <math>a = 2\sqrt{120}</math>. As a direct result, <math>a+b+c = | ||
+ | 6\sqrt{120}</math>. | ||
+ | |||
+ | Apply this to the formula <math>\frac{\sqrt{(bc)(b+c-a)}}{\sqrt{a+b+c}}</math> listed above to get <math>2Rr = 156 = \frac{bc}{3}</math>, so <math>bc = 468</math>. We're done. - sepehr2010 | ||
+ | |||
+ | ==Solution 8== | ||
+ | |||
+ | Let the intersection of the <math>A</math>-angle bisector and the circumcircle be <math>M</math>, and denote the <math>A</math>-excenter as <math>I_A</math>. Denote the tangent to the incircle from <math>AC</math> as <math>E</math> and the tangent to the excircle from <math>AC</math> as <math>E_A</math>. | ||
+ | |||
+ | Notice that our perpendicular condition implies <math>AI = IM</math>, and Incenter-Excenter gives <math>IM = MI_A</math>. Thus we have <math>AI_A = 3AI</math>. From similar triangles we get <math>3(s-a) = 3AE = AE_A = s</math>. This implies <math>a = \frac23 S</math>. | ||
+ | |||
+ | Using areas we have that <math>\frac{abc}{4R} = rs</math>. Substituting gives <math>\frac{sbc}{6R} = rs \implies bc = 6Rr = \boxed{468}</math> and we're done. - thoom | ||
==Video Solution== | ==Video Solution== | ||
Line 126: | Line 399: | ||
~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com) | ~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com) | ||
+ | ==Video Solution== | ||
+ | |||
+ | https://www.youtube.com/watch?v=pPBPfpo12j4 | ||
+ | |||
+ | ~MathProblemSolvingSkills.com | ||
==See also== | ==See also== |
Latest revision as of 16:09, 13 October 2024
Contents
Problem
Let have circumcenter and incenter with , circumradius , and inradius . Find .
Solution 1 (Similar Triangles and PoP)
Start off by (of course) drawing a diagram! Let and be the incenter and circumcenters of triangle , respectively. Furthermore, extend to meet at and the circumcircle of triangle at .
We'll tackle the initial steps of the problem in two different manners, both leading us to the same final calculations.
Solution 1.1
Since is the incenter, . Furthermore, and are both subtended by the same arc , so Therefore by AA similarity, . From this we can say that
Since is a chord of the circle and is a perpendicular from the center to that chord, must bisect . This can be seen by drawing and recognizing that this creates two congruent right triangles. Therefore,
We have successfully represented in terms of and . Solution 1.2 will explain an alternate method to get a similar relationship, and then we'll rejoin and finish off the solution.
Solution 1.2
by vertical angles and because both are subtended by arc . Thus .
Thus
Symmetrically, we get , so
Substituting, we get
Lemma 1: BD = CD = ID
Proof:
We commence angle chasing: we know . Therefore . Looking at triangle , we see that , and . Therefore because the sum of the angles must be , . Now is a straight line, so . Since , triangle is isosceles and thus .
A similar argument should suffice to show by symmetry, so thus .
Now we regroup and get
Now note that and are part of the same chord in the circle, so we can use Power of a point to express their product differently.
Solution 1 (Continued)
Now we have some sort of expression for in terms of and . Let's try to find first.
Drop an altitude from to , to , and to :
Since and , .
Furthermore, we know and , so . Since we have two right similar triangles and the corresponding sides are equal, these two triangles are actually congruent: this implies that since is the inradius.
Now notice that because of equal vertical angles and right angles. Furthermore, is the inradius so it's length is , which equals the length of . Therefore these two triangles are congruent, so .
Since , . Furthermore, .
We can now plug back into our initial equations for :
From ,
Alternatively, from ,
Now all we need to do is find .
The problem now becomes very simple if one knows Euler's Formula for the distance between the incenter and the circumcenter of a triangle. This formula states that , where is the circumradius and is the inradius. We will prove this formula first, but if you already know the proof, skip this part.
Theorem: in any triangle, let be the distance from the circumcenter to the incenter of the triangle. Then , where is the circumradius of the triangle and is the inradius of the triangle.
Proof:
Construct the following diagram:
Let , , . By the Power of a Point, .
and , so
Now consider . Since all three points lie on the circumcircle of , the two triangles have the same circumcircle. Thus we can apply law of sines and we get . This implies
Also, , and is right. Therefore
Plugging in, we have
Thus
Now we can finish up our solution. We know that . Since , . Since is right, we can apply the pythagorean theorem: .
Plugging in from Euler's formula, .
Thus .
Finally .
~KingRavi
Solution 2 (Excenters)
By Euler's formula , we have . Thus, by the Pythagorean theorem, . Let ; notice is isosceles and which is enough to imply that is the midpoint of , and itself is the midpoint of where is the -excenter of . Therefore, and
Note that this problem is extremely similar to 2019 CIME I/14.
Solution 3
Denote . By the given condition, , where is the area of .
Moreover, since , the second intersection of the line and is the reflection of about , denote that as . By the incenter-excenter lemma with Ptolemy's Theorem, .
Thus, we have . Now, we have
~Bluesoul
Solution 4 (Trig)
Denote by and the circumradius and inradius, respectively.
First, we have
Second, because , \begin{align*} AI & = AO \cos \angle IAO \\ & = AO \cos \left( 90^\circ - C - \frac{A}{2} \right) \\ & = AO \sin \left( C + \frac{A}{2} \right) \\ & = R \sin \left( C + \frac{180^\circ - B - C}{2} \right) \\ & = R \cos \frac{B - C}{2} . \end{align*}
Thus, \begin{align*} r & = AI \sin \frac{A}{2} \\ & = R \sin \frac{A}{2} \cos \frac{B-C}{2} \hspace{1cm} (2) \end{align*}
Taking , we get \[ 4 \sin \frac{B}{2} \sin \frac{C}{2} = \cos \frac{B-C}{2} . \]
We have \begin{align*} 2 \sin \frac{B}{2} \sin \frac{C}{2} & = - \cos \frac{B+C}{2} + \cos \frac{B-C}{2} . \end{align*}
Plugging this into the above equation, we get \[ \cos \frac{B-C}{2} = 2 \cos \frac{B+C}{2} . \hspace{1cm} (3) \]
Now, we analyze Equation (2). We have \begin{align*} \frac{r}{R} & = \sin \frac{A}{2} \cos \frac{B-C}{2} \\ & = \sin \frac{180^\circ - B - C}{2} \cos \frac{B-C}{2} \\ & = \cos \frac{B+C}{2} \cos \frac{B-C}{2} \hspace{1cm} (4) \end{align*}
Solving Equations (3) and (4), we get \[ \cos \frac{B+C}{2} = \sqrt{\frac{r}{2R}}, \hspace{1cm} \cos \frac{B-C}{2} = \sqrt{\frac{2r}{R}} . \hspace{1cm} (5) \]
Now, we compute . We have \begin{align*} AB \cdot AC & = 2R \sin C \cdot 2R \sin B \\ & = 2 R^2 \left( - \cos \left( B + C \right) + \cos \left( B - C \right) \right) \\ & = 2 R^2 \left( - \left( 2 \left( \cos \frac{B+C}{2} \right)^2 - 1 \right) + \left( 2 \left( \cos \frac{B-C}{2} \right)^2 - 1 \right) \right) \\ & = 6 R r \\ & = \boxed{\textbf{(468) }} \end{align*} where the first equality follows from the law of sines, the fourth equality follows from (5).
~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com)
Solution 5 (Trig)
Firstly, we can construct the triangle by drawing the circumcirlce (centered at with radius ) and incircle (centered at with radius ). Next, from , construct tangent lines to the incircle meeting the circumcirlce at point and , say, as shown in the diagram. By Euler's theorem (relating the distance between and to the circumradius and inradius), we have
This leads to
Let be the point of tangency where the incircle meets the side . Now we denote
Notice that . Finally, the crux move is to recognize
since is the circumcenter. Then multiply these two expressions and apply the compound-angle formula to get
\begin{aligned}
AB \cdot AC
&= 4R^2 \cos(\theta - \phi) \cos(\theta + \phi) \\[0.3em]
&= 4R^2\left(
\cos^2\theta \cos^2\phi - \sin^2\theta \sin^2\phi
\right) \\[0.3em]
&= 4\cos^2\theta(\underbrace{R\cos\phi}_{AI \, = \, \sqrt{156}})^2 - 4\sin^2\theta(\underbrace{R\sin\phi}_{OI \, = \, \sqrt{13}})^2 \\[0.3em]
&= 52 (12\cos^2\theta - \sin^2 \theta) \\[0.3em]
AB \cdot AC
&= 52 (12 - 13\sin^2\theta),
\end{aligned}
where in the last equality, we make use of the substitution . Looking at , we learn that which means . Hence we have
This completes the solution
-- VensL.
Solution 6 (Close to Solution 3)
Denote is inradius. It is known that
vladimir.shelomovskii@gmail.com, vvsss
Solution 7
Call side , and similarly label the other sides. Note that . Also note that , so by the right angle, . However, we can double Angle Bisector theorem. The length of the angle bisector from A is . As a direct result, the length AI simplifies down to .
Draw the incircle and call the tangent to side AB F. Then, . But this length, by Pythagorean, is , so .
Also note that the area of the triangle is , by . By the incircle, we know that , and similarly, . By double-angle, . But the area of the triangle is simply , which is also . But we know this is from above, so . As a direct result, .
Apply this to the formula listed above to get , so . We're done. - sepehr2010
Solution 8
Let the intersection of the -angle bisector and the circumcircle be , and denote the -excenter as . Denote the tangent to the incircle from as and the tangent to the excircle from as .
Notice that our perpendicular condition implies , and Incenter-Excenter gives . Thus we have . From similar triangles we get . This implies .
Using areas we have that . Substituting gives and we're done. - thoom
Video Solution
~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com)
Video Solution
https://www.youtube.com/watch?v=pPBPfpo12j4
~MathProblemSolvingSkills.com
See also
2024 AIME II (Problems • Answer Key • Resources) | ||
Preceded by Problem 9 |
Followed by Problem 11 | |
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.