Difference between revisions of "2024 AIME II Problems/Problem 10"
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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> | ||
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~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com) | ~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com) | ||
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+ | |||
+ | |||
+ | |||
+ | ==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 | ||
+ | |||
+ | ==Solution 9== | ||
+ | |||
+ | We know that the area of <math>\triangle{ABC}</math> is equal to <math>\frac{abc}{4R}</math>, but is also equal to <math>\frac{a+b+c}{2}r</math>, where R is the circumcircle and r is the incircle. So, <math>abc = 156(a+b+c)</math>. Let's extend <math>AI</math> so it intersects the circumcircle of <math>\triangle{ABC}</math> at <math>P</math>. Something that we see is that <math>\triangle{AIO}</math> is congruent to <math>\triangle{PIO}</math>. Something else that we notice that since <math>AI</math> is the angle bisector of <math>\angle{A}</math>, <math>P</math> is the midpoint of arc <math>BC</math>. Now, let's try calculating <math>AI</math>. By Euler's Theorem, <math>OI^{2} = R^{2} - 2Rr</math> where R is the circumcircle and r is the incircle, so <math>OI = \sqrt{13}</math>. Using Pythagorean Theorem on <math>\triangle{AOI}</math> gives us <math>AI = 3\sqrt{39}</math> as we know that <math>AO</math> is 13. | ||
+ | |||
+ | However, since <math>\triangle{AOI}</math> is congruent to <math>\triangle{POI}</math>, <math>PI = 3\sqrt{39}</math>. Since we know that <math>P</math> is the midpoint of arc <math>BC</math>, we can apply the Incenter-Excenter Lemma to get that <math>BP = 3\sqrt{39}</math> and <math>CP = 3\sqrt{39}</math>. Now, we can use Ptolemy's Theorem on quadrilateral ABPC: | ||
+ | |||
+ | <math>(b+c)(3\sqrt{39}) = a \times 6\sqrt{39}</math> | ||
+ | |||
+ | However, we know that <math>abc = 156(a+b+c)</math>, so we can solve for a! So, <math>abc - 156c = 156a + 156b</math>. Dividing gives us <math>a = \frac{156b + 156c}{bc - 156}</math>. Substituting and cancelling into our equation, | ||
+ | |||
+ | <math>b+c = 2\frac{156b+156c}{bc-156}</math>. | ||
+ | |||
+ | Multiplying, <math>(b+c)(bc-156) = 2 \times 156(b+c).</math> | ||
+ | |||
+ | So, <math>(bc-156)</math> = 312. Our answer is 312 + 156 = <math>\boxed{468}</math>. | ||
+ | |||
+ | ~aleyang | ||
+ | |||
+ | ==Solution 10== | ||
+ | We know by Euler's theorem <math>OI^2=R^2-2Rr.</math> Since <math>AO=R,</math> we have <math>AI=\sqrt{2Rr}.</math> Now, extend <math>AI</math> to meet <math>BC</math> at <math>A'</math> and the circumcircle of <math>\Delta ABC</math> at <math>L.</math> By the Incenter-Excenter lemma, <math>BL=CL=IL=r_a.</math> (Note that <math>OI \perp AL \rightarrow AI=IL=r_a\rightarrow r_a=\sqrt{2Rr}.</math>) Using Ptolemy in the cyclic quadrilateral <math>ABLC,</math> we have <math>c\cdot r_a+b\cdot r_a=2r_a\cdot a \iff \frac{b+c}{a}=2.</math> Also using the angle-bisector theorem we get, <math>\frac{c}{A'B}=\frac{b}{A'C}=\frac{b+c}{a}=2,</math> so call <math>c=2m, b=2n, A'B=m, A'C=n.</math> Since <math>\Delta AA'B \sim \Delta CA'L,</math> <math>\frac{AB}{r_a}=\frac{A'B}{A'L}\rightarrow LA'=\frac{r_a}{2}.</math> Thus, <math>AA'=\frac{3r_a}{2}</math> (as <math>AL=2r_a</math>), and <math>mn=AA'\cdot LA'=\frac{3r_a^2}{4}=\frac{3Rr}{2}.</math> In this problem, we want to find <math>4mn=6Rr,</math> yielding an answer of <math>\boxed{468}.</math> | ||
+ | |||
+ | ~anduran | ||
==Video Solution== | ==Video Solution== |
Latest revision as of 03:22, 1 December 2024
Contents
[hide]- 1 Problem
- 2 Solution 1 (Similar Triangles and PoP)
- 3 Solution 1.1
- 4 Solution 1.2
- 5 Solution 1 (Continued)
- 6 Solution 2 (Excenters)
- 7 Solution 3
- 8 Solution 4 (Trig)
- 9 Solution 5 (Trig)
- 10 Solution 6 (Close to Solution 3)
- 11 Solution 7
- 12 Solution 8
- 13 Solution 9
- 14 Solution 10
- 15 Video Solution
- 16 Video Solution
- 17 See also
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 ,
Thus,
Taking , we get \[ 4 \sin \frac{B}{2} \sin \frac{C}{2} = \cos \frac{B-C}{2} . \]
We have
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
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
~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
-- 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
Solution 9
We know that the area of is equal to , but is also equal to , where R is the circumcircle and r is the incircle. So, . Let's extend so it intersects the circumcircle of at . Something that we see is that is congruent to . Something else that we notice that since is the angle bisector of , is the midpoint of arc . Now, let's try calculating . By Euler's Theorem, where R is the circumcircle and r is the incircle, so . Using Pythagorean Theorem on gives us as we know that is 13.
However, since is congruent to , . Since we know that is the midpoint of arc , we can apply the Incenter-Excenter Lemma to get that and . Now, we can use Ptolemy's Theorem on quadrilateral ABPC:
However, we know that , so we can solve for a! So, . Dividing gives us . Substituting and cancelling into our equation,
.
Multiplying,
So, = 312. Our answer is 312 + 156 = .
~aleyang
Solution 10
We know by Euler's theorem Since we have Now, extend to meet at and the circumcircle of at By the Incenter-Excenter lemma, (Note that ) Using Ptolemy in the cyclic quadrilateral we have Also using the angle-bisector theorem we get, so call Since Thus, (as ), and In this problem, we want to find yielding an answer of
~anduran
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 |
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