Difference between revisions of "2024 AIME II Problems/Problem 10"
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(→Solution 5 (Trig)) |
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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. | 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> and <math>I</math> to <math> | + | 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> | <asy> | ||
Line 91: | Line 91: | ||
pair A = (c/3,8.65*c/10); | pair A = (c/3,8.65*c/10); | ||
pair F = (2*c/3-0.14, 4-0.29); | pair F = (2*c/3-0.14, 4-0.29); | ||
+ | pair G = (c/2-0.68,0); | ||
draw(circumcircle(A,B,C)); | draw(circumcircle(A,B,C)); | ||
pair I=incenter(A,B,C); | pair I=incenter(A,B,C); | ||
pair O=circumcenter(A,B,C); | pair O=circumcenter(A,B,C); | ||
pair L=extension(A,I,C,B); | pair L=extension(A,I,C,B); | ||
− | dot(I^^O^^A^^B^^C^^D^^L^^E^^F); | + | dot(I^^O^^A^^B^^C^^D^^L^^E^^F^^G); |
draw(A--L); | draw(A--L); | ||
draw(A--D); | draw(A--D); | ||
draw(D--E); | 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;} | 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(C--B--D--cycle); | ||
Line 114: | Line 117: | ||
label("$E$",E,N); | label("$E$",E,N); | ||
label("$F$",F,NE); | 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(A,B,I)),fontsize(8)); | ||
label("$\alpha$",B,5*dir(midangle(I,B,C)),fontsize(8)); | label("$\alpha$",B,5*dir(midangle(I,B,C)),fontsize(8)); | ||
Line 126: | Line 130: | ||
draw(rightanglemark(A,I,O)); | draw(rightanglemark(A,I,O)); | ||
draw(rightanglemark(B,E,D)); | draw(rightanglemark(B,E,D)); | ||
+ | draw(rightanglemark(I,F,A)); | ||
+ | draw(rightanglemark(I,G,L)); | ||
</asy> | </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> | ||
+ | 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^^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;} | ||
+ | draw(C--B--D--cycle); | ||
+ | draw(A--C--B); | ||
+ | draw(A--B); | ||
+ | draw(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,S); | ||
+ | label("$F$",F,NE); | ||
+ | label("$\gamma$",A,5*dir(midangle(B,A,I)),fontsize(8)); | ||
+ | label("$\gamma$",A,5*dir(midangle(I,A,C)),fontsize(8)); | ||
+ | |||
+ | 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> | ||
+ | |||
+ | |||
+ | 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 141: | 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>. | ||
Line 151: | 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 217: | 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. | ||
==Video Solution== | ==Video Solution== | ||
Line 224: | Line 371: | ||
~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 12:31, 18 July 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,
.
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.
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.