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Difference between revisions of "2024 AMC 8 Problems/Problem 15"

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==Problem==
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==Problem 15==
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Let <math>D</math> be an interior point of the acute triangle <math>ABC</math> with <math>AB > AC</math> so that <math>\angle DAB= \angle CAD</math>. The point <math>E</math> on the segment <math>AC</math> satisfies <math>\angle ADE= \angle BCD</math>, the point <math>F</math> on the segment <math>AB</math> satisfies <math>\angle FDA= \angle DBC</math>, and the point <math>X</math> on the line <math>AC</math> satisfies <math>CX=BX</math>. Let <math>O_1</math> and <math>O_2</math> be the circumcentres of the triangles <math>ADC</math> and <math>EXD</math> respectively. Prove that the lines <math>BC</math>, <math>EF</math>, and <math>O_1 O_2</math> are concurrent. (source: 2021 IMO)
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Let the letters <math>F</math>,<math>L</math>,<math>Y</math>,<math>B</math>,<math>U</math>,<math>G</math> represent distinct digits. Suppose <math>\underline{F}~\underline{L}~\underline{Y}~\underline{F}~\underline{L}~\underline{Y}</math> is the greatest number that satisfies the equation
  
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now go do this problem as a punishment for trying to cheat
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<cmath>8\cdot\underline{F}~\underline{L}~\underline{Y}~\underline{F}~\underline{L}~\underline{Y}=\underline{B}~\underline{U}~\underline{G}~\underline{B}~\underline{U}~\underline{G}.</cmath>
  
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==Solution==
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What is the value of <math>\underline{F}~\underline{L}~\underline{Y}+\underline{B}~\underline{U}~\underline{G}</math>?
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[[File:2021 IMO 3j.png|450px|right]] [[File:2021 IMO 3i.png|450px|right]] We prove that circles <math>ACD, EXD</math> and <math>\Omega_0</math> centered at <math>P</math> (the intersection point <math>BC</math> and <math>EF)</math> have a common chord. Let <math>P</math> be the intersection point of the tangent to the circle <math>\omega_2 = BDC</math> at the point <math>D</math> and the line <math>BC, A'</math> is inverse to <math>A</math> with respect to the circle <math>\Omega_0</math> centered at <math>P</math> with radius <math>PD.</math> Then the pairs of points <math>F</math> and <math>E, B</math> and <math>C</math>  are inverse with respect to <math>\Omega_0</math>, so the points <math>F, E,</math> and <math>P</math> are collinear. Quadrilaterals containing the pairs of inverse points <math>B</math> and <math>C, E</math> and <math>F, A</math> and <math>A'</math> are inscribed, <math>FE</math> is antiparallel to <math>BC</math> with respect to angle <math>A</math> (see <math>\boldsymbol{Claim}</math>). Consider the circles <math>\omega = ACD</math> centered at <math>O_1, \omega' = A'BD,</math> <math>\omega_1 = ABC, \Omega = EXD</math> centered at <math>O_2 , \Omega_1 = A'BX,</math> and <math>\Omega_0.</math> Denote <math>\angle ACB = \gamma</math>. Then <math>\angle BXC =  \angle BXE = \pi – 2\gamma,</math> <math>\angle AA'B = \gamma (AA'CB</math> is cyclic), <math>\angle AA'E =  \pi –  \angle AFE = \pi – \gamma (AA'EF</math> is cyclic, <math>FE</math> is antiparallel), <math>\angle BA'E =  \angle AA'E –  \angle AA'B = \pi – 2\gamma =  \angle BXE \implies</math> <math>\hspace{13mm}E</math> is the point of the circle <math>\Omega_1.</math> Let the point <math>Y</math> be the radical center of the circles <math>\omega, \omega', \omega_1.</math> It has the same power <math>\nu</math> with respect to these circles. The common chords of the pairs of circles <math>A'B, AC, DT,</math> where <math>T = \omega \cap \omega',</math>  intersect at this point. <math>Y</math> has power <math>\nu</math> with respect to <math>\Omega_1</math> since <math>A'B</math> is the radical axis of <math>\omega', \omega_1, \Omega_1.</math> <math>Y</math> has power <math>\nu</math> with respect to <math>\Omega</math> since <math>XE</math> containing <math>Y</math> is the radical axis of <math>\Omega</math> and <math>\Omega_1.</math> Hence <math>Y</math> has power <math>\nu</math> with respect to <math>\omega, \omega', \Omega.</math> Let <math>T'</math> be the point of intersection <math>\omega \cap \Omega.</math> Since the circles <math>\omega</math> and <math>\omega'</math> are inverse with respect to <math>\Omega_0,</math> then <math>T</math> lies on <math>\Omega_0,</math> and <math>P</math> lies on the perpendicular bisector of <math>DT.</math> The power of a point <math>Y</math> with respect to the circles  <math>\omega, \omega',</math> and <math>\Omega</math> are the same, <math>DY \cdot YT = DY \cdot YT' \implies</math> the points <math>T</math> and <math>T'</math> coincide. The centers of the circles  <math>\omega</math> and  <math>\Omega</math> (<math>O_1</math> and <math>O_2</math>) are located on the perpendicular bisector <math>DT'</math>, the point <math>P</math> is located on the perpendicular bisector <math>DT</math> and, therefore, the points <math>P, O_1,</math> and <math>O_2</math> lie on a line, that is, the lines <math>BC, EF,</math> and <math>O_1 O_2</math> are concurrent. [[File:2021 IMO 3.png|450px|right]] [[File:2021 IMO 3j.png|450px|right]] <math>\boldsymbol{Claim}</math> Let <math>AK</math> be bisector of the triangle <math>ABC</math>, point <math>D</math> lies on <math>AK.</math> The point <math>E</math> on the segment <math>AC</math> satisfies <math>\angle ADE= \angle BCD</math>. The point <math>F</math> on the segment <math>AB</math> satisfies <math>\angle ADF= \angle CBD.</math> Let <math>P</math> be the intersection point of the tangent to the circle <math>BDC</math> at the point <math>D</math> and the line <math>BC.</math> Let the circle <math>\Omega_0</math> be centered at <math>P</math> and has the radius <math>PD.</math> Then the pairs of points <math>F</math> and <math>E, B</math> and <math>C</math>  are inverse with respect to <math>\Omega_0</math> and <math>EF</math> and <math>BC</math> are antiparallel with respect to the sides of an angle <math>A.</math> <math>\boldsymbol{Proof}</math> Let the point <math>E'</math> is symmetric to <math>E</math> with respect to  bisector <math>AK, E'L || BC.</math> Symmetry of points <math>E</math> and <math>E'</math> implies <math>\angle AEL = \angle AE'L.</math> <cmath>\angle DCK = \angle E'DL,  \angle DKC = \angle E'LD \implies</cmath> <cmath> \triangle DCK \sim \triangle E'DL \implies \frac {E'L}{KD}=  \frac {DL}{KC}.</cmath> <cmath>\triangle ALE' \sim \triangle AKB \implies \frac {E'L}{BK}=  \frac {AL}{AK}\implies</cmath> <cmath> \frac {AL}{DL} = \frac {AK \cdot DK}{BK \cdot KC}.</cmath> Similarly, we prove that <math>FL</math> and <math>BC</math> are antiparallel with respect to angle <math>A,</math> and the points <math>L</math> in triangles <math>\triangle EDL</math> and <math>\triangle FDL</math> coincide. Hence, <math>FE</math> and <math>BC</math> are antiparallel and <math>BCEF</math> is cyclic. Note that <math>\angle DFE =  \angle DLE –  \angle FDL =  \angle AKC –  \angle CBD</math> and <math>\angle PDE = 180^o –  \angle CDK –  \angle CDP –  \angle LDE = 180^o – (180^o –  \angle AKC –  \angle BCD) –  \angle CBD –  \angle BCD</math> <math>\angle PDE  =  \angle AKC –  \angle CBD = \angle DFE,</math> so <math>PD</math> is tangent to the circle <math>DEF.</math> <math>PD^2 = PC \cdot PB = PE \cdot PF,</math> that is, the points <math>B</math> and <math>C, E</math> and <math>F</math> are inverse with respect to the circle <math>\Omega_0.</math> '''vladimir.shelomovskii@gmail.com, vvsss'''
 
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(This was on the page of the 2021 IMO problem 3) Yes this person actually solved it. I just copy pasted for people that want to know. -Multpi12)
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<math>\textbf{(A)}\ 1089 \qquad \textbf{(B)}\ 1098 \qquad \textbf{(C)}\ 1107 \qquad \textbf{(D)}\ 1116 \qquad \textbf{(E)}\ 1125</math>
 +
 
 +
==Solution 1==
 +
The highest that <math>FLYFLY</math> can be would have to be <math>124124</math>, and it cannot be higher than that, because then it would be <math>125125</math>, and <math>125125</math> multiplied by 8 is <math>1000000</math>, and then it would exceed the <math>6</math> - digit limit set on <math>BUGBUG</math>.
 +
 
 +
So, if we start at <math>124124\cdot8</math>, we get <math>992992</math>, which would be wrong because both <math>B  \&  U</math> would be <math>9</math>, and the numbers cannot be repeated between different letters.
 +
 
 +
If we move on to the next highest, <math>123123</math>, and multiply by <math>8</math>, we get <math>984984</math>. All the digits are different, so <math>FLY+BUG</math> would be <math>123+984</math>, which is <math>1107</math>. So, the answer is <math>\boxed{\textbf{(C)}1107}</math>.
 +
 
 +
- Akhil Ravuri of John Adams Middle School
 +
 
 +
 
 +
- Aryan Varshney of John Adams Middle School (minor edits... props to Akhil for the main/full answer :D)
 +
 
 +
~ cxsmi (minor formatting edits)
 +
 
 +
~Alice of Evergreen Middle School
 +
 
 +
==Solution 2==
 +
 
 +
Notice that <math>\underline{F}~\underline{L}~\underline{Y}~\underline{F}~\underline{L}~\underline{Y} = 1000(\underline{F}~\underline{L}~\underline{Y}) + \underline{F}~\underline{L}~\underline{Y}</math>.
 +
 
 +
Likewise, <math>\underline{B}~\underline{U}~\underline{G}~\underline{B}~\underline{U}~\underline{G} = 1000(\underline{B}~\underline{U}~\underline{G}) + \underline{B}~\underline{U}~\underline{G}</math>.
 +
 
 +
Therefore, we have the following equation:
 +
 
 +
<math>8 \times 1001(\underline{F}~\underline{L}~\underline{Y}) = 1001(\underline{B}~\underline{U}~\underline{G})</math>.
 +
 
 +
Simplifying the equation gives
 +
 
 +
<math>8(\underline{F}~\underline{L}~\underline{Y}) = (\underline{B}~\underline{U}~\underline{G})</math>.
 +
 
 +
We can now use our equation to test each answer choice.
 +
 
 +
We have that <math>123123 \times 8 = 984984</math>, so we can find the sum:
 +
 
 +
<math>\underline{F}~\underline{L}~\underline{Y}+\underline{B}~\underline{U}~\underline{G} = 123 + 984 = 1107</math>.
 +
 
 +
So, the correct answer is <math>\boxed{\textbf{(C)}\ 1107}</math>.
 +
 
 +
- C. Ren
 +
 
 +
==Solution 3 (Answer Choices)==
 +
Note that <math>FLY+BUG = 9 \cdot FLY</math>. Thus, we can check the answer choices and find <math>FLY</math> through each of the answer choices, we find the 1107 works, so the answer is <math>\boxed{\textbf{(C)}\ 1107}</math>.
 +
 
 +
~andliu766
 +
 
 +
 
 +
==Video Solution by Math-X (Apply this simple strategy that works every time!!!)==
 +
https://youtu.be/BaE00H2SHQM?si=8CKIFksKvhOWABN3&t=3485
 +
 
 +
~MATH-x
 +
 
 +
==Video Solution (A Clever Explanation You’ll Get Instantly)==
 +
https://youtu.be/5ZIFnqymdDQ?si=rxqPhk-xiKjmbhNF&t=1683
 +
 
 +
~hsnacademy
 +
 
 +
==Video Solution 2 (easy to digest) by Power Solve==
 +
https://youtu.be/TKBVYMv__Bg
 +
 
 +
==Video Solution 3 (2 minute solve, fast) by MegaMath==
 +
https://www.youtube.com/watch?v=QvJ1b0TzCTc
 +
 
 +
==Video Solution 4 by SpreadTheMathLove==
 +
https://www.youtube.com/watch?v=RRTxlduaDs8
 +
==Video Solution by NiuniuMaths (Easy to understand!)==
 +
https://www.youtube.com/watch?v=V-xN8Njd_Lc
 +
 
 +
~NiuniuMaths
 +
 
 +
== Video Solution by CosineMethod==
 +
 
 +
https://www.youtube.com/watch?v=77UBBu1bKxk
 +
don't recommend but its quite clean, learn what you must- Orion 2010
 +
minor edits by Fireball9746
 +
 
 +
==Video Solution by Interstigation==
 +
https://youtu.be/ktzijuZtDas&t=1585
 +
 
 +
==Video Solution by Dr. David==
 +
https://youtu.be/kMkps16-xwE
 +
 
 +
==Video Solution by WhyMath==
 +
https://youtu.be/GJcAdyDYpQk
 +
 
 +
==See Also==
 +
{{AMC8 box|year=2024|num-b=14|num-a=16}}
 +
{{MAA Notice}}
 +
[[Category:Introductory Number Theory Problems]]

Latest revision as of 06:46, 15 November 2024

Problem 15

Let the letters $F$,$L$,$Y$,$B$,$U$,$G$ represent distinct digits. Suppose $\underline{F}~\underline{L}~\underline{Y}~\underline{F}~\underline{L}~\underline{Y}$ is the greatest number that satisfies the equation

\[8\cdot\underline{F}~\underline{L}~\underline{Y}~\underline{F}~\underline{L}~\underline{Y}=\underline{B}~\underline{U}~\underline{G}~\underline{B}~\underline{U}~\underline{G}.\]

What is the value of $\underline{F}~\underline{L}~\underline{Y}+\underline{B}~\underline{U}~\underline{G}$?

$\textbf{(A)}\ 1089 \qquad \textbf{(B)}\ 1098 \qquad \textbf{(C)}\ 1107 \qquad \textbf{(D)}\ 1116 \qquad \textbf{(E)}\ 1125$

Solution 1

The highest that $FLYFLY$ can be would have to be $124124$, and it cannot be higher than that, because then it would be $125125$, and $125125$ multiplied by 8 is $1000000$, and then it would exceed the $6$ - digit limit set on $BUGBUG$.

So, if we start at $124124\cdot8$, we get $992992$, which would be wrong because both $B \& U$ would be $9$, and the numbers cannot be repeated between different letters.

If we move on to the next highest, $123123$, and multiply by $8$, we get $984984$. All the digits are different, so $FLY+BUG$ would be $123+984$, which is $1107$. So, the answer is $\boxed{\textbf{(C)}1107}$.

- Akhil Ravuri of John Adams Middle School


- Aryan Varshney of John Adams Middle School (minor edits... props to Akhil for the main/full answer :D)

~ cxsmi (minor formatting edits)

~Alice of Evergreen Middle School

Solution 2

Notice that $\underline{F}~\underline{L}~\underline{Y}~\underline{F}~\underline{L}~\underline{Y} = 1000(\underline{F}~\underline{L}~\underline{Y}) + \underline{F}~\underline{L}~\underline{Y}$.

Likewise, $\underline{B}~\underline{U}~\underline{G}~\underline{B}~\underline{U}~\underline{G} = 1000(\underline{B}~\underline{U}~\underline{G}) + \underline{B}~\underline{U}~\underline{G}$.

Therefore, we have the following equation:

$8 \times 1001(\underline{F}~\underline{L}~\underline{Y}) = 1001(\underline{B}~\underline{U}~\underline{G})$.

Simplifying the equation gives

$8(\underline{F}~\underline{L}~\underline{Y}) = (\underline{B}~\underline{U}~\underline{G})$.

We can now use our equation to test each answer choice.

We have that $123123 \times 8 = 984984$, so we can find the sum:

$\underline{F}~\underline{L}~\underline{Y}+\underline{B}~\underline{U}~\underline{G} = 123 + 984 = 1107$.

So, the correct answer is $\boxed{\textbf{(C)}\ 1107}$.

- C. Ren

Solution 3 (Answer Choices)

Note that $FLY+BUG = 9 \cdot FLY$. Thus, we can check the answer choices and find $FLY$ through each of the answer choices, we find the 1107 works, so the answer is $\boxed{\textbf{(C)}\ 1107}$.

~andliu766


Video Solution by Math-X (Apply this simple strategy that works every time!!!)

https://youtu.be/BaE00H2SHQM?si=8CKIFksKvhOWABN3&t=3485

~MATH-x

Video Solution (A Clever Explanation You’ll Get Instantly)

https://youtu.be/5ZIFnqymdDQ?si=rxqPhk-xiKjmbhNF&t=1683

~hsnacademy

Video Solution 2 (easy to digest) by Power Solve

https://youtu.be/TKBVYMv__Bg

Video Solution 3 (2 minute solve, fast) by MegaMath

https://www.youtube.com/watch?v=QvJ1b0TzCTc

Video Solution 4 by SpreadTheMathLove

https://www.youtube.com/watch?v=RRTxlduaDs8

Video Solution by NiuniuMaths (Easy to understand!)

https://www.youtube.com/watch?v=V-xN8Njd_Lc

~NiuniuMaths

Video Solution by CosineMethod

https://www.youtube.com/watch?v=77UBBu1bKxk don't recommend but its quite clean, learn what you must- Orion 2010 minor edits by Fireball9746

Video Solution by Interstigation

https://youtu.be/ktzijuZtDas&t=1585

Video Solution by Dr. David

https://youtu.be/kMkps16-xwE

Video Solution by WhyMath

https://youtu.be/GJcAdyDYpQk

See Also

2024 AMC 8 (Problems • Answer Key • Resources)
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Problem 14 		Followed by
Problem 16
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