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Difference between revisions of "2025 AIME II Problems/Problem 1"

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== Problem ==
 
== Problem ==
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Six points <math>A, B, C, D, E,</math> and <math>F</math> lie in a straight line in that order. Suppose that <math>G</math> is a point not on the line and that <math>AC=26, BD=22, CE=31, DF=33, AF=73, CG=40,</math> and <math>DG=30.</math> Find the area of <math>\triangle BGE.</math>
 
Six points <math>A, B, C, D, E,</math> and <math>F</math> lie in a straight line in that order. Suppose that <math>G</math> is a point not on the line and that <math>AC=26, BD=22, CE=31, DF=33, AF=73, CG=40,</math> and <math>DG=30.</math> Find the area of <math>\triangle BGE.</math>
  
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==Solution 1==
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== Solution 1 ==
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<asy>
 
<asy>
 
pair A,B,C,D,E,F,G;
 
pair A,B,C,D,E,F,G;
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Let <math>AB=a</math>, <math>BC=b</math>, <math>CD=c</math>, <math>DE=d</math> and <math>EF=e</math>. Then we know that <math>a+b+c+d+e=73</math>, <math>a+b=26</math>, <math>b+c=22</math>, <math>c+d=31</math> and <math>d+e=33</math>. From this we can easily deduce <math>c=14</math> and <math>a+e=34</math> thus <math>b+c+d=39</math>. Using Heron's formula we can calculate the area of <math>\triangle{CGD}</math> to be <math>\sqrt{(42)(28)(12)(2)}=168</math>, and since the base of <math>\triangle{BGE}</math> is <math>\frac{39}{14}</math> of that of <math>\triangle{CGD}</math>, we calculate the area of <math>\triangle{BGE}</math> to be <math>168\times \frac{39}{14}=\boxed{468}</math>.
 
Let <math>AB=a</math>, <math>BC=b</math>, <math>CD=c</math>, <math>DE=d</math> and <math>EF=e</math>. Then we know that <math>a+b+c+d+e=73</math>, <math>a+b=26</math>, <math>b+c=22</math>, <math>c+d=31</math> and <math>d+e=33</math>. From this we can easily deduce <math>c=14</math> and <math>a+e=34</math> thus <math>b+c+d=39</math>. Using Heron's formula we can calculate the area of <math>\triangle{CGD}</math> to be <math>\sqrt{(42)(28)(12)(2)}=168</math>, and since the base of <math>\triangle{BGE}</math> is <math>\frac{39}{14}</math> of that of <math>\triangle{CGD}</math>, we calculate the area of <math>\triangle{BGE}</math> to be <math>168\times \frac{39}{14}=\boxed{468}</math>.
  
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~quick Asymptote fix by eevee9406
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~ Quick Asymptote Fix by [[User:Eevee9406|eevee9406]], edited by [[User:Aoum|aoum]]
 +
 
 +
== Solution 2 (Law of Cosines) ==
 +
 
 +
We need to solve for the lengths of <math>AB</math>, <math>BC</math>, <math>CD</math>, <math>DE</math>, and <math>EF</math>. 
 +
Let <math>AB = a</math>, <math>BC = b</math>, <math>CD = c</math>, <math>DE = d</math>, and <math>EF = e</math>. 
 +
We are given the following system of equations:
 +
 
 +
<cmath>a + b = 26, \quad b + c = 22, \quad c + d = 31, \quad d + e = 33, \quad a + b + c + d + e = 73.</cmath>
 +
 
 +
Substituting <math>a + b = 26</math> and <math>d + e = 33</math> into the equation <math>a + b + c + d + e = 73</math>, we get:
 +
 
 +
<cmath>c = 14.</cmath>
 +
 
 +
Thus, we have:
 +
 
 +
<cmath>a = 18, \quad b = 8, \quad c = 14, \quad d = 17, \quad e = 16.</cmath>
 +
 
 +
Next, consider triangle <math>CDG</math>, where <math>CD = 14</math>, <math>CG = 40</math>, and <math>DG = 30</math>. 
 +
By the Law of Cosines, we have:
  
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== Solution 2 (Law of Cosines)==
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<cmath>CD^2 = CG^2 + DG^2 - 2 \times CG \times DG \times \cos(\angle CGD).</cmath>
  
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We need to solve the lengths of AB, BC, CD, DE and EF.
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Substituting the known values:
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Let AB = a, BC = b, CD = c, DE = d and EF = e
 
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a + b = 26, b + c = 22, c + d = 31, d + e = 33, a + b + c + d + e = 73
 
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Substituting a + b = 26 and d + e = 33 into a + b + c + d + e = 73, c = 14
 
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Therefore, we get a = 18, b = 8, c = 14, d = 17 and e = 16
 
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Noting that triangle CDG has CD = 14, CG = 40 and DG = 30,
 
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14^2 = 40^2 + 30^2 - 2 * 40 * 30 * cos(angle CGD)
 
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2 * 40 * 30 * cos(angle CGD) = 2500 - 196 = 2304
 
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cos(angle CGD) = 24/25, therefore sin(angle CGD)
 
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The area for triangle CDG = 1/2 * 40 * 30 * 7/25 = 168
 
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Noting that the height of the triangle CDG would be the exact same height as the triangle as triangle BGE, the ratio of the length would be the ratio of area. Therefore, the answer would be given by 168 * 39 / 14 = 468
 
  
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(Feel free to correct any latex and formats)
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<cmath>14^2 = 40^2 + 30^2 - 2 \times 40 \times 30 \times \cos(\angle CGD).</cmath>
−
~Mitsuihisashi14
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 +
Simplifying:
 +
 
 +
<cmath>196 = 1600 + 900 - 2400 \cos(\angle CGD).</cmath>
 +
 
 +
<cmath>2400 \cos(\angle CGD) = 2500 - 196 = 2304.</cmath>
 +
 
 +
<cmath>\cos(\angle CGD) = \frac{24}{25}.</cmath>
 +
 
 +
Therefore, we can find <math>\sin(\angle CGD)</math> using the identity <math>\sin^2 \theta + \cos^2 \theta = 1</math>:
 +
 
 +
<cmath>\sin(\angle CGD) = \sqrt{1 - \left(\frac{24}{25}\right)^2} = \frac{7}{25}.</cmath>
 +
 
 +
Now, the area of triangle <math>CDG</math> is
 +
 
 +
<cmath>\frac{1}{2} \times 40 \times 30 \times \frac{7}{25} = 168.</cmath>
 +
 
 +
Noting that the height of triangle <math>CDG</math> is the same as the height of triangle <math>BGE</math>, the ratio of the areas of the two triangles will be the same as the ratio of their corresponding lengths. Therefore, the answer is
 +
 
 +
<cmath>\frac{168 \times 39}{14} = \boxed{468}.</cmath>
 +
 
 +
(Feel free to add or correct any LaTeX and formatting)
 +
 
 +
~ Mitsuihisashi14, edited by [[User:Aoum|aoum]]
 +
 
 +
==Video Solution by Mathletes Corner==
 +
 
 +
https://www.youtube.com/watch?v=cLIMq3LBQE0
 +
 
 +
~GP102
 +
 
 +
== See also ==
  
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==See also==
 
 
{{AIME box|year=2025|before=First Problem|num-a=2|n=II}}
 
{{AIME box|year=2025|before=First Problem|num-a=2|n=II}}
−
 
 
{{MAA Notice}}
 
{{MAA Notice}}

Latest revision as of 19:19, 16 April 2025

Problem

Six points $A, B, C, D, E,$ and $F$ lie in a straight line in that order. Suppose that $G$ is a point not on the line and that $AC=26, BD=22, CE=31, DF=33, AF=73, CG=40,$ and $DG=30.$ Find the area of $\triangle BGE.$

Solution 1

[asy] pair A,B,C,D,E,F,G; A=(0,0); label("$A$", A, S); B=(1.5,0); label("$B$", B, S); C=(2.9,0); label("$C$", C, S); D=(4.2,0); label("$D$", D, S); E=(5.3,0); label("$E$", E, S); F=(6.5,0); label("$F$", F, S); G=(3.7,3); label("$G$", G, N); draw(A--B--C--D--E--F); draw(C--G--D); draw(B--G--E); [/asy]

Let $AB=a$, $BC=b$, $CD=c$, $DE=d$ and $EF=e$. Then we know that $a+b+c+d+e=73$, $a+b=26$, $b+c=22$, $c+d=31$ and $d+e=33$. From this we can easily deduce $c=14$ and $a+e=34$ thus $b+c+d=39$. Using Heron's formula we can calculate the area of $\triangle{CGD}$ to be $\sqrt{(42)(28)(12)(2)}=168$, and since the base of $\triangle{BGE}$ is $\frac{39}{14}$ of that of $\triangle{CGD}$, we calculate the area of $\triangle{BGE}$ to be $168\times \frac{39}{14}=\boxed{468}$.

~ Quick Asymptote Fix by eevee9406, edited by aoum

Solution 2 (Law of Cosines)

We need to solve for the lengths of $AB$, $BC$, $CD$, $DE$, and $EF$. Let $AB = a$, $BC = b$, $CD = c$, $DE = d$, and $EF = e$. We are given the following system of equations:

\[a + b = 26, \quad b + c = 22, \quad c + d = 31, \quad d + e = 33, \quad a + b + c + d + e = 73.\]

Substituting $a + b = 26$ and $d + e = 33$ into the equation $a + b + c + d + e = 73$, we get:

\[c = 14.\]

Thus, we have:

\[a = 18, \quad b = 8, \quad c = 14, \quad d = 17, \quad e = 16.\]

Next, consider triangle $CDG$, where $CD = 14$, $CG = 40$, and $DG = 30$. By the Law of Cosines, we have:

\[CD^2 = CG^2 + DG^2 - 2 \times CG \times DG \times \cos(\angle CGD).\]

Substituting the known values:

\[14^2 = 40^2 + 30^2 - 2 \times 40 \times 30 \times \cos(\angle CGD).\]

Simplifying:

\[196 = 1600 + 900 - 2400 \cos(\angle CGD).\]

\[2400 \cos(\angle CGD) = 2500 - 196 = 2304.\]

\[\cos(\angle CGD) = \frac{24}{25}.\]

Therefore, we can find $\sin(\angle CGD)$ using the identity $\sin^2 \theta + \cos^2 \theta = 1$:

\[\sin(\angle CGD) = \sqrt{1 - \left(\frac{24}{25}\right)^2} = \frac{7}{25}.\]

Now, the area of triangle $CDG$ is

\[\frac{1}{2} \times 40 \times 30 \times \frac{7}{25} = 168.\]

Noting that the height of triangle $CDG$ is the same as the height of triangle $BGE$, the ratio of the areas of the two triangles will be the same as the ratio of their corresponding lengths. Therefore, the answer is

\[\frac{168 \times 39}{14} = \boxed{468}.\]

(Feel free to add or correct any LaTeX and formatting)

~ Mitsuihisashi14, edited by aoum

Video Solution by Mathletes Corner

https://www.youtube.com/watch?v=cLIMq3LBQE0

~GP102

See also

2025 AIME II (Problems • Answer Key • Resources)
Preceded by
First Problem 		Followed by
Problem 2
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15
All AIME Problems and Solutions

These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions. AMC logo.png

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