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2025 AIME II Problems/Problem 1: Difference between revisions

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== Problem ==
== Problem ==
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>


==Solution 1==
== Solution 1 ==
 
<asy>
<asy>
pair A,B,C,D,E,F,G;
pair A,B,C,D,E,F,G;
Line 27: Line 29:
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>.


~ Quick Asymptote Fix by [https://artofproblemsolving.com/wiki/index.php/User:Eevee9406 eevee9406]
~ Quick Asymptote Fix by [[User:Eevee9406|eevee9406]], edited by [[User:Aoum|aoum]]
 
~ Edited by [https://artofproblemsolving.com/wiki/index.php/User:Aoum aoum]


== Solution 2 (Law of Cosines) ==
== Solution 2 (Law of Cosines) ==
Line 37: Line 37:
We are given the following system of equations:
We are given the following system of equations:


<cmath>
<cmath>a + b = 26, \quad b + c = 22, \quad c + d = 31, \quad d + e = 33, \quad a + b + c + d + e = 73.</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:
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>
<cmath>c = 14.</cmath>
c = 14.
</cmath>


Thus, we have:
Thus, we have:


<cmath>
<cmath>a = 18, \quad b = 8, \quad c = 14, \quad d = 17, \quad e = 16.</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>.   
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:
By the Law of Cosines, we have:


<cmath>
<cmath>CD^2 = CG^2 + DG^2 - 2 \times CG \times DG \times \cos(\angle CGD).</cmath>
CD^2 = CG^2 + DG^2 - 2 \times CG \times DG \times \cos(\angle CGD).
</cmath>


Substituting the known values:
Substituting the known values:


<cmath>
<cmath>14^2 = 40^2 + 30^2 - 2 \times 40 \times 30 \times \cos(\angle CGD).</cmath>
14^2 = 40^2 + 30^2 - 2 \times 40 \times 30 \times \cos(\angle CGD).
</cmath>


Simplifying:
Simplifying:


<cmath>
<cmath>196 = 1600 + 900 - 2400 \cos(\angle CGD).</cmath>
196 = 1600 + 900 - 2400 \cos(\angle CGD).
</cmath>


<cmath>
<cmath>2400 \cos(\angle CGD) = 2500 - 196 = 2304.</cmath>
2400 \cos(\angle CGD) = 2500 - 196 = 2304.
</cmath>


<cmath>
<cmath>\cos(\angle CGD) = \frac{24}{25}.</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>:
Therefore, we can find <math>\sin(\angle CGD)</math> using the identity <math>\sin^2 \theta + \cos^2 \theta = 1</math>:


<cmath>
<cmath>\sin(\angle CGD) = \sqrt{1 - \left(\frac{24}{25}\right)^2} = \frac{7}{25}.</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:
Now, the area of triangle <math>CDG</math> is


<cmath>
<cmath>\frac{1}{2} \times 40 \times 30 \times \frac{7}{25} = 168.</cmath>
\text{Area of triangle } CDG = \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:
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>
<cmath>\frac{168 \times 39}{14} = \boxed{\textbf{468}}.</cmath>
\frac{168 \times 39}{14} = \boxed{\textbf{468}}.
</cmath>


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


~ Mitsuihisashi14
~ Mitsuihisashi14, edited by [[User:Aoum|aoum]]


~ Edited by [https://artofproblemsolving.com/wiki/index.php/User:Aoum aoum]
== See also ==


==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}}

Revision as of 14:34, 15 February 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

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

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

~ Mitsuihisashi14, edited by aoum

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

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