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

(→‎Solution 2 (Law of Cosines))
(→‎Solution 2 (Law of Cosines))
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~ Edited by [https://artofproblemsolving.com/wiki/index.php/User:Aoum aoum]
 
~ Edited by [https://artofproblemsolving.com/wiki/index.php/User:Aoum aoum]
  
−
== Solution 2 (Law of Cosines)==
+
== Solution 2 (Law of Cosines) ==
  
−
We need to solve the lengths of <math>AB</math>, <math>BC</math>, <math>CD</math>, <math>DE</math> and <math>EF</math>.
+
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>
+
Let <math>AB = a</math>, <math>BC = b</math>, <math>CD = c</math>, <math>DE = d</math>, and <math>EF = e</math>. 
−
a + b = 26, b + c = 22, c + d = 31, d + e = 33, a + b + c + d + e = 73
+
We are given the following system of equations:
−
Substituting a + b = 26 and d + e = 33 into a + b + c + d + e = 73, c = 14
+
 
−
Therefore, we get a = 18, b = 8, c = 14, d = 17 and e = 16
+
<cmath>
−
Noting that triangle CDG has CD = 14, CG = 40 and DG = 30,  
+
a + b = 26, \quad b + c = 22, \quad c + d = 31, \quad d + e = 33, \quad a + b + c + d + e = 73.
−
14^2 = 40^2 + 30^2 - 2 * 40 * 30 * cos(angle CGD)
+
</cmath>
−
2 * 40 * 30 * cos(angle CGD) = 2500 - 196 = 2304
+
 
−
cos(angle CGD) = 24/25, therefore sin(angle CGD)
+
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:
−
The area for triangle CDG = 1/2 * 40 * 30 * 7/25 = 168
+
 
−
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
+
<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:
 +
 
 +
<cmath>
 +
CD^2 = CG^2 + DG^2 - 2 \times CG \times DG \times \cos(\angle CGD).
 +
</cmath>
 +
 
 +
Substituting the known values:
 +
 
 +
<cmath>
 +
14^2 = 40^2 + 30^2 - 2 \times 40 \times 30 \times \cos(\angle CGD).
 +
</cmath>
 +
 
 +
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>
 +
\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:
 +
 
 +
<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.)

Revision as of 13:24, 14 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:

\[\text{Area of triangle } CDG = \frac{1}{2} \times 40 \times 30 \times \frac{7}{25} = 168.\]

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, as requested by Mitsuihisashi14

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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