Difference between revisions of "2025 AIME II Problems/Problem 1"
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== Solution 2 (Law of Cosines)== | == Solution 2 (Law of Cosines)== | ||
| − | We need to solve the lengths of AB, BC, CD, DE and EF. | + | We need to solve the lengths of <math>AB</math>, <math>BC</math>, <math>CD</math>, <math>DE</math> and <math>EF</math>. |
| − | Let AB = a, BC = b, CD = c, DE = d and EF = e | + | 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 | a + b = 26, b + c = 22, c + d = 31, d + e = 33, a + b + c + d + e = 73 | ||
Substituting a + b = 26 and d + e = 33 into a + b + c + d + e = 73, c = 14 | Substituting a + b = 26 and d + e = 33 into a + b + c + d + e = 73, c = 14 | ||
| Line 45: | Line 45: | ||
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 | 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 | ||
| − | (Feel free to correct any | + | (Feel free to add or correct any LATEX and formatting.) |
| − | ~Mitsuihisashi14 | + | |
| + | ~ Mitsuihisashi14 | ||
==See also== | ==See also== | ||
Revision as of 13:11, 14 February 2025
Problem
Six points 
and 
lie in a straight line in that order. Suppose that 
is a point not on the line and that 
and 
Find the area of 
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]](http://latex.artofproblemsolving.com/b/b/a/bba5a80ea1caae7fb0677811375ba2bbe0e7f2f7.png)
Let 
, 
, 
, 
and 
. Then we know that 
, 
, 
, 
and 
. From this we can easily deduce 
and 
thus 
. Using Heron's formula we can calculate the area of 
to be 
, and since the base of 
is 
of that of , we calculate the area of to be 
.
~ Quick Asymptote Fix by eevee9406
~ Edited by aoum
Solution 2 (Law of Cosines)
We need to solve the lengths of 
, 
, 
, 
and 
.
Let 
, 
, 
, 
, and 
a + b = 26, b + c = 22, c + d = 31, d + e = 33, a + b + c + d + e = 73
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
Noting that triangle CDG has CD = 14, CG = 40 and DG = 30,
14^2 = 40^2 + 30^2 - 2 * 40 * 30 * cos(angle CGD)
2 * 40 * 30 * cos(angle CGD) = 2500 - 196 = 2304
cos(angle CGD) = 24/25, therefore sin(angle CGD)
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
(Feel free to add or correct any LATEX and formatting.)
~ 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 | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
