Difference between revisions of "2025 AIME II Problems/Problem 1"
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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> | ||
− | == Solution == | + | ==Solution 1== |
+ | <asy> | ||
+ | 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 <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</math>=73, <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}=468</math>. | ||
+ | |||
+ | ==See also== | ||
+ | {{AIME box|year=2025|before=First Problem|num-a=2|n=II}} | ||
+ | |||
+ | {{MAA Notice}} |
Revision as of 22:50, 13 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
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); (Error making remote request. Unknown error_msg)
Let ,
,
,
and
. Then we know that
=73,
,
,
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
.
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.