Difference between revisions of "2025 AIME II Problems/Problem 1"
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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</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>. | + | 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}=\boxed{468}</math>. |
==See also== | ==See also== | ||
Revision as of 22:51, 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 | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
