Difference between revisions of "2005 AMC 10A Problems/Problem 25"
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D = (6*A + 19*B)/25; | D = (6*A + 19*B)/25; | ||
E = (28*A + 14*C)/42; | E = (28*A + 14*C)/42; | ||
− | F = (4 | + | F = (4*E + 24*C)/28; |
+ | |||
draw(A--B--C--cycle); | draw(A--B--C--cycle); | ||
draw(D--E); | draw(D--E); |
Revision as of 22:37, 28 December 2019
Contents
Problem
In we have
,
, and
. Points
and
are on
and
respectively, with
and
. What is the ratio of the area of triangle
to the area of the quadrilateral
?
Solution 1(no trig)
We have that
But , so
Solution 2(no trig)
We can let .
Since
,
.
So,
.
This means that
.
Thus,
-Conantwiz2023
Solution 3(trig)
Using this formula:
Since the area of is equal to the area of
minus the area of
,
.
Therefore, the desired ratio is
Note: was not used in this problem
Solution 4
Diagram borrowed from Solution 1.
unitsize(0.15 cm); pair A, B, C, D, E; A = (191/39,28*sqrt(1166)/39); B = (0,0); C = (39,0); D = (6*A + 19*B)/25; E = (28*A + 14*C)/42; F = (4*E + 24*C)/28; draw(A--B--C--cycle); draw(D--E); draw(B--F); label("$A$", A, N); label("$B$", B, SW); label("$C$", C, SE); label("$D$", D, W); label("$E$", E, NE); label("$19$", (A + D)/2, W); label("$6$", (B + D)/2, W); label("$14$", (A + E)/2, NE); label("$28$", (C + E)/2, NE); (Error making remote request. Unknown error_msg)
See also
2005 AMC 10A (Problems • Answer Key • Resources) | ||
Preceded by Problem 24 |
Followed by Last Problem | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 | ||
All AMC 10 Problems and Solutions |
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