Difference between revisions of "2004 AMC 10B Problems/Problem 20"
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<asy> | <asy> | ||
− | unitsize( | + | unitsize(1.5 cm); |
− | + | ||
− | + | pair A, B, C, D, E, F, T; | |
+ | |||
+ | A = (0,0); | ||
+ | B = (3,3); | ||
+ | C = (4.5,0); | ||
+ | D = (2*C + B)/3; | ||
+ | E = (5*C + 2*A)/7; | ||
+ | T = extension(A,D,B,E); | ||
+ | F = extension(D, D + A - C, B, E); | ||
+ | |||
draw(A--B--C--cycle); | draw(A--B--C--cycle); | ||
draw(A--D); | draw(A--D); | ||
draw(B--E); | draw(B--E); | ||
− | + | ||
− | label("$A$",A,SW); | + | |
− | label("$B$",B,N); | + | label("$A$", A, SW); |
− | label("$C$",C,SE); | + | label("$B$", B, N); |
− | label("$D$",D,NE); | + | label("$C$", C, SE); |
− | label("$E$",E,S); | + | label("$D$", D, NE); |
− | label("$T$",T, | + | label("$E$", E, S); |
+ | label("$T$", T, SE); | ||
+ | |||
</asy> | </asy> | ||
+ | |||
+ | |||
+ | <math> \mathrm{(A) \ } \frac{1}{8} \qquad \mathrm{(B) \ } \frac{2}{9} \qquad \mathrm{(C) \ } \frac{3}{10} \qquad \mathrm{(D) \ } \frac{4}{11} \qquad \mathrm{(E) \ } \frac{5}{12} </math> | ||
== Solution 1 (Triangle Areas) == | == Solution 1 (Triangle Areas) == | ||
Line 30: | Line 43: | ||
WLOG, we can assume <math>[\triangle BTD] = 1</math>. Then <math>[\triangle BTA] = 3</math>, and <math>[\triangle ATE] = 3/4</math>. We have <math>CD/BD = [\triangle ACD]/[\triangle ABD]</math>, so we need to find the area of quadrilateral <math>TDCE</math>. | WLOG, we can assume <math>[\triangle BTD] = 1</math>. Then <math>[\triangle BTA] = 3</math>, and <math>[\triangle ATE] = 3/4</math>. We have <math>CD/BD = [\triangle ACD]/[\triangle ABD]</math>, so we need to find the area of quadrilateral <math>TDCE</math>. | ||
− | Draw the line segment <math>TC</math> to form the two triangles <math>\triangle TDC</math> and <math>\triangle TEC</math>. Let <math>x = [\triangle TDC]</math>, and <math>y = [\triangle TEC]</math>. By considering triangles <math>\triangle BTC</math> and <math>\triangle ETC</math>, we obtain <math>(1+x)/y | + | Draw the line segment <math>TC</math> to form the two triangles <math>\triangle TDC</math> and <math>\triangle TEC</math>. Let <math>x = [\triangle TDC]</math>, and <math>y = [\triangle TEC]</math>. By considering triangles <math>\triangle BTC</math> and <math>\triangle ETC</math>, we obtain <math>(1+x)/4=y</math>, and by considering triangles <math>\triangle ATC</math> and <math>\triangle DTC</math>, we obtain <math>(3/4+y)/3=x</math>. Solving, we get <math>x=4/11</math>, <math>y=15/44</math>, so the area of quadrilateral <math>TDEC</math> is <math>x+y=31/44</math>. |
Therefore <math>\frac{CD}{BD}=\frac{\frac{3}{4}+\frac{31}{44}}{3+1}=\boxed{\textbf{(D)} \frac{4}{11}}</math> | Therefore <math>\frac{CD}{BD}=\frac{\frac{3}{4}+\frac{31}{44}}{3+1}=\boxed{\textbf{(D)} \frac{4}{11}}</math> |
Latest revision as of 17:59, 29 July 2024
Contents
Problem
In points
and
lie on
and
, respectively. If
and
intersect at
so that
and
, what is
?
Solution 1 (Triangle Areas)
We use the square bracket notation to denote area.
WLOG, we can assume . Then
, and
. We have
, so we need to find the area of quadrilateral
.
Draw the line segment to form the two triangles
and
. Let
, and
. By considering triangles
and
, we obtain
, and by considering triangles
and
, we obtain
. Solving, we get
,
, so the area of quadrilateral
is
.
Therefore
Solution 2 (Triangle Areas, Alternate Approach)
We observe that . The proof is that if
and
are the feet of the altitudes from
and
, respectively, to
, then both sides of that equation are equal to
.
From there, we can easily finish:
and thus
.
Solution 3 (Mass points)
The presence of only ratios in the problem essentially cries out for mass points.
As per the problem, we assign a mass of to point
, and a mass of
to
. Then, to balance
and
on
,
has a mass of
.
Now, were we to assign a mass of to
and a mass of
to
, we'd have
. Scaling this down by
(to get
, which puts
and
in terms of the masses of
and
), we assign a mass of
to
and a mass of
to
.
Now, to balance and
on
, we must give
a mass of
.
Finally, the ratio of to
is given by the ratio of the mass of
to the mass of
, which is
.
Solution 4 (Coordinates)
Affine transformations preserve ratios of distances, and for any pair of triangles, there is an affine transformation that maps the first one onto the second one. This is why the answer is the same for any , and we just need to compute it for any single triangle.
We can choose the points ,
, and
. This way we will have
, and
. The situation is
shown in the picture below:
The point is the intersection of the lines
and
. The points on the first line have the form
, the points on the second line have the form
. Solving for
we get
, hence
.
The ratio can now be computed simply by observing the
coordinates of
,
, and
:
Solution 5 (Menelaus)
By Menelaus on triangle , we have that
giving
. Therefore,
.
See also
2004 AMC 10B (Problems • Answer Key • Resources) | ||
Preceded by Problem 19 |
Followed by Problem 21 | |
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.