Difference between revisions of "2004 AMC 10B Problems/Problem 20"
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− | In <math>\triangle ABC</math> points <math>D</math> and <math>E</math> lie on <math>BC</math> and <math>AC</math>, respectively. If <math>AD</math> and <math>BE</math> intersect at <math>T</math> so that <math>AT | + | In <math>\triangle ABC</math> points <math>D</math> and <math>E</math> lie on <math>BC</math> and <math>AC</math>, respectively. If <math>AD</math> and <math>BE</math> intersect at <math>T</math> so that <math>\frac{AT}{DT}=3</math> and <math>\frac{BT}{ET}=4</math>, what is <math>\frac{CD}{BD}</math>? |
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</asy> | </asy> | ||
− | == Solution | + | == Solution 1a (Triangle Areas) == |
We use the square bracket notation <math>[\cdot]</math> to denote area. | We use the square bracket notation <math>[\cdot]</math> to denote area. | ||
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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> | ||
+ | |||
+ | == Solution 1b (Triangle Areas, Alternate Approach) == | ||
+ | |||
+ | We observe that <math>\frac{BC}{CD} = \frac{[BAE]}{[DAE]}</math>. The proof is that if <math>B_H</math> and <math>D_H</math> are the feet of the altitudes from <math>B</math> and <math>D</math>, respectively, to <math>AC</math>, then both sides of that equation are equal to <math>\frac{BB_H}{DD_H}</math>. | ||
+ | |||
+ | From there, we can easily finish: | ||
+ | <cmath>\frac{BC}{CD} = \frac{[BAE]}{[DAE]} = \frac{[BAT]\cdot\frac{5}{4}}{[EAT]\cdot \frac{4}{3}} = 4\cdot\frac{5}{4}\cdot\frac{3}{4} = \frac{15}{4}</cmath> | ||
+ | and thus <math>\frac{CD}{BD} = \boxed{\textbf{(D)}\frac 4{11}}</math>. | ||
== Solution 2 (Mass points) == | == Solution 2 (Mass points) == | ||
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\frac{CD}{BD} = \frac{15/11 - 1}{1 - 0} = \boxed{\textbf{(D)}\frac 4{11}} | \frac{CD}{BD} = \frac{15/11 - 1}{1 - 0} = \boxed{\textbf{(D)}\frac 4{11}} | ||
</cmath> | </cmath> | ||
+ | == Solution 4 (Menelaus) == | ||
+ | By Menelaus on triangle <math>BDT</math>, we have that <cmath>\begin{align*}1&=\dfrac{DA}{TA}\cdot\dfrac{TE}{BE}\cdot\dfrac{BC}{DC} \\ &= \dfrac43 \cdot \dfrac 15 \cdot \dfrac {BC}{DC},\end{align*}</cmath>giving <math>\dfrac{BC}{DC} = \dfrac{15}{4}</math>. Therefore, <math>\dfrac{CD}{BD} = \boxed{\textbf{(D)}\dfrac{4}{11}}</math>. | ||
== See also == | == See also == |
Latest revision as of 00:27, 2 January 2020
Contents
Problem
In points and lie on and , respectively. If and intersect at so that and , what is ?
Solution 1a (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 1b (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 2 (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 3 (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 4 (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 |
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