Difference between revisions of "2000 AMC 12 Problems/Problem 19"
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<cmath>\frac{DE}{BC}[ABC]=\frac{\frac{1}{2}}{14}\cdot 84=3</cmath> | <cmath>\frac{DE}{BC}[ABC]=\frac{\frac{1}{2}}{14}\cdot 84=3</cmath> | ||
Therefore, the answer is <math>\mathrm{C}</math>. | Therefore, the answer is <math>\mathrm{C}</math>. | ||
+ | |||
+ | |||
+ | ==Solution 3== | ||
+ | |||
+ | <asy> | ||
+ | pair A,B,C,D,E; | ||
+ | B=(0,0); | ||
+ | C=(14;0); | ||
+ | A=intersectionpoint(arc(B,13,0,90),arc(C,15,90,180)); | ||
+ | draw(A--B--C--cycle); | ||
+ | D=(7,0); | ||
+ | E=(6.5,0); | ||
+ | draw(A--E); | ||
+ | draw(A--D); | ||
+ | label("$A$",A,N); | ||
+ | label("$B$",B,S); | ||
+ | label("$C$",C,S); | ||
+ | label("$E$",E,S); | ||
+ | label("$D$",D,S); | ||
+ | label("$13$",A--B,NW); | ||
+ | label("$15$",A--C,NE); | ||
+ | label("$14$",B--C,S); | ||
+ | </asy> | ||
+ | |||
+ | Let's find the area of <math>\Delta ABC</math> by Heron, | ||
+ | |||
+ | <math>s=\frac{a+b+c}{2}\\\\s=\frac{14+15+13}{2}\to\boxed{s=21}</math> | ||
+ | |||
+ | Then, | ||
+ | |||
+ | <math>A^2=s(s-a)(s-b)(s-c)\\\\A^2=21(21-14)(21-15)(21-13)\\\\\boxed{A=84}</math> | ||
+ | |||
+ | Knowing that D is the midpoint of BC, thus the area of <math>\Delta ADC=\Delta ABD</math>, which is half the area of <math>\Delta ABC</math>, so <math>\Delta ADC=\Delta ABD=42</math>. | ||
+ | |||
+ | By [[Angle Bisector Theorem]] we know that: | ||
+ | |||
+ | <math>\frac{13}{BE}=\frac{15}{CE}</math> | ||
+ | |||
+ | <math>\boxed{CE=14-BE}</math> | ||
+ | |||
+ | <math>\frac{13}{BE}=\frac{15}{14-BE}</math> | ||
+ | |||
+ | <math>\boxed{BE=6.5}\Rightarrow CE=7.5</math> | ||
+ | |||
+ | Also, we know that: | ||
+ | |||
+ | <math>\frac{A_{\Delta ADE}}{A_{\Delta ABC}}=\frac{ED}{BC}</math> | ||
+ | |||
+ | And, we can easily see that <math>DE=0.5</math>, so, | ||
+ | |||
+ | <math>\frac{A_{\Delta ADE}}{84}=\frac{0.5}{14}</math> | ||
+ | |||
+ | <math>\boxed{A_{\Delta ADE}=3}</math> | ||
== See also == | == See also == |
Revision as of 17:15, 7 August 2017
Problem
In triangle , , , . Let denote the midpoint of and let denote the intersection of with the bisector of angle . Which of the following is closest to the area of the triangle ?
Solution 1
The answer is exactly , choice . We can find the area of triangle by using the simple formula . Dropping an altitude from , we see that it has length ( we can split the large triangle into a and a triangle). Then we can apply the Angle Bisector Theorem on triangle to solve for . Solving , we get that . is the midpoint of so . Thus we get the base of triangle , to be units long. Applying the formula , we get .
Solution 2
The area of is where is the height of triangle . Using Angle Bisector Theorem, we find , which we solve to get . is the midpoint of so . Thus we get the base of triangle , to be units long. We can now use Heron's Formula on . Therefore, the answer is .
Solution 3
pair A,B,C,D,E; B=(0,0); C=(14;0); A=intersectionpoint(arc(B,13,0,90),arc(C,15,90,180)); draw(A--B--C--cycle); D=(7,0); E=(6.5,0); draw(A--E); draw(A--D); label("$A$",A,N); label("$B$",B,S); label("$C$",C,S); label("$E$",E,S); label("$D$",D,S); label("$13$",A--B,NW); label("$15$",A--C,NE); label("$14$",B--C,S); (Error compiling LaTeX. 32c2c358ed0750fe240d2c14d3cb08587d95c9a3.asy: 7.6: syntax error error: could not load module '32c2c358ed0750fe240d2c14d3cb08587d95c9a3.asy')
Let's find the area of by Heron,
Then,
Knowing that D is the midpoint of BC, thus the area of , which is half the area of , so .
By Angle Bisector Theorem we know that:
Also, we know that:
And, we can easily see that , so,
See also
2000 AMC 12 (Problems • Answer Key • Resources) | |
Preceded by Problem 18 |
Followed by Problem 20 |
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 12 Problems and Solutions |
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