Difference between revisions of "2016 AMC 10A Problems/Problem 21"
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==Solution== | ==Solution== | ||
+ | <asy> | ||
+ | size(250); | ||
+ | defaultpen(linewidth(0.4)); | ||
+ | //Variable Declarations | ||
+ | pair P,Q,R,Pp,Qp,Rp; | ||
+ | pair A,B; | ||
+ | //Variable Definitions | ||
+ | A=(-5, 0); | ||
+ | B=(8, 0); | ||
+ | P=(-2.828,1); | ||
+ | Q=(0,2); | ||
+ | R=(4.899,3); | ||
+ | Pp=foot(P,A,B); | ||
+ | Qp=foot(Q,A,B); | ||
+ | Rp=foot(R,A,B); | ||
+ | path PQR = P--Q--R--cycle; | ||
+ | //Initial Diagram | ||
+ | dot(P); | ||
+ | dot(Q); | ||
+ | dot(R); | ||
+ | dot(Pp); | ||
+ | dot(Qp); | ||
+ | dot(Rp); | ||
+ | draw(Circle(P, 1), linewidth(0.8)); | ||
+ | draw(Circle(Q, 2), linewidth(0.8)); | ||
+ | draw(Circle(R, 3), linewidth(0.8)); | ||
+ | draw(A--B,Arrows); | ||
+ | label("$P$",P,N); | ||
+ | label("$Q$",Q,N); | ||
+ | label("$R$",R,N); | ||
+ | label("$P'$",Pp,S); | ||
+ | label("$Q'$",Qp,S); | ||
+ | label("$R'$",Rp,S); | ||
+ | label("$l$",B,E); | ||
+ | |||
+ | //Added lines | ||
+ | draw(PQR); | ||
+ | draw(P--Pp); | ||
+ | draw(Q--Qp); | ||
+ | draw(R--Rp); | ||
+ | |||
+ | //Angle marks | ||
+ | draw(rightanglemark(P,Pp,B)); | ||
+ | draw(rightanglemark(Q,Qp,B)); | ||
+ | draw(rightanglemark(R,Rp,B)); | ||
+ | </asy> | ||
Notice that we can find <math>[P'PQRR']</math> in two different ways: <math>[P'PQQ']+[Q'QRR']</math> and <math>[PQR]+[P'PRR']</math>, so <math>[P'PQQ']+[Q'QRR']=[PQR]+[P'PRR']</math> | Notice that we can find <math>[P'PQRR']</math> in two different ways: <math>[P'PQQ']+[Q'QRR']</math> and <math>[PQR]+[P'PRR']</math>, so <math>[P'PQQ']+[Q'QRR']=[PQR]+[P'PRR']</math> | ||
<math>\break</math> | <math>\break</math> |
Revision as of 12:27, 4 February 2016
Circles with centers and , having radii and , respectively, lie on the same side of line and are tangent to at and , respectively, with between and . The circle with center is externally tangent to each of the other two circles. What is the area of triangle ?
Solution
Notice that we can find in two different ways: and , so
Thus, these are equal. . Additionally, . Therefore, . Similarly, . We can calculate easily because . .
Plugging into first equation, the two sums of areas, .
.
See Also
2016 AMC 10A (Problems • Answer Key • Resources) | ||
Preceded by Problem 20 |
Followed by Problem 22 | |
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.