Difference between revisions of "2016 AMC 10A Problems/Problem 21"
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We solve the nested root. We want to turn <math>8 - 4\sqrt{3}</math> into the square of something. If we have <math>(a - b) ^ 2 = 8 - 4\sqrt{3}</math>, then we get <cmath>\begin{cases} a^2 + b^2 = 8 \\ ab = 2\sqrt{3} \end{cases}</cmath> Solving the system of equations, we get <math>a = \sqrt{6}</math> and <math>b = \sqrt{2}</math>. Alternatively, you can square all the possible solutions until you find one that is equal to <math>8 - 4\sqrt{3}</math>. <cmath>A = \sqrt{8 - 4\sqrt{3}} = \sqrt{(\sqrt{6} - \sqrt{2})^2} = \sqrt{6} - \sqrt{2} \rightarrow \fbox{D}</cmath> | We solve the nested root. We want to turn <math>8 - 4\sqrt{3}</math> into the square of something. If we have <math>(a - b) ^ 2 = 8 - 4\sqrt{3}</math>, then we get <cmath>\begin{cases} a^2 + b^2 = 8 \\ ab = 2\sqrt{3} \end{cases}</cmath> Solving the system of equations, we get <math>a = \sqrt{6}</math> and <math>b = \sqrt{2}</math>. Alternatively, you can square all the possible solutions until you find one that is equal to <math>8 - 4\sqrt{3}</math>. <cmath>A = \sqrt{8 - 4\sqrt{3}} = \sqrt{(\sqrt{6} - \sqrt{2})^2} = \sqrt{6} - \sqrt{2} \rightarrow \fbox{D}</cmath> | ||
~ZericH | ~ZericH | ||
+ | |||
+ | ==Solution 5== | ||
+ | <asy> | ||
+ | // Initial Pen Sizing | ||
+ | size(250); | ||
+ | defaultpen(linewidth(0.4)); | ||
+ | defaultpen(fontsize(10pt)); | ||
+ | |||
+ | // Variable Declarations | ||
+ | pair P,Q,R,Pp,Qp,Rp,X,Y,Z,A,B; | ||
+ | |||
+ | // Variable Definitions | ||
+ | A=(-5, 0); | ||
+ | B=(8, 0); | ||
+ | P=(-2.828,1); | ||
+ | Q=(0,2); | ||
+ | R=(4.899,3); | ||
+ | X=(0,1); | ||
+ | Y=(4.899,1); | ||
+ | Z=(4.899,2); | ||
+ | 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); | ||
+ | dot(X); | ||
+ | dot(Y); | ||
+ | dot(Z); | ||
+ | draw(Circle(P, 1), linewidth(0.3)); | ||
+ | draw(Circle(Q, 2), linewidth(0.3)); | ||
+ | draw(Circle(R, 3), linewidth(0.3)); | ||
+ | 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); | ||
+ | label("$X$",X,NE); | ||
+ | label("$Y$",Y,E); | ||
+ | label("$Z$",Z,E); | ||
+ | |||
+ | //Added lines | ||
+ | filldraw(PQR,gray(0.8)); | ||
+ | draw(P--Pp,linetype("8 8")); | ||
+ | draw(Q--Qp,linetype("8 8")); | ||
+ | draw(R--Rp,linetype("8 8")); | ||
+ | draw(P--Y,linetype("8 8")); | ||
+ | draw(Q--Z,linetype("8 8")); | ||
+ | |||
+ | //Angle marks | ||
+ | draw(rightanglemark(R,Y,P)); | ||
+ | draw(rightanglemark(Q,X,P)); | ||
+ | draw(rightanglemark(R,Z,Q)); | ||
+ | |||
+ | //Length labeling | ||
+ | label("$2\sqrt{2}$",P--X,fontsize(8pt)); | ||
+ | label("$2\sqrt{6}$",X--Y,fontsize(8pt)); | ||
+ | label("$2\sqrt{6}$",Q--Z,fontsize(8pt)); | ||
+ | label("$1$",R--Z,E,fontsize(8pt)); | ||
+ | label("$1$",Z--Y,E,fontsize(8pt)); | ||
+ | label("$3$",(-2.828,1.3)--Q,W,fontsize(8pt)); | ||
+ | label("$5$",Q--R,N,fontsize(8pt)); | ||
+ | </asy> | ||
+ | The above diagram can be achieved relatively simply using basic knowledge of the Pythagorean theorem and the fact that the radius from the center to the point of tangency is perpendicular to the tangent line. From there, observe that <math>[PQRY]</math> can be calculated in two ways: <math>[\triangle PQX] + [QZYX] + [\triangle QRZ]</math> and <math>[\triangle PRY] + [\triangle PQR]</math>. Solving, we get: | ||
+ | <cmath> | ||
+ | \begin{aligned} | ||
+ | [\triangle PQR] &= [PQRY] - [\triangle PRY] \\ &= [\triangle PQX] + [QZYX] + [\triangle QRZ] - [\triangle PRY] \\ &= \sqrt{2} + 2\sqrt{6} + \sqrt{6}- 2\sqrt{2}-2\sqrt{6} \\ &= \boxed{\textbf{(D)} \sqrt{6}-\sqrt{2}} | ||
+ | \end{aligned} | ||
+ | </cmath> | ||
==See Also== | ==See Also== |
Revision as of 23:01, 12 June 2020
Problem
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 1
Notice that we can find in two different ways: and , so
. Additionally, . Therefore, . Similarly, . We can calculate easily because . .
Plugging into first equation, the two sums of areas, .
.
Solution 2
Use the Shoelace Theorem.
Let the center of the first circle of radius 1 be at .
Draw the trapezoid and using the Pythagorean Theorem, we get that so the center of the second circle of radius 2 is at .
Draw the trapezoid and using the Pythagorean Theorem, we get that so the center of the third circle of radius 3 is at .
Now, we may use the Shoelace Theorem!
.
Solution 3
and because they are the sum of two radii. and , the difference of the radii. Using pythagorean theorem, we find that and are and , .
Draw a perpendicular from to line , then we can use the Pythagorean theorem to find . . We get
To make our calculations easier, let . The semi-perimeter of our triangle is . Symbolize the area of the triangle with . Using Heron's formula, we have We can remove the outer root of a.
We solve the nested root. We want to turn into the square of something. If we have , then we get Solving the system of equations, we get and . Alternatively, you can square all the possible solutions until you find one that is equal to . ~ZericH
Solution 5
The above diagram can be achieved relatively simply using basic knowledge of the Pythagorean theorem and the fact that the radius from the center to the point of tangency is perpendicular to the tangent line. From there, observe that can be calculated in two ways: and . Solving, we get:
\begin{aligned} [\triangle PQR] &= [PQRY] - [\triangle PRY] \\ &= [\triangle PQX] + [QZYX] + [\triangle QRZ] - [\triangle PRY] \\ &= \sqrt{2} + 2\sqrt{6} + \sqrt{6}- 2\sqrt{2}-2\sqrt{6} \\ &= \boxed{\textbf{(D)} \sqrt{6}-\sqrt{2}} \end{aligned} (Error compiling LaTeX. ! Package amsmath Error: \begin{aligned} allowed only in math mode.)
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 |
2016 AMC 12A (Problems • Answer Key • Resources) | |
Preceded by Problem 14 |
Followed by Problem 16 |
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.