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Difference between revisions of "2016 AMC 10A Problems/Problem 21"

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- ColtsFan10, diagram partially borrowed from Solution 1
 
- ColtsFan10, diagram partially borrowed from Solution 1
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==Solution 5 (Heron’s)==
 +
 +
We can use the Pythagorean theorem to find that the lengths are <math>3,5,2\sqrt3</math>. If we apply Heron’s, we know that it must be the sum (or difference) of two or more square roots, by instinct. This means that <math>\boxed{\textbf{(D)} \sqrt{6}-\sqrt{2}}</math> is the answer.
  
 
==See Also==
 
==See Also==

Revision as of 15:17, 1 January 2021

Problem

Circles with centers $P, Q$ and $R$, having radii $1, 2$ and $3$, respectively, lie on the same side of line $l$ and are tangent to $l$ at $P', Q'$ and $R'$, respectively, with $Q'$ between $P'$ and $R'$. The circle with center $Q$ is externally tangent to each of the other two circles. What is the area of triangle $PQR$?

$\textbf{(A) } 0\qquad \textbf{(B) } \sqrt{6}/3\qquad\textbf{(C) } 1\qquad\textbf{(D) } \sqrt{6}-\sqrt{2}\qquad\textbf{(E) }\sqrt{6}/2$

Solution 1

[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 $[P'PQRR'Q']$ in two different ways: $[P'PQQ']+[Q'QRR']$ and $[PQR]+[P'PRR']$, so $[P'PQQ']+[Q'QRR']=[PQR]+[P'PRR']$ $\break$

$P'Q'=\sqrt{PQ^2-(QQ'-PP')^2}=\sqrt{9-1}=\sqrt{8}=2\sqrt{2}$. Additionally, $Q'R'=\sqrt{QR^2-(RR'-QQ')^2}=\sqrt{5^2-1^2}=\sqrt{24}=2\sqrt{6}$. Therefore, $[P'PQQ']=\frac{P'P+Q'Q}{2}*2\sqrt{2}=\frac{1+2}{2}*2\sqrt{2}=3\sqrt{2}$. Similarly, $[Q'QRR']=5\sqrt6$. We can calculate $[P'PRR']$ easily because $P'R'=P'Q'+Q'R'=2\sqrt{2}+2\sqrt{6}$. $[P'PRR']=4\sqrt{2}+4\sqrt{6}$. $\newline$

Plugging into first equation, the two sums of areas, $3\sqrt{2}+5\sqrt{6}=4\sqrt{2}+4\sqrt{6}+[PQR]$. $\newline$

$[PQR]=\sqrt{6}-\sqrt{2}\rightarrow \fbox{D}$.

Solution 2

Use the Shoelace Theorem.

Let the center of the first circle of radius 1 be at $(0, 1)$.

Draw the trapezoid $PQQ'P'$ and using the Pythagorean Theorem, we get that $P'Q' = 2\sqrt{2}$ so the center of the second circle of radius 2 is at $(2\sqrt{2}, 2)$.

Draw the trapezoid $QRR'Q'$ and using the Pythagorean Theorem, we get that $Q'R' = 2\sqrt{6}$ so the center of the third circle of radius 3 is at $(2\sqrt{2}+2\sqrt{6}, 3)$.

Now, we may use the Shoelace Theorem!

$(0,1)$

$(2\sqrt{2}, 2)$

$(2\sqrt{2}+2\sqrt{6}, 3)$

$\frac{1}{2}|(2\sqrt{2}+4\sqrt{2}+4\sqrt{6})-(6\sqrt{2}+2\sqrt{2}+2\sqrt{6})|$

$= \sqrt{6}-\sqrt{2}$ $\fbox{D}$.

Solution 3

$PQ = 3$ and $QR = 5$ because they are the sum of two radii. $QQ' - PP' = 1$ and $RR' - QQ' = 1$, the difference of the radii. Using pythagorean theorem, we find that $P'Q'$ and $Q'R'$ are $\sqrt{8}$ and $\sqrt{24}$, $P'R' = \sqrt{8} + \sqrt{24}$.

Draw a perpendicular from $P$ to line $RR'$, then we can use the Pythagorean theorem to find $PR$. $RR' - PP' = 2$. We get \[PR^2 = (\sqrt{8} + \sqrt{24})^2 + 4 = 36 + 16\sqrt{3} \Rightarrow PR = \sqrt{36 + 16\sqrt{3}} = 2\sqrt{9 + 4\sqrt{3}}\]

To make our calculations easier, let $\sqrt{9 + 4\sqrt{3}} = a$. The semi-perimeter of our triangle is $\frac{3 + 5 + 2a}{2} = 4 + a$. Symbolize the area of the triangle with $A$. Using Heron's formula, we have \[A^2 = (4 + a)(4 + a - 2a)(4 + a - 3)(4 + a - 5) = (4 + a)(4 - a)(a + 1)(a - 1) = (16 - a^2)(a^2 - 1)\] We can remove the outer root of a. \[A^2 = (16 - 9 - 4\sqrt{3})(9 + 4\sqrt{3} - 1) = (7 - 4\sqrt{3})(8 + 4\sqrt{3}) = 8 - 4\sqrt{3} \rightarrow A = \sqrt{8 - 4\sqrt{3}}\]

We solve the nested root. We want to turn $8 - 4\sqrt{3}$ into the square of something. If we have $(a - b) ^ 2 = 8 - 4\sqrt{3}$, then we get \[\begin{cases} a^2 + b^2 = 8 \\ ab = 2\sqrt{3} \end{cases}\] Solving the system of equations, we get $a = \sqrt{6}$ and $b = \sqrt{2}$. Alternatively, you can square all the possible solutions until you find one that is equal to $8 - 4\sqrt{3}$. \[A = \sqrt{8 - 4\sqrt{3}} = \sqrt{(\sqrt{6} - \sqrt{2})^2} = \sqrt{6} - \sqrt{2} \rightarrow \fbox{D}\] ~ZericH

Solution 4

[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 $[PQRY]$ can be calculated in two ways: $[\triangle PQX] + [QZYX] + [\triangle QRZ]$ and $[\triangle PRY] + [\triangle PQR]$. Solving, we get: \begin{align*} [\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{align*}

- ColtsFan10, diagram partially borrowed from Solution 1

Solution 5 (Heron’s)

We can use the Pythagorean theorem to find that the lengths are $3,5,2\sqrt3$. If we apply Heron’s, we know that it must be the sum (or difference) of two or more square roots, by instinct. This means that $\boxed{\textbf{(D)} \sqrt{6}-\sqrt{2}}$ is the answer.

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

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