area is denoted by x.
[asy]
size(125);
pair O1=(0,0),O2=(5,0);
pair A=(1,0),B=(1,3);
pair C=(4,0),D=(0.75,-0.661);
pair E=(4.8,2.4);
fill(O1--B--D--cycle,gray(0.7)); fill(O2--C--E--cycle,gray(0.7));
draw(Circle(O1,1)^^Circle(O2,4));
draw((0,-1)--(0,3),Arrows); draw((5,-1)--(5,3),Arrows);
label("$p$",O1+(1.2,-0.7)); label("$q$",O2+(3,-0.7));
draw(-1,B--O1--D, dashed); draw((-1,-0.661)--(5,2.767), dashed);
draw((0,-0.661)--(4,-0.661),dashed);
draw(A--B^^O1--A^^C--E^^O2--C);
label("$A$",A,dir(-90)); label("$B$",B,dir(90));
label("$C$",C,dir(-90)); label("$D$",D,dir(-120));
label("$E$",E,dir(30)); label("$O_1$",O1,dir(-150));
label("$O_2$",O2,dir(-180)); label("$x$",(1.5,0.3));
[/asy]
Let $O_1$ and $O_2$ be the centers of $p$ and $q$, respectively. Also let $BC$ intersect $O_1O_2$ at $F$. It is clear that $O_1FO_2$ is a line parallel to $BC$, and thus $\triangle BDA \sim \triangle CEA$ (since $\triangle BFA \sim \triangle CFA$). Therefore, $\frac{BD}{CE} = \frac{DA}{EA}$. Note that $BD = BO_1 - DO_1 = 1 - DA$ and $CE = CO_2 + EO_2 = 4 - EA$. Substituting these expressions and simplifying, we get $\frac{1 - DA}{4 - EA} = \frac{DA}{EA} \Rightarrow DA = 4\left(\frac{EA}{5}\right)$.
Let the perpendicular from $O_1$ to $BC$ be $T_1$ and the perpendicular from $O_2$ to $BC$ be $T_2$. Then $T_1O_1 = 1$ and $T_2O_2 = 4$. Further, we have $AT_1 = T_1D$ and $AT_2 = T_2E$. Therefore, $AD = AT_1 + T_1D = AT_2 + T_2E = AE$, which means that $\triangle BDA$ and $\triangle CEA$ are actually congruent. Let their common area be $x$.
Since $\triangle BDA$ and $\triangle CEA$ are both isosceles triangles, their heights are the perpendiculars from $A$ to $BD$ and $CE$, respectively. These heights are equal to $T_1O_1$ and $T_2O_2$, respectively. Therefore, we have:
$2x = BD \cdot T_1O_1 = (1 - DA) \cdot 1 = \frac{1}{5} \cdot 4 = \frac{4}{5}$
$2x = CE \cdot T_2O_2 = (4 - EA) \cdot 4 = \frac{16}{5} \cdot \frac{4}{5} = \frac{64}{25}$
Equating these two expressions for $x$, we get $\frac{4}{5} = \frac{64}{25} \Rightarrow x = \boxed{\frac{16}{25}}$.
(5) circles p and q have radii 1 and 4, respectively, and are externally tangent at point a. point b is on p and point c is on q so that line bc is a common external tangent of the two circles. a line l through a intersects p again at d and intersects q again at e. points b and c lie on the same side of l, and the areas of △dba and △ace are equal. this common
1 answer
1 answer