Difference between revisions of "2022 AIME II Problems/Problem 15"
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circle w1=circle(O1,9),w2=circle(O2,6),o=circle(O1,O2,B); | circle w1=circle(O1,9),w2=circle(O2,6),o=circle(O1,O2,B); | ||
point A=intersectionpoints(o,w2)[1],D=intersectionpoints(o,w2)[0],C=intersectionpoints(o,w1)[0]; | point A=intersectionpoints(o,w2)[1],D=intersectionpoints(o,w2)[0],C=intersectionpoints(o,w1)[0]; | ||
− | filldraw(A--B--O1--C--D--O2--cycle,0.2* | + | filldraw(A--B--O1--C--D--O2--cycle,0.2*fuchsia+white,black); |
draw(w1); | draw(w1); | ||
draw(w2); | draw(w2); | ||
Line 33: | Line 33: | ||
</asy> | </asy> | ||
− | ==Solution== | + | ==Solution 1== |
+ | First observe that <math>AO_2 = O_2D</math> and <math>BO_1 = O_1C</math>. Let points <math>A'</math> and <math>B'</math> be the reflections of <math>A</math> and <math>B</math>, respectively, about the perpendicular bisector of <math>\overline{O_1O_2}</math>. Then quadrilaterals <math>ABO_1O_2</math> and <math>B'A'O_2O_1</math> are congruent, so hexagons <math>ABO_1CDO_2</math> and <math>A'B'O_1CDO_2</math> have the same area. Furthermore, triangles <math>DO_2A'</math> and <math>B'O_1C</math> are congruent, so <math>A'D = B'C</math> and quadrilateral <math>A'B'CD</math> is an isosceles trapezoid. | ||
+ | <asy> | ||
+ | import olympiad; | ||
+ | size(180); | ||
+ | defaultpen(linewidth(0.7)); | ||
+ | pair Ap = dir(105), Bp = dir(75), O1 = dir(25), C = dir(320), D = dir(220), O2 = dir(175); | ||
+ | draw(unitcircle^^Ap--Bp--O1--C--D--O2--cycle); | ||
+ | label("$A'$",Ap,dir(origin--Ap)); | ||
+ | label("$B'$",Bp,dir(origin--Bp)); | ||
+ | label("$O_1$",O1,dir(origin--O1)); | ||
+ | label("$C$",C,dir(origin--C)); | ||
+ | label("$D$",D,dir(origin--D)); | ||
+ | label("$O_2$",O2,dir(origin--O2)); | ||
+ | draw(O2--O1,linetype("4 4")); | ||
+ | draw(Ap--D^^Bp--C,linetype("2 2")); | ||
+ | </asy> | ||
+ | Next, remark that <math>B'O_1 = DO_2</math>, so quadrilateral <math>B'O_1DO_2</math> is also an isosceles trapezoid; in turn, <math>B'D = O_1O_2 = 15</math>, and similarly <math>A'C = 15</math>. Thus, Ptolmey's theorem on <math>A'B'CD</math> yields <math>A'D\cdot B'C + 2\cdot 16 = 15^2</math>, whence <math>A'D = B'C = \sqrt{193}</math>. Let <math>\alpha = \angle A'B'D</math>. The Law of Cosines on triangle <math>A'B'D</math> yields | ||
+ | <cmath>\cos\alpha = \frac{15^2 + 2^2 - (\sqrt{193})^2}{2\cdot 2\cdot 15} = \frac{36}{60} = \frac 35,</cmath> and hence <math>\sin\alpha = \tfrac 45</math>. Thus the distance between bases <math>A’B’</math> and <math>CD</math> is <math>12</math> (in fact, <math>\triangle A'B'D</math> is a <math>9-12-15</math> triangle with a <math>7-12-\sqrt{193}</math> triangle removed), which implies the area of <math>A'B'CD</math> is <math>\tfrac12\cdot 12\cdot(2+16) = 108</math>. | ||
+ | |||
+ | Now let <math>O_1C = O_2A' = r_1</math> and <math>O_2D = O_1B' = r_2</math>; the tangency of circles <math>\omega_1</math> and <math>\omega_2</math> implies <math>r_1 + r_2 = 15</math>. Furthermore, angles <math>A'O_2D</math> and <math>A'B'D</math> are opposite angles in cyclic quadrilateral <math>B'A'O_2D</math>, which implies the measure of angle <math>A'O_2D</math> is <math>180^\circ - \alpha</math>. Therefore, the Law of Cosines applied to triangle <math>\triangle A'O_2D</math> yields | ||
+ | <cmath>\begin{align*} | ||
+ | 193 &= r_1^2 + r_2^2 - 2r_1r_2(-\tfrac 35) = (r_1^2 + 2r_1r_2 + r_2^2) - \tfrac45r_1r_2\\ | ||
+ | &= (r_1+r_2)^2 - \tfrac45 r_1r_2 = 225 - \tfrac45r_1r_2. | ||
+ | \end{align*}</cmath> | ||
+ | |||
+ | Thus <math>r_1r_2 = 40</math>, and so the area of triangle <math>A'O_2D</math> is <math>\tfrac12r_1r_2\sin\alpha = 16</math>. | ||
+ | |||
+ | Thus, the area of hexagon <math>ABO_{1}CDO_{2}</math> is <math>108 + 2\cdot 16 = \boxed{140}</math>. | ||
+ | |||
+ | ~djmathman | ||
+ | |||
+ | ==Solution 2== | ||
+ | |||
+ | Denote by <math>O</math> the center of <math>\Omega</math>. | ||
+ | Denote by <math>r</math> the radius of <math>\Omega</math>. | ||
+ | |||
+ | We have <math>O_1</math>, <math>O_2</math>, <math>A</math>, <math>B</math>, <math>C</math>, <math>D</math> are all on circle <math>\Omega</math>. | ||
+ | |||
+ | Denote <math>\angle O_1 O O_2 = 2 \theta</math>. | ||
+ | Denote <math>\angle O_1 O B = \alpha</math>. | ||
+ | Denote <math>\angle O_2 O A = \beta</math>. | ||
+ | |||
+ | Because <math>B</math> and <math>C</math> are on circles <math>\omega_1</math> and <math>\Omega</math>, <math>BC</math> is a perpendicular bisector of <math>O_1 O</math>. Hence, <math>\angle O_1 O C = \alpha</math>. | ||
+ | |||
+ | Because <math>A</math> and <math>D</math> are on circles <math>\omega_2</math> and <math>\Omega</math>, <math>AD</math> is a perpendicular bisector of <math>O_2 O</math>. Hence, <math>\angle O_2 O D = \beta</math>. | ||
+ | |||
+ | In <math>\triangle O O_1 O_2</math>, | ||
+ | <cmath> | ||
+ | \[ | ||
+ | O_1 O_2 = 2 r \sin \theta . | ||
+ | \] | ||
+ | </cmath> | ||
+ | |||
+ | Hence, | ||
+ | <cmath> | ||
+ | \[ | ||
+ | 2 r \sin \theta = 15 . | ||
+ | \] | ||
+ | </cmath> | ||
+ | |||
+ | In <math>\triangle O AB</math>, | ||
+ | <cmath> | ||
+ | \begin{align*} | ||
+ | AB & = 2 r \sin \frac{2 \theta - \alpha - \beta}{2} \\ | ||
+ | & = 2 r \sin \theta \cos \frac{\alpha + \beta}{2} | ||
+ | - 2 r \cos \theta \sin \frac{\alpha + \beta}{2} \\ | ||
+ | & = 15 \cos \frac{\alpha + \beta}{2} | ||
+ | - 2 r \cos \theta \sin \frac{\alpha + \beta}{2} . | ||
+ | \end{align*} | ||
+ | </cmath> | ||
+ | |||
+ | Hence, | ||
+ | <cmath> | ||
+ | \[ | ||
+ | 15 \cos \frac{\alpha + \beta}{2} | ||
+ | - 2 r \cos \theta \sin \frac{\alpha + \beta}{2} | ||
+ | = 2 . \hspace{1cm} (1) | ||
+ | \] | ||
+ | </cmath> | ||
+ | |||
+ | |||
+ | In <math>\triangle O CD</math>, | ||
+ | <cmath> | ||
+ | \begin{align*} | ||
+ | CD & = 2 r \sin \frac{360^\circ - 2 \theta - \alpha - \beta}{2} \\ | ||
+ | & = 2 r \sin \left( \theta + \frac{\alpha + \beta}{2} \right) \\ | ||
+ | & = 2 r \sin \theta \cos \frac{\alpha + \beta}{2} | ||
+ | + 2 r \cos \theta \sin \frac{\alpha + \beta}{2} \\ | ||
+ | & = 15 \cos \frac{\alpha + \beta}{2} | ||
+ | + 2 r \cos \theta \sin \frac{\alpha + \beta}{2} . | ||
+ | \end{align*} | ||
+ | </cmath> | ||
+ | |||
+ | Hence, | ||
+ | <cmath> | ||
+ | \[ | ||
+ | 15 \cos \frac{\alpha + \beta}{2} | ||
+ | + 2 r \cos \theta \sin \frac{\alpha + \beta}{2} | ||
+ | = 16 . \hspace{1cm} (2) | ||
+ | \] | ||
+ | </cmath> | ||
+ | |||
+ | Taking <math>\frac{(1) + (2)}{30}</math>, we get <math>\cos \frac{\alpha + \beta}{2} = \frac{3}{5}</math>. | ||
+ | Thus, <math>\sin \frac{\alpha + \beta}{2} = \frac{4}{5}</math>. | ||
+ | |||
+ | Taking these into (1), we get <math>2 r \cos \theta = \frac{35}{4}</math>. | ||
+ | Hence, | ||
+ | <cmath> | ||
+ | \begin{align*} | ||
+ | 2 r & = \sqrt{ \left( 2 r \sin \theta \right)^2 + \left( 2 r \cos \theta \right)^2} \\ | ||
+ | & = \frac{5}{4} \sqrt{193} . | ||
+ | \end{align*} | ||
+ | </cmath> | ||
+ | |||
+ | Hence, <math>\cos \theta = \frac{7}{\sqrt{193}}</math>. | ||
+ | |||
+ | In <math>\triangle O O_1 B</math>, | ||
+ | <cmath> | ||
+ | \[ | ||
+ | O_1 B = 2 r \sin \frac{\alpha}{2} . | ||
+ | \] | ||
+ | </cmath> | ||
+ | |||
+ | In <math>\triangle O O_2 A</math>, by applying the law of sines, we get | ||
+ | <cmath> | ||
+ | \[ | ||
+ | O_2 A = 2 r \sin \frac{\beta}{2} . | ||
+ | \] | ||
+ | </cmath> | ||
+ | |||
+ | Because circles <math>\omega_1</math> and <math>\omega_2</math> are externally tangent, <math>B</math> is on circle <math>\omega_1</math>, <math>A</math> is on circle <math>\omega_2</math>, | ||
+ | <cmath> | ||
+ | \begin{align*} | ||
+ | O_1 O_2 & = O_1 B + O_2 A \\ | ||
+ | & = 2 r \sin \frac{\alpha}{2} + 2 r \sin \frac{\beta}{2} \\ | ||
+ | & = 2 r \left( \sin \frac{\alpha}{2} + \sin \frac{\beta}{2} \right) . | ||
+ | \end{align*} | ||
+ | </cmath> | ||
+ | |||
+ | Thus, <math>\sin \frac{\alpha}{2} + \sin \frac{\beta}{2} = \frac{12}{\sqrt{193}}</math>. | ||
+ | |||
+ | Now, we compute <math>\sin \alpha</math> and <math>\sin \beta</math>. | ||
+ | |||
+ | Recall <math>\cos \frac{\alpha + \beta}{2} = \frac{3}{5}</math> and <math>\sin \frac{\alpha + \beta}{2} = \frac{4}{5}</math>. | ||
+ | Thus, <math>e^{i \frac{\alpha}{2}} e^{i \frac{\beta}{2}} = e^{i \frac{\alpha + \beta}{2}} = \frac{3}{5} + i \frac{4}{5}</math>. | ||
+ | |||
+ | We also have | ||
+ | <cmath> | ||
+ | \begin{align*} | ||
+ | \sin \frac{\alpha}{2} + \sin \frac{\beta}{2} | ||
+ | & = \frac{1}{2i} \left( e^{i \frac{\alpha}{2}} - e^{-i \frac{\alpha}{2}} | ||
+ | + e^{i \frac{\beta}{2}} - e^{-i \frac{\beta}{2}} \right) \\ | ||
+ | & = \frac{1}{2i} \left( 1 - \frac{1}{e^{i \frac{\alpha + \beta}{2}} } \right) | ||
+ | \left( e^{i \frac{\alpha}{2}} + e^{i \frac{\beta}{2}} \right) | ||
+ | . | ||
+ | \end{align*} | ||
+ | </cmath> | ||
+ | |||
+ | Thus, | ||
+ | <cmath> | ||
+ | \begin{align*} | ||
+ | \sin \alpha + \sin \beta | ||
+ | & = \frac{1}{2i} \left( e^{i \alpha} - e^{-i \alpha} | ||
+ | + e^{i \beta} - e^{-i \beta} \right) \\ | ||
+ | & = \frac{1}{2i} \left( 1 - \frac{1}{e^{i \left( \alpha + \beta \right)}} \right) | ||
+ | \left( e^{i \alpha} + e^{i \beta} \right) \\ | ||
+ | & = \frac{1}{2i} \left( 1 - \frac{1}{e^{i \left( \alpha + \beta \right)}} \right) | ||
+ | \left( | ||
+ | \left( e^{i \frac{\alpha}{2}} + e^{i \frac{\beta}{2}} \right)^2 | ||
+ | - 2 e^{i \frac{\alpha + \beta}{2}} | ||
+ | \right) \\ | ||
+ | & = \frac{1}{2i} \left( 1 - \frac{1}{e^{i \left( \alpha + \beta \right)}} \right) | ||
+ | \left( | ||
+ | \left( \frac{2 i \left( \sin \frac{\alpha}{2} + \sin \frac{\beta}{2} \right)}{1 - \frac{1}{e^{i \frac{\alpha + \beta}{2}} }} \right)^2 | ||
+ | - 2 e^{i \frac{\alpha + \beta}{2}} | ||
+ | \right) \\ | ||
+ | & = - \frac{1}{i} \left( e^{i \frac{\alpha + \beta}{2}} - e^{-i \frac{\alpha + \beta}{2}} \right) | ||
+ | \left( | ||
+ | \frac{2 \left( \sin \frac{\alpha}{2} + \sin \frac{\beta}{2} \right)^2} | ||
+ | {e^{i \frac{\alpha + \beta}{2}} + e^{-i \frac{\alpha + \beta}{2}} - 2 } | ||
+ | + 1 | ||
+ | \right) \\ | ||
+ | & = \frac{167 \cdot 8}{193 \cdot 5 } . | ||
+ | \end{align*} | ||
+ | </cmath> | ||
+ | |||
+ | Therefore, | ||
+ | <cmath> | ||
+ | \begin{align*} | ||
+ | {\rm Area} \ ABO_1CDO_2 | ||
+ | & = {\rm Area} \ \triangle O_3 AB + {\rm Area} \ \triangle O_3 BO_1 | ||
+ | + {\rm Area} \ \triangle O_3 O_1 C \\ | ||
+ | & \quad + {\rm Area} \ \triangle O_3 C D | ||
+ | + {\rm Area} \ \triangle O_3 D O_2 + {\rm Area} \ \triangle O_3 O_2 A \\ | ||
+ | & = \frac{1}{2} r^2 \left( | ||
+ | \sin \left( 2 \theta - \alpha - \beta \right) + \sin \alpha + \sin \alpha | ||
+ | + \sin \left( 360^\circ - 2 \theta - \alpha - \beta \right) | ||
+ | + \sin \beta + \sin \beta \right) \\ | ||
+ | & = \frac{1}{2} r^2 \left( | ||
+ | \sin \left( 2 \theta - \alpha - \beta \right) | ||
+ | - \sin \left( 2 \theta + \alpha + \beta \right) | ||
+ | + 2 \sin \alpha + 2 \sin \beta | ||
+ | \right) \\ | ||
+ | & = r^2 \left( | ||
+ | - \cos 2 \theta \sin \left( \alpha + \beta \right) | ||
+ | + \sin \alpha + \sin \beta | ||
+ | \right) \\ | ||
+ | & = r^2 \left( \left( 1 - 2 \cos^2 \theta \right) 2 \sin \frac{\alpha + \beta}{2} \cos \frac{\alpha + \beta}{2} | ||
+ | + \sin \alpha + \sin \beta \right) \\ | ||
+ | & = \boxed{\textbf{(140) }} . | ||
+ | \end{align*} | ||
+ | </cmath> | ||
+ | |||
+ | ~Steven Chen (www.professorchenedu.com) | ||
+ | |||
+ | ==Solution 3== | ||
+ | [[File:AIME-II-2022-15.png|400px|right]] | ||
+ | Let points <math>A'</math> and <math>B'</math> be the reflections of <math>A</math> and <math>B,</math> respectively, about the perpendicular bisector of <math>O_1 O_2.</math> | ||
+ | <cmath>B'O_2 = BO_1 = O_1 P = O_1 C,</cmath> | ||
+ | <cmath>A'O_1 = AO_2 = O_2 P = O_2 D.</cmath> | ||
+ | We establish the equality of the arcs and conclude that the corresponding chords are equal | ||
+ | <cmath>\overset{\Large\frown} {CO_1} + \overset{\Large\frown} {A'O_1} +\overset{\Large\frown} {A'B'} = \overset{\Large\frown} {B'O_2} +\overset{\Large\frown} {A'O_1} +\overset{\Large\frown} {A'B'} =\overset{\Large\frown} {B'O_2} +\overset{\Large\frown} {DO_2} +\overset{\Large\frown} {A'B'}</cmath> | ||
+ | <cmath> \implies A'D = B'C = O_1 O_2 = 15.</cmath> | ||
+ | Similarly <math>A'C = B'D \implies \triangle A'CO_1 = \triangle B'DO_2.</math> | ||
+ | |||
+ | Ptolemy's theorem on <math>A'CDB'</math> yields <cmath>B'D \cdot A'C + A'B' \cdot CD = A'D \cdot B'C \implies</cmath> | ||
+ | <cmath> B'D^2 + 2 \cdot 16 = 15^2 \implies B'D = A'C = \sqrt{193}.</cmath> | ||
+ | The area of the trapezoid <math>A'CDB'</math> is equal to the area of an isosceles triangle with sides | ||
+ | <math>A'D = B'C = 15</math> and <math>A'B' + CD = 18.</math> | ||
+ | |||
+ | The height of this triangle is <math>\sqrt{15^2-9^2} = 12.</math> The area of <math>A'CDB'</math> is <math>108.</math> | ||
+ | |||
+ | <cmath>\sin \angle B'CD = \frac{12}{15} = \frac{4}{5},</cmath> | ||
+ | <cmath>\angle B'CD + \angle B'O_2 D = 180^o \implies \sin \angle B'O_2 D = \frac{4}{5}.</cmath> | ||
+ | |||
+ | Denote <math>\angle B'O_2 D = 2\alpha.</math> | ||
+ | <math>\angle B'O_2 D > \frac{\pi}{2},</math> hence <math>\cos \angle B'O_2 D = \cos 2\alpha = -\frac{3}{5}.</math> | ||
+ | <cmath>\tan \alpha =\frac { \sin 2 \alpha}{1+\cos 2 \alpha} = \frac {4/5}{1 - 3/5}=2.</cmath> | ||
+ | |||
+ | Semiperimeter of <math>\triangle B'O_2 D </math> is <math>s = \frac {15 + \sqrt{193}}{2}.</math> | ||
+ | |||
+ | The distance from the vertex <math>O_2</math> to the tangent points of the inscribed circle of the triangle <math>B'O_2 D</math> is equal <math>s – B'D = \frac{15 – \sqrt{193}}{2}.</math> | ||
+ | |||
+ | The radius of the inscribed circle is <math>r = (s – B'D) \tan \alpha.</math> | ||
+ | |||
+ | The area of triangle <math>B'O_2 D</math> is <math>[B'O_2 D] = sr = s (s – B'D) \tan \alpha = \frac {15^2 – 193}{2} = 16.</math> | ||
+ | |||
+ | The hexagon <math>ABO_1 CDO_2</math> has the same area as hexagon <math>B'A'O_1 CDO_2.</math> | ||
+ | |||
+ | The area of hexagon <math>B'A'O_1 CDO_2</math> is equal to the sum of the area of the trapezoid <math>A'CDB'</math> and the areas of two equal triangles <math>B'O_2 D</math> and <math>A'O_1 C,</math> so the area of the hexagon <math>ABO_1 CDO_2</math> is <cmath>108 + 16 + 16 = \boxed{140}.</cmath> | ||
+ | |||
+ | '''vladimir.shelomovskii@gmail.com, vvsss''' | ||
+ | |||
+ | ==Solution 4== | ||
+ | |||
+ | |||
+ | [[File:AIME 2022 II Prob15.jpg|400px|top]] | ||
+ | |||
+ | |||
+ | Let circle <math>O_1</math>'s radius be <math>r</math>, then the radius of circle <math>O_2</math> is <math>15-r</math>. Based on Brahmagupta's Formula, | ||
+ | |||
+ | |||
+ | the hexagon's Area <math>= \sqrt{14(1)(16-r)(1+r)} + \sqrt{7(8)(23-r)(8+r)}</math>. | ||
+ | |||
+ | |||
+ | Now we only need to find the <math>r</math>. | ||
+ | |||
+ | |||
+ | Connect <math>O_1</math> and <math>A</math> , <math>O_1</math> and <math>D</math> , and let <math>X</math> be the point of intersection between <math>O_1D</math> and circle <math>O_2</math> , based on the " 2 Non-Congruent Triangles of 'SSA' Scenario " , we can immediately see <math>O_1X = O_1A</math> and therefore get an equation from the "Power of A Point Theorem: | ||
+ | |||
+ | |||
+ | <math>(O_1A)(O_1D) = r(15+15-r) = 15r + r(15-r)</math> (1). | ||
+ | |||
+ | |||
+ | Similarly, | ||
+ | |||
+ | |||
+ | <math>(O_2B)(O_2C) = (15-r)(15+r) = 15(15-r) + r(15-r)</math> (2). | ||
+ | |||
+ | |||
+ | We can also get two other equations about these 4 segments from Ptolemy's Theorem: | ||
+ | |||
+ | |||
+ | <math>(O_1A)(O_2B) = 30 + r(15-r)</math> (3) | ||
+ | |||
+ | |||
+ | <math>(O_1D)(O_2C) = 240 + r(15-r)</math> (4) | ||
+ | |||
+ | |||
+ | Multiply equations (1) and (2), and equations (3) and (4) respectively, we will get a very simple and nice equation of <math>r</math>: | ||
+ | |||
+ | |||
+ | <math>2(15^2)r(15-r) = 7200 + 270r(15-r)</math>, | ||
+ | |||
+ | |||
+ | then: | ||
+ | |||
+ | |||
+ | <math>r(15-r) = 40</math>. | ||
+ | |||
+ | |||
+ | This result is good enough for us to find the hexagon's area, which: | ||
+ | |||
+ | <math>= \sqrt{14(1)(16-r)(1+r)} + \sqrt{7(8)(23-r)(8+r)}</math> | ||
+ | <math>= \sqrt{14(1)(1+15-r)(1+r)} + \sqrt{7(8)(8+15-r)(8+r)}</math> | ||
+ | <math>= \sqrt{14(1)(1+15+40)} + \sqrt{7(8)(64+8(15)+40)}</math> | ||
+ | <math>= 28 + 112 = \boxed{\textbf{140}}</math>. | ||
+ | |||
+ | eJMaSc | ||
+ | |||
+ | |||
+ | ==Video Solution== | ||
+ | https://youtu.be/TeAWS5H5bcc | ||
+ | |||
+ | ~MathProblemSolvingSkills.com | ||
+ | |||
==See Also== | ==See Also== | ||
− | {{AIME box|year=2022|n= | + | {{AIME box|year=2022|n=II|num-b=14|after=Last Problem}} |
+ | |||
+ | [[Category:Intermediate Geometry Problems]] | ||
{{MAA Notice}} | {{MAA Notice}} |
Latest revision as of 00:28, 29 June 2023
Problem
Two externally tangent circles and
have centers
and
, respectively. A third circle
passing through
and
intersects
at
and
and
at
and
, as shown. Suppose that
,
,
, and
is a convex hexagon. Find the area of this hexagon.
Solution 1
First observe that and
. Let points
and
be the reflections of
and
, respectively, about the perpendicular bisector of
. Then quadrilaterals
and
are congruent, so hexagons
and
have the same area. Furthermore, triangles
and
are congruent, so
and quadrilateral
is an isosceles trapezoid.
Next, remark that
, so quadrilateral
is also an isosceles trapezoid; in turn,
, and similarly
. Thus, Ptolmey's theorem on
yields
, whence
. Let
. The Law of Cosines on triangle
yields
and hence
. Thus the distance between bases
and
is
(in fact,
is a
triangle with a
triangle removed), which implies the area of
is
.
Now let and
; the tangency of circles
and
implies
. Furthermore, angles
and
are opposite angles in cyclic quadrilateral
, which implies the measure of angle
is
. Therefore, the Law of Cosines applied to triangle
yields
Thus , and so the area of triangle
is
.
Thus, the area of hexagon is
.
~djmathman
Solution 2
Denote by the center of
.
Denote by
the radius of
.
We have ,
,
,
,
,
are all on circle
.
Denote .
Denote
.
Denote
.
Because and
are on circles
and
,
is a perpendicular bisector of
. Hence,
.
Because and
are on circles
and
,
is a perpendicular bisector of
. Hence,
.
In ,
Hence,
In ,
Hence,
In ,
Hence,
Taking , we get
.
Thus,
.
Taking these into (1), we get .
Hence,
Hence, .
In ,
In , by applying the law of sines, we get
Because circles and
are externally tangent,
is on circle
,
is on circle
,
Thus, .
Now, we compute and
.
Recall and
.
Thus,
.
We also have
Thus,
Therefore,
~Steven Chen (www.professorchenedu.com)
Solution 3
Let points and
be the reflections of
and
respectively, about the perpendicular bisector of
We establish the equality of the arcs and conclude that the corresponding chords are equal
Similarly
Ptolemy's theorem on yields
The area of the trapezoid
is equal to the area of an isosceles triangle with sides
and
The height of this triangle is The area of
is
Denote
hence
Semiperimeter of is
The distance from the vertex to the tangent points of the inscribed circle of the triangle
is equal
The radius of the inscribed circle is
The area of triangle is
The hexagon has the same area as hexagon
The area of hexagon is equal to the sum of the area of the trapezoid
and the areas of two equal triangles
and
so the area of the hexagon
is
vladimir.shelomovskii@gmail.com, vvsss
Solution 4
Let circle 's radius be
, then the radius of circle
is
. Based on Brahmagupta's Formula,
the hexagon's Area .
Now we only need to find the .
Connect and
,
and
, and let
be the point of intersection between
and circle
, based on the " 2 Non-Congruent Triangles of 'SSA' Scenario " , we can immediately see
and therefore get an equation from the "Power of A Point Theorem:
(1).
Similarly,
(2).
We can also get two other equations about these 4 segments from Ptolemy's Theorem:
(3)
(4)
Multiply equations (1) and (2), and equations (3) and (4) respectively, we will get a very simple and nice equation of :
,
then:
.
This result is good enough for us to find the hexagon's area, which:
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.
eJMaSc
Video Solution
~MathProblemSolvingSkills.com
See Also
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