Difference between revisions of "2022 AIME I Problems/Problem 11"
(→Solution 1 (No trig)) |
m (→Video Solution by Punxsutawney Phil) |
||
(61 intermediate revisions by 13 users not shown) | |||
Line 1: | Line 1: | ||
== Problem == | == Problem == | ||
− | Let <math>ABCD</math> be a parallelogram with <math>\angle BAD < 90^ | + | Let <math>ABCD</math> be a parallelogram with <math>\angle BAD < 90^\circ.</math> A circle tangent to sides <math>\overline{DA},</math> <math>\overline{AB},</math> and <math>\overline{BC}</math> intersects diagonal <math>\overline{AC}</math> at points <math>P</math> and <math>Q</math> with <math>AP < AQ,</math> as shown. Suppose that <math>AP=3,</math> <math>PQ=9,</math> and <math>QC=16.</math> Then the area of <math>ABCD</math> can be expressed in the form <math>m\sqrt{n},</math> where <math>m</math> and <math>n</math> are positive integers, and <math>n</math> is not divisible by the square of any prime. Find <math>m+n.</math> |
<asy> | <asy> | ||
Line 23: | Line 23: | ||
dot(A^^B^^C^^D^^P^^Q); | dot(A^^B^^C^^D^^P^^Q); | ||
</asy> | </asy> | ||
+ | |||
+ | ==Video Solution by Punxsutawney Phil (Currently Privated)== | ||
+ | https://www.youtube.com/watch?v=1m3pqCgwLFE | ||
==Solution 1 (No trig)== | ==Solution 1 (No trig)== | ||
Line 29: | Line 32: | ||
<asy> | <asy> | ||
− | + | size(10cm); | |
− | size( | + | pair A,B,C,D,EE,F,P,Q,O; |
− | pair A,B,C,D,P,Q,O; | ||
A=(0,0); | A=(0,0); | ||
+ | EE = (24,15); | ||
+ | F = (30,0); | ||
O = (10.5,7.5); | O = (10.5,7.5); | ||
label("$A$", A, SW); | label("$A$", A, SW); | ||
Line 50: | Line 54: | ||
dot(A^^B^^C^^D^^P^^Q); | dot(A^^B^^C^^D^^P^^Q); | ||
dot(O); | dot(O); | ||
− | label("$O$",O, | + | label("$O$",O,W); |
draw((10.5,15)--(10.5,0)); | draw((10.5,15)--(10.5,0)); | ||
draw(D--(24,15),dashed); | draw(D--(24,15),dashed); | ||
draw(C--(30,0),dashed); | draw(C--(30,0),dashed); | ||
draw(D--(30,0)); | draw(D--(30,0)); | ||
+ | dot(EE); | ||
+ | dot(F); | ||
label("$3$", midpoint(A--P), S); | label("$3$", midpoint(A--P), S); | ||
Line 66: | Line 72: | ||
label("$14+x$", (17.25,0), S); | label("$14+x$", (17.25,0), S); | ||
label("$6-x$", (27,15), N); | label("$6-x$", (27,15), N); | ||
− | label("$6+x$", (27,7.5), | + | label("$6+x$", (27,7.5), S); |
− | label("$6\sqrt{3}$", (30,7.5),E); | + | label("$6\sqrt{3}$", (30,7.5), E); |
label("$T_1$", (10.5,15), N); | label("$T_1$", (10.5,15), N); | ||
label("$T_2$", (10.5,0), S); | label("$T_2$", (10.5,0), S); | ||
− | label("$T_3$", (4.5,11.25),W); | + | label("$T_3$", (4.5,11.25), W); |
+ | label("$E$", EE, N); | ||
+ | label("$F$", F, S); | ||
</asy> | </asy> | ||
− | We obviously see that we must use power of a point since they've given us lengths in a circle and there are intersection points. By the secant power of a point, the power of <math>A = 3 \cdot (3+9) = 36</math>. Then | + | We obviously see that we must use power of a point since they've given us lengths in a circle and there are intersection points. Let <math>T_1, T_2, T_3</math> be our tangents from the circle to the parallelogram. By the secant power of a point, the power of <math>A = 3 \cdot (3+9) = 36</math>. Then <math>AT_2 = AT_3 = \sqrt{36} = 6</math>. Similarly, the power of <math>C = 16 \cdot (16+9) = 400</math> and <math>CT_1 = \sqrt{400} = 20</math>. We let <math>BT_3 = BT_1 = x</math> and label the diagram accordingly. |
+ | |||
+ | Notice that because <math>BC = AD, 20+x = 6+DT_2 \implies DT_2 = 14+x</math>. Let <math>O</math> be the center of the circle. Since <math>OT_1</math> and <math>OT_2</math> intersect <math>BC</math> and <math>AD</math>, respectively, at right angles, we have <math>T_2T_1CD</math> is a right-angled trapezoid and more importantly, the diameter of the circle is the height of the triangle. Therefore, we can drop an altitude from <math>D</math> to <math>BC</math> and <math>C</math> to <math>AD</math>, and both are equal to <math>2r</math>. Since <math>T_1E = T_2D</math>, <math>20 - CE = 14+x \implies CE = 6-x</math>. Since <math>CE = DF, DF = 6-x</math> and <math>AF = 6+14+x+6-x = 26</math>. We can now use Pythagorean theorem on <math>\triangle ACF</math>; we have <math>26^2 + (2r)^2 = (3+9+16)^2 \implies 4r^2 = 784-676 \implies 4r^2 = 108 \implies 2r = 6\sqrt{3}</math> and <math>r^2 = 27</math>. | ||
+ | |||
+ | We know that <math>CD = 6+x</math> because <math>ABCD</math> is a parallelogram. Using Pythagorean theorem on <math>\triangle CDF</math>, <math>(6+x)^2 = (6-x)^2 + 108 \implies (6+x)^2-(6-x)^2 = 108 \implies 12 \cdot 2x = 108 \implies 2x = 9 \implies x = \frac{9}{2}</math>. Therefore, base <math>BC = 20 + \frac{9}{2} = \frac{49}{2}</math>. Thus the area of the parallelogram is the base times the height, which is <math>\frac{49}{2} \cdot 6\sqrt{3} = 147\sqrt{3}</math> and the answer is <math>\boxed{150}</math> | ||
+ | |||
+ | |||
+ | ~KingRavi | ||
==Solution 2== | ==Solution 2== | ||
Line 80: | Line 95: | ||
Let the circle tangent to <math>BC,AD,AB</math> at <math>P,Q,M</math> separately, denote that <math>\angle{ABC}=\angle{D}=\alpha</math> | Let the circle tangent to <math>BC,AD,AB</math> at <math>P,Q,M</math> separately, denote that <math>\angle{ABC}=\angle{D}=\alpha</math> | ||
− | Using POP, it is very clear that <math>PC=20,AQ=AM=6</math>, let <math>BM=BP=x,QD=14+x</math>, using LOC in <math>\triangle{ABP}</math>,<math>x^2+(x+6)^2-2x(x+6)\cos\alpha=36+PQ^2</math>, similarly, use LOC in <math>\triangle{DQC}</math>, getting that <math>(14+x)^2+(6+x)^2-2(6+x)(14+x)\cos\alpha=400+PQ^2</math>. We use the second equation to minus the first equation, getting that <math>28x+196-(2x+12) | + | Using POP, it is very clear that <math>PC=20,AQ=AM=6</math>, let <math>BM=BP=x,QD=14+x</math>, using LOC in <math>\triangle{ABP}</math>,<math>x^2+(x+6)^2-2x(x+6)\cos\alpha=36+PQ^2</math>, similarly, use LOC in <math>\triangle{DQC}</math>, getting that <math>(14+x)^2+(6+x)^2-2(6+x)(14+x)\cos\alpha=400+PQ^2</math>. We use the second equation to minus the first equation, getting that <math>28x+196-(2x+12)\times14\times\cos\alpha=364</math>, we can get <math>\cos\alpha=\frac{2x-12}{2x+12}</math>. |
− | Now applying LOC in <math>\triangle{ADC}</math>, getting <math>(6+x)^2+ | + | Now applying LOC in <math>\triangle{ADC}</math>, getting <math>(6+x)^2+(20+x)^2-2(6+x)\times(20+x)\times\frac{2x-12}{2x+12}=(3+9+16)^2</math>, solving this equation to get <math>x=\frac{9}{2}</math>, then <math>\cos\alpha=-\frac{1}{7}</math>, <math>\sin\alpha=\frac{4\sqrt{3}}{7}</math>, the area is <math>\frac{21}{2}\cdot\frac{49}{2}\cdot\frac{4\sqrt{3}}{7}=147\sqrt{3}</math> leads to <math>\boxed{150}</math> |
− | ~bluesoul | + | ~bluesoul,HarveyZhang |
==Solution 3== | ==Solution 3== | ||
Line 103: | Line 118: | ||
Hence, <math>\left( 6 + x \right)^2 = 4 r^2 + \left( 6 - x \right)^2</math>. | Hence, <math>\left( 6 + x \right)^2 = 4 r^2 + \left( 6 - x \right)^2</math>. | ||
This can be simplified as | This can be simplified as | ||
− | \[ | + | <cmath>\[ |
6 x = r^2 . \hspace{1cm} (1) | 6 x = r^2 . \hspace{1cm} (1) | ||
− | \] | + | \]</cmath> |
In <math>\triangle ACB</math>, by applying the law of cosines, we have | In <math>\triangle ACB</math>, by applying the law of cosines, we have | ||
− | \begin{align*} | + | <cmath>\begin{align*} |
AC^2 & = AB^2 + CB^2 - 2 AB \cdot CB \cos B \\ | AC^2 & = AB^2 + CB^2 - 2 AB \cdot CB \cos B \\ | ||
& = AB^2 + CB^2 + 2 AB \cdot CB \cos A \\ | & = AB^2 + CB^2 + 2 AB \cdot CB \cos A \\ | ||
Line 115: | Line 130: | ||
& = \left( 6 + x \right)^2 + \left( 20 + x \right)^2 + 2 \left( 20 + x \right) \left( 6 - x \right) \\ | & = \left( 6 + x \right)^2 + \left( 20 + x \right)^2 + 2 \left( 20 + x \right) \left( 6 - x \right) \\ | ||
& = 24 x + 676 . | & = 24 x + 676 . | ||
− | \end{align*} | + | \end{align*}</cmath> |
Because <math>AC = AP + PQ + QC = 28</math>, we get <math>x = \frac{9}{2}</math>. | Because <math>AC = AP + PQ + QC = 28</math>, we get <math>x = \frac{9}{2}</math>. | ||
Line 121: | Line 136: | ||
Therefore, | Therefore, | ||
− | \begin{align*} | + | <cmath>\begin{align*} |
{\rm Area} \ ABCD & = CB \cdot EF \\ | {\rm Area} \ ABCD & = CB \cdot EF \\ | ||
& = \left( 20 + x \right) \cdot 2r \\ | & = \left( 20 + x \right) \cdot 2r \\ | ||
& = 147 \sqrt{3} . | & = 147 \sqrt{3} . | ||
− | \end{align*} | + | \end{align*}</cmath> |
Therefore, the answer is <math>147 + 3 = \boxed{\textbf{(150) }}</math>. | Therefore, the answer is <math>147 + 3 = \boxed{\textbf{(150) }}</math>. | ||
Line 136: | Line 151: | ||
1. By the half-base-height formula, <math>[ABC]=r(20+BX)</math>. | 1. By the half-base-height formula, <math>[ABC]=r(20+BX)</math>. | ||
− | 2. We can drop altitudes from the center <math>O</math> of <math>\omega</math> to <math>AB</math>, <math>BC</math>, and <math>AC</math>, which have lengths <math>r</math>, <math>r</math>, and <math>\sqrt{r^2-81 | + | 2. We can drop altitudes from the center <math>O</math> of <math>\omega</math> to <math>AB</math>, <math>BC</math>, and <math>AC</math>, which have lengths <math>r</math>, <math>r</math>, and <math>\sqrt{r^2-\frac{81}{4}}</math>. Thus, <math>[ABC]=[OAB]+[OBC]+[OAC]=r(BX+13)+14\sqrt{r^2-\frac{81}{4}}</math>. |
Equating the two expressions for <math>[ABC]</math> and solving for <math>r</math> yields <math>r=3\sqrt{3}</math>. | Equating the two expressions for <math>[ABC]</math> and solving for <math>r</math> yields <math>r=3\sqrt{3}</math>. | ||
Line 143: | Line 158: | ||
~ Leo.Euler | ~ Leo.Euler | ||
+ | |||
+ | ==Solution 5== | ||
+ | [[File:AIME-I-2022-11.png|530px|right]] | ||
+ | Let <math>\omega</math> be the circle, let <math>r</math> be the radius of <math>\omega</math>, and let the points at which <math>\omega</math> is tangent to <math>AB</math>, <math>BC</math>, and <math>AD</math> be <math>H</math>, <math>K</math>, and <math>T</math>, respectively. PoP on <math>A</math> and <math>C</math> with respect to <math>\omega</math> yields <cmath>AT=6, CK=20.</cmath> | ||
+ | |||
+ | Let <math>TG = AC, CG||AT.</math> | ||
+ | |||
+ | In <math>\triangle KGT</math> <math>KT \perp BC,</math> | ||
+ | <math>KT = \sqrt{GT^2 – (KC + AT)^2} = 6 \sqrt{3}=2r.</math> | ||
+ | |||
+ | <math>\angle AOB = 90^{\circ}, OH \perp AB, OH = r = \frac{KT}{2},</math> | ||
+ | <cmath>OH^2 = AH \cdot BH \implies BH = \frac {9}{2}.</cmath> | ||
+ | |||
+ | Area is <cmath>(BK + KC) \cdot KT = (BH + KC) \cdot 2r = \frac{49}{2} \cdot 6\sqrt{3} = 147 \sqrt{3} \implies 147+3 = \boxed{\textbf{150}}.</cmath> | ||
+ | |||
+ | '''vladimir.shelomovskii@gmail.com, vvsss''' | ||
+ | |||
+ | ==Solution 6 (Short and Sweet)== | ||
+ | |||
+ | |||
+ | |||
+ | Let <math>O</math> be the center of the circle. Let points <math>M, N</math> and <math>L</math> be the tangent points of lines <math>BC, AD</math> and <math>AB</math> respectively to the circle. By Power of a Point, <math>({MC})^2=16\cdot{25} \Longrightarrow MC=20</math>. Similarly, <math>({AL})^2=3\cdot{12} \Longrightarrow AL=6</math>. Notice that <math>AL=AN=6</math> since quadrilateral <math>LONA</math> is symmetrical. Let <math>AC</math> intersect <math>MN</math> at <math>I</math>. Then, <math>\bigtriangleup{IMC}</math> is similar to <math>\bigtriangleup{AIN}</math>. Therefore, <math>\frac{CI}{MC}=\frac{AI}{AN}</math>. Let the length of <math>PI=l</math>, then <math>\frac{25-l}{20}=\frac{3+l}{6}</math>. Solving we get <math>l=\frac{45}{13}</math>. Doing the Pythagorean theorem on triangles <math>IMC</math> and <math>AIN</math> for sides <math>MI</math> and <math>IN</math> respectively, we obtain the equation <math>\sqrt{(\frac{280}{13})^2-400} +\sqrt{(\frac{84}{13})^2-36}=MN=2r_1</math> where <math>r_1</math> denotes the radius of the circle. Solving, we get <math>MN=6\sqrt{3}</math>. Additionally, quadrilateral <math>OLBM</math> is symmetrical so <math>OL=OM</math>. Let <math>OL=OM=x</math> and extend a perpendicular foot from <math>B</math> to <math>AD</math> and call it <math>R</math>. Then, <math>\bigtriangleup{ABR}</math> is right with <math>AR=6-x</math>, <math>AB=6+x</math>, and <math>RB=2r_1=MN=6\sqrt{3}</math>. Taking the difference of squares, we get <math>108=24x \Longrightarrow x=\frac{9}{2}</math>. The area of <math>ABCD</math> is <math>MN\cdot{BC}=(20+x)\cdot{MN} \Longrightarrow \frac{49}{2}\cdot{6\sqrt{3}}=147\sqrt{3}</math>. Therefore, the answer is <math>147+3=\boxed{150}</math> | ||
+ | |||
+ | ~[https://artofproblemsolving.com/wiki/index.php/User:Magnetoninja Magnetoninja] | ||
+ | |||
+ | ==Solution 7 (Intuitive, no trig, no weird auxiliary lines)== | ||
+ | |||
+ | Say that <math>BC</math> is tangent to the circle at <math>X</math> and <math>AD</math> tangent at <math>Y</math>. Also, <math>H</math> is the intersection of <math>XY</math> (diameter) and <math>AC</math> (diagonal). Then by power of a point with given info on <math>A</math> and <math>C</math> we get that <math>AY=6</math> and <math>CX=20</math>. Note that <math>HAY \sim HCX</math>, and since <math>\frac{AY}{CX}=\frac{3}{10}</math> we note that <cmath>\frac{AH}{CH} = \frac{AP+PH}{CQ+QH} = \frac{3+PH}{16+QH} =\frac{AY}{CX}=\frac{3}{10}</cmath>. Since <math>PH+HQ=9</math>, we get that <math>PH=\frac{45}{13}</math> and <math>QH=\frac{72}{13}</math>. This is the length information within the circle. | ||
+ | The same triangle similarity also means that <math>\frac{YH}{XH}=\frac{3}{10}</math>, so if the radius of the circle is <math>r</math> then we have <math>XH=\frac{20}{13}r</math> and <math>YH = \frac{6}{13}r</math>. | ||
+ | By power of a point on H, we can figure out <math>r</math>: | ||
+ | <cmath>XH\cdot YH = PH \cdot QG</cmath> | ||
+ | <cmath>\frac{20}{13}r \cdot \frac{6}{13}r = \frac{45}{13} \cdot \frac{72}{13}</cmath> | ||
+ | and we get that <math>r = 3 \sqrt 3</math>. Thus, we have that the height of the parallelogram is <math>2r=6 \sqrt 3</math> and we want to find <math>BC</math>. If <math>AB</math> is tangent to the circle at <math>E</math>, then set <math>a = BX = BE</math>. Using pythagorean theorem, <math>AO^2+BO^2=AB^2</math> and we can plug in diagram values: <cmath>(AY^2+OY^2)+(BX^2+OX)^2=AB^2</cmath> <cmath>(6^2+(3 \sqrt 3)^2) + (a^2+(3 \sqrt 3)^2)=(a+6)^2.</cmath> Solving, we get <math>a=\frac{9}{2}</math> | ||
+ | Finally, we have <math>[ABCD]=XY \cdot BC = 6 \sqrt 3 \cdot (20+\frac{9}{2}) \rightarrow \boxed{150}</math> | ||
+ | |||
+ | ~ Brocolimanx | ||
+ | |||
+ | ==Solution 8 (Ptolemy's Theorem + Power of Point + Pythagorean Theorem)== | ||
+ | Let <math>E</math>, <math>F</math>, <math>G</math> be the circle's point of tangency with sides <math>AD</math>, <math>AB</math>, and <math>BC</math>, respectively. Let <math>O</math> be the center of the inscribed circle. | ||
+ | |||
+ | By Power of a Point, <math>AE^2 = AP \cdot AQ = 3(3+9) = 36</math>, so <math>AE = 6</math>. Similarly, <math>GC^2 = CQ \cdot CP = 16(16+9) = 400</math>, so <math>GC = 20</math>. | ||
+ | |||
+ | Construct <math>GE</math>, and let <math>I</math> be the point of intersection of <math>GE</math> and <math>AC</math>. <math>GE \perp BC</math> and <math>GE \perp AD</math>. By AA, <math>\triangle IGC \sim \triangle IEA</math>, and we have <math>\frac{AI}{IC} = \frac{AE}{GC} = \frac{3}{10}</math>. We also know <math>AI + IC = AC = 28</math>, so <math>AI = \frac{84}{13}</math> and <math>IC = \frac{280}{13}</math>. | ||
+ | |||
+ | Using Pythagorean Theorem on <math>\triangle IEA</math> and <math>\triangle CIG</math>, we find that <math>EI = \frac{18\sqrt{3}}{13}</math> and <math>IG = \frac{60\sqrt{3}}{13}</math>. Thus, <math>GE = EI + IG = 6\sqrt{3}</math>, and the radius of the circle is <math>3\sqrt{3}</math>. | ||
+ | |||
+ | Construct <math>EF</math>, <math>FG</math>. <math>\angle AFO = \angle AEO = 90^{\circ}</math>, so <math>AEOF</math> is cyclic. Similarly, <math>BFOG</math> is cyclic. | ||
+ | |||
+ | Now, we attempt to set up Ptolemy. Using Pythagorean Theorem on <math>\triangle AEO</math>, we find that <math>AO = 3\sqrt{7}</math>. By Ptolemy's Theorem, <math>(AE)(FO) + (AF)(EO) = (AO)(FE)</math>, from which we have <math>(6)(3\sqrt{3}) + (6)(3\sqrt{3}) = (3\sqrt{7})(FE)</math> and <math>FE = 12\frac{\sqrt{3}}{\sqrt{7}}</math>. From Thales' Circle, <math>\triangle FGE</math> is a right triangle, and <math>EF^2 + FG^2 = GE^2</math>, so <math>FG = \frac{18}{\sqrt{7}}</math>. | ||
+ | |||
+ | Set <math>BF = BG = s</math>. <math>BO = \sqrt{s^2 + (3\sqrt{3})^2} = \sqrt{s^2+27}</math>, so by Ptolemy's Theorem on <math>BFOG</math>, we have | ||
+ | |||
+ | <cmath> | ||
+ | (BF)(GO) + (BG)(FO) = (FG)(BO) | ||
+ | </cmath> | ||
+ | <cmath> | ||
+ | (3\sqrt{3})(s) + (3\sqrt{3})(s) = (\frac{18}{\sqrt{7}})(\sqrt{s^2+27}) | ||
+ | </cmath> | ||
+ | Solving yields <math>s = \frac{9}{2}</math>. | ||
+ | |||
+ | We know that <math>BC = BG + GC = 20 + \frac{9}{2} = \frac{49}{2}</math>, so the area of <math>ABCD = (\frac{49}{2})(6\sqrt{3}) = 147\sqrt{3}</math>. The requested answer is <math>147 + 3 = \boxed{150}</math>. | ||
+ | |||
+ | ~ adam_zheng | ||
==Video Solution== | ==Video Solution== | ||
Line 152: | Line 230: | ||
==Video Solution 2 (Mathematical Dexterity)== | ==Video Solution 2 (Mathematical Dexterity)== | ||
https://www.youtube.com/watch?v=1nDKQkr9NaU | https://www.youtube.com/watch?v=1nDKQkr9NaU | ||
+ | |||
+ | == Video Solution 3 by OmegaLearn == | ||
+ | https://youtu.be/LpOegT0fKy8?t=740 | ||
+ | |||
+ | ~ pi_is_3.14 | ||
==See Also== | ==See Also== | ||
{{AIME box|year=2022|n=I|num-b=10|num-a=12}} | {{AIME box|year=2022|n=I|num-b=10|num-a=12}} | ||
+ | |||
+ | [[Category:Intermediate Geometry Problems]] | ||
{{MAA Notice}} | {{MAA Notice}} |
Latest revision as of 20:22, 31 January 2024
Contents
- 1 Problem
- 2 Video Solution by Punxsutawney Phil (Currently Privated)
- 3 Solution 1 (No trig)
- 4 Solution 2
- 5 Solution 3
- 6 Solution 4
- 7 Solution 5
- 8 Solution 6 (Short and Sweet)
- 9 Solution 7 (Intuitive, no trig, no weird auxiliary lines)
- 10 Solution 8 (Ptolemy's Theorem + Power of Point + Pythagorean Theorem)
- 11 Video Solution
- 12 Video Solution 2 (Mathematical Dexterity)
- 13 Video Solution 3 by OmegaLearn
- 14 See Also
Problem
Let be a parallelogram with
A circle tangent to sides
and
intersects diagonal
at points
and
with
as shown. Suppose that
and
Then the area of
can be expressed in the form
where
and
are positive integers, and
is not divisible by the square of any prime. Find
Video Solution by Punxsutawney Phil (Currently Privated)
https://www.youtube.com/watch?v=1m3pqCgwLFE
Solution 1 (No trig)
Let's redraw the diagram, but extend some helpful lines.
We obviously see that we must use power of a point since they've given us lengths in a circle and there are intersection points. Let be our tangents from the circle to the parallelogram. By the secant power of a point, the power of
. Then
. Similarly, the power of
and
. We let
and label the diagram accordingly.
Notice that because . Let
be the center of the circle. Since
and
intersect
and
, respectively, at right angles, we have
is a right-angled trapezoid and more importantly, the diameter of the circle is the height of the triangle. Therefore, we can drop an altitude from
to
and
to
, and both are equal to
. Since
,
. Since
and
. We can now use Pythagorean theorem on
; we have
and
.
We know that because
is a parallelogram. Using Pythagorean theorem on
,
. Therefore, base
. Thus the area of the parallelogram is the base times the height, which is
and the answer is
~KingRavi
Solution 2
Let the circle tangent to at
separately, denote that
Using POP, it is very clear that , let
, using LOC in
,
, similarly, use LOC in
, getting that
. We use the second equation to minus the first equation, getting that
, we can get
.
Now applying LOC in , getting
, solving this equation to get
, then
,
, the area is
leads to
~bluesoul,HarveyZhang
Solution 3
Denote by the center of the circle. Denote by
the radius of the circle.
Denote by
,
,
the points that the circle meets
,
,
at, respectively.
Because the circle is tangent to ,
,
,
,
,
,
.
Because ,
,
,
are collinear.
Following from the power of a point, . Hence,
.
Following from the power of a point, . Hence,
.
Denote . Because
and
are tangents to the circle,
.
Because is a right trapezoid,
.
Hence,
.
This can be simplified as
In , by applying the law of cosines, we have
Because , we get
.
Plugging this into Equation (1), we get
.
Therefore,
Therefore, the answer is .
~Steven Chen (www.professorchenedu.com)
Solution 4
Let be the circle, let
be the radius of
, and let the points at which
is tangent to
,
, and
be
,
, and
, respectively. Note that PoP on
and
with respect to
yields
and
. We can compute the area of
in two ways:
1. By the half-base-height formula, .
2. We can drop altitudes from the center of
to
,
, and
, which have lengths
,
, and
. Thus,
.
Equating the two expressions for and solving for
yields
.
Let . By the Parallelogram Law,
. Solving for
yields
. Thus,
, for a final answer of
.
~ Leo.Euler
Solution 5
Let be the circle, let
be the radius of
, and let the points at which
is tangent to
,
, and
be
,
, and
, respectively. PoP on
and
with respect to
yields
Let
In
Area is
vladimir.shelomovskii@gmail.com, vvsss
Solution 6 (Short and Sweet)
Let be the center of the circle. Let points
and
be the tangent points of lines
and
respectively to the circle. By Power of a Point,
. Similarly,
. Notice that
since quadrilateral
is symmetrical. Let
intersect
at
. Then,
is similar to
. Therefore,
. Let the length of
, then
. Solving we get
. Doing the Pythagorean theorem on triangles
and
for sides
and
respectively, we obtain the equation
where
denotes the radius of the circle. Solving, we get
. Additionally, quadrilateral
is symmetrical so
. Let
and extend a perpendicular foot from
to
and call it
. Then,
is right with
,
, and
. Taking the difference of squares, we get
. The area of
is
. Therefore, the answer is
Solution 7 (Intuitive, no trig, no weird auxiliary lines)
Say that is tangent to the circle at
and
tangent at
. Also,
is the intersection of
(diameter) and
(diagonal). Then by power of a point with given info on
and
we get that
and
. Note that
, and since
we note that
. Since
, we get that
and
. This is the length information within the circle.
The same triangle similarity also means that
, so if the radius of the circle is
then we have
and
.
By power of a point on H, we can figure out
:
and we get that
. Thus, we have that the height of the parallelogram is
and we want to find
. If
is tangent to the circle at
, then set
. Using pythagorean theorem,
and we can plug in diagram values:
Solving, we get
Finally, we have
~ Brocolimanx
Solution 8 (Ptolemy's Theorem + Power of Point + Pythagorean Theorem)
Let ,
,
be the circle's point of tangency with sides
,
, and
, respectively. Let
be the center of the inscribed circle.
By Power of a Point, , so
. Similarly,
, so
.
Construct , and let
be the point of intersection of
and
.
and
. By AA,
, and we have
. We also know
, so
and
.
Using Pythagorean Theorem on and
, we find that
and
. Thus,
, and the radius of the circle is
.
Construct ,
.
, so
is cyclic. Similarly,
is cyclic.
Now, we attempt to set up Ptolemy. Using Pythagorean Theorem on , we find that
. By Ptolemy's Theorem,
, from which we have
and
. From Thales' Circle,
is a right triangle, and
, so
.
Set .
, so by Ptolemy's Theorem on
, we have
Solving yields
.
We know that , so the area of
. The requested answer is
.
~ adam_zheng
Video Solution
https://www.youtube.com/watch?v=FeM_xXiJj0c&t=1s
~Steven Chen (www.professorchenedu.com)
Video Solution 2 (Mathematical Dexterity)
https://www.youtube.com/watch?v=1nDKQkr9NaU
Video Solution 3 by OmegaLearn
https://youtu.be/LpOegT0fKy8?t=740
~ pi_is_3.14
See Also
2022 AIME I (Problems • Answer Key • Resources) | ||
Preceded by Problem 10 |
Followed by Problem 12 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
All AIME Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.