Difference between revisions of "2020 AMC 12B Problems/Problem 12"
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<math>\textbf{(A)}\ 96 \qquad\textbf{(B)}\ 98 \qquad\textbf{(C)}\ 44\sqrt5 \qquad\textbf{(D)}\ 70\sqrt2 \qquad\textbf{(E)}\ 100</math> | <math>\textbf{(A)}\ 96 \qquad\textbf{(B)}\ 98 \qquad\textbf{(C)}\ 44\sqrt5 \qquad\textbf{(D)}\ 70\sqrt2 \qquad\textbf{(E)}\ 100</math> | ||
+ | |||
+ | ==Diagram== | ||
+ | <asy> | ||
+ | /* Made by Shihan; edited by MRENTHUSIASM */ | ||
+ | size(250); | ||
+ | |||
+ | pair O, A, B, C, D, E; | ||
+ | O = origin; | ||
+ | A = (-5*sqrt(2),0); | ||
+ | B = (5*sqrt(2),0); | ||
+ | E = (5*sqrt(2)-2*sqrt(5),0); | ||
+ | path p; | ||
+ | p = Circle(O,5*sqrt(2)); | ||
+ | C = intersectionpoint(p,E--E+10*dir(135)); | ||
+ | D = intersectionpoint(p,E--E+10*dir(-45)); | ||
+ | draw(p); | ||
+ | dot(O,linewidth(4)); | ||
+ | dot("$A$",A,1.5*dir(A),linewidth(4)); | ||
+ | dot("$B$",B,1.5*dir(B),linewidth(4)); | ||
+ | dot("$E$",E,1.5*dir(135/2),linewidth(4)); | ||
+ | dot("$C$",C,1.5*dir(C),linewidth(4)); | ||
+ | dot("$D$",D,1.5*dir(D),linewidth(4)); | ||
+ | draw(A--B^^C--D); | ||
+ | label("$45^\circ$",E,3.5*dir(155.5),red+fontsize(10)); | ||
+ | label("$5\sqrt2$",midpoint(A--O),S); | ||
+ | label("$2\sqrt5$",midpoint(E--B),S); | ||
+ | </asy> | ||
+ | ~Shihan ~MRENTHUSIASM | ||
==Solution 1 (Pythagorean Theorem) == | ==Solution 1 (Pythagorean Theorem) == | ||
− | Let <math>O</math> be the center of the circle, and <math>X</math> be the midpoint of <math>\overline{CD}</math>. Let <math>CX=a</math> and <math>EX=b</math>. This implies that <math>DE = a - b</math>. Since <math>CE = CX + EX = a + b</math>, we now want to find <math>(a+b)^2+(a-b)^2=2(a^2+b^2)</math>. Since <math>\angle CXO</math> is a right angle, by Pythagorean theorem <math>a^2 + b^2 = CX^2 + OX^2 = (5\sqrt{2})^2=50</math>. Thus, our answer is <math>2\times50=\boxed{\textbf{(E) } 100}</math>. | + | Let <math>O</math> be the center of the circle, and <math>X</math> be the midpoint of <math>\overline{CD}</math>. Let <math>CX=a</math> and <math>EX=b</math>. This implies that <math>DE = a - b</math>. Since <math>CE = CX + EX = a + b</math>, we now want to find <math>(a+b)^2+(a-b)^2=2(a^2+b^2)</math>. Since <math>\angle CXO</math> is a right angle, by Pythagorean theorem <math>a^2 + b^2 = CX^2 + OX^2 = (5\sqrt{2})^2=50</math>. Thus, our answer is <math>2\times50=\boxed{\textbf{(E)}\ 100}</math>. |
~JHawk0224 | ~JHawk0224 | ||
Line 18: | Line 46: | ||
By Power of a Point, <math>CE \cdot DE = AE \cdot BE = 2\sqrt5\cdot(10\sqrt2 - 2\sqrt5) = 20\sqrt{10} - 20</math>, so <math>2 \cdot CE \cdot DE = 40\sqrt{10} - 40</math>. | By Power of a Point, <math>CE \cdot DE = AE \cdot BE = 2\sqrt5\cdot(10\sqrt2 - 2\sqrt5) = 20\sqrt{10} - 20</math>, so <math>2 \cdot CE \cdot DE = 40\sqrt{10} - 40</math>. | ||
− | Finally, <math>CE^2 + DE^2 = (CE+ED)^2-2\cdot CE \cdot DE=(60 + 40\sqrt{10}) - (40\sqrt{10} - 40) = \boxed{(E) 100}</math>. | + | Finally, <math>CE^2 + DE^2 = (CE+ED)^2-2\cdot CE \cdot DE=(60 + 40\sqrt{10}) - (40\sqrt{10} - 40) = \boxed{\textbf{(E)}\ 100}</math>. |
==Solution 3 (Law of Cosines)== | ==Solution 3 (Law of Cosines)== | ||
− | Let <math>O</math> be the center of the circle. Notice how <math>OC = OD = r</math>, where <math>r</math> is the radius of the circle. By applying the law of cosines on triangle <math>OCE</math>, < | + | Let <math>O</math> be the center of the circle. Notice how <math>OC = OD = r</math>, where <math>r</math> is the radius of the circle. By applying the law of cosines on triangle <math>OCE</math>, <cmath>r^2=CE^2+OE^2-2(CE)(OE)\cos{45}=CE^2+OE^2-(CE)(OE)\sqrt{2}.</cmath> |
+ | |||
+ | Similarly, by applying the law of cosines on triangle <math>ODE</math>, <cmath>r^2=DE^2+OE^2-2(DE)(OE)\cos{135}=DE^2+OE^2+(DE)(OE)\sqrt{2}.</cmath> | ||
+ | |||
+ | By subtracting these two equations, we get <cmath>CE^2-DE^2-(CE)(OE)\sqrt{2}-(DE)(OE)\sqrt{2}=0.</cmath> We can rearrange it to get <cmath>CE^2-DE^2=(CE)(OE)\sqrt{2}+(DE)(OE)\sqrt{2}=(CE+DE)(OE\sqrt{2}).</cmath> | ||
− | ~Math_Wiz_3.14 | + | Because both <math>CE</math> and <math>DE</math> are both positive, we can safely divide both sides by <math>(CE+DE)</math> to obtain <math>CE-DE=OE\sqrt{2}</math>. Because <math>OE = OB - BE = 5\sqrt{2} - 2\sqrt{5}</math>, <cmath>(CE-DE)^2 = CE^2+DE^2 - 2(CE)(DE) = (OE\sqrt{2})^2 =2(5\sqrt{2} - 2\sqrt{5})^2 = 140 - 40\sqrt{10}.</cmath> |
+ | |||
+ | Through power of a point, we can find out that <math>(CE)(DE)=20\sqrt{10} - 20</math>, so <cmath>CE^2+DE^2 = (CE-DE)^2+ 2(CE)(DE)= (140 - 40\sqrt{10}) + 2(20\sqrt{10} - 20) = \boxed{\textbf{(E)}\ 100}.</cmath> | ||
+ | |||
+ | ~Math_Wiz_3.14 (legibility changes by eagleye) | ||
==Solution 4 (Reflections)== | ==Solution 4 (Reflections)== | ||
<asy> | <asy> | ||
− | + | /* Made by sofas103; edited by MRENTHUSIASM */ | |
− | + | size(250); | |
− | + | ||
− | + | pair O, A, B, C, D, E, D1; | |
− | + | O = origin; | |
− | + | A = (-5*sqrt(2),0); | |
− | + | B = (5*sqrt(2),0); | |
− | + | E = (5*sqrt(2)-2*sqrt(5),0); | |
− | + | path p; | |
− | + | p = Circle(O,5*sqrt(2)); | |
− | + | C = intersectionpoint(p,E--E+10*dir(135)); | |
− | + | D = intersectionpoint(p,E--E+10*dir(-45)); | |
− | dot(( | + | D1 = (D.x,-D.y); |
− | + | draw(p); | |
− | + | dot("$O$",O,1.5*S,linewidth(4)); | |
− | + | dot("$A$",A,1.5*dir(A),linewidth(4)); | |
− | + | dot("$B$",B,1.5*dir(B),linewidth(4)); | |
− | + | dot("$E$",E,1.5*dir(180+135/2),linewidth(4)); | |
− | draw( | + | dot("$C$",C,1.5*dir(C),linewidth(4)); |
− | + | dot("$D$",D,1.5*dir(D),linewidth(4)); | |
− | + | dot("$D'$",D1,1.5*dir(D1),linewidth(4)); | |
+ | draw(A--B^^C--D^^C--D1--O--cycle^^D1--E); | ||
</asy> | </asy> | ||
Let <math>O</math> be the center of the circle. By reflecting <math>D</math> across the line <math>AB</math> to produce <math>D'</math>, we have that <math>\angle BED'=45</math>. Since <math>\angle AEC=45</math>, <math>\angle CED'=90</math>. Since <math>DE=ED'</math>, by the Pythagorean Theorem, our desired solution is just <math>CD'^2</math>. | Let <math>O</math> be the center of the circle. By reflecting <math>D</math> across the line <math>AB</math> to produce <math>D'</math>, we have that <math>\angle BED'=45</math>. Since <math>\angle AEC=45</math>, <math>\angle CED'=90</math>. Since <math>DE=ED'</math>, by the Pythagorean Theorem, our desired solution is just <math>CD'^2</math>. | ||
Looking next to circle arcs, we know that <math>\angle AEC=\frac{\overarc{AC}+\overarc{BD}}{2}=45</math>, so <math>\overarc{AC}+\overarc{BD}=90</math>. Since <math>\overarc{BD'}=\overarc{BD}</math>, and <math>\overarc{AC}+\overarc{BD'}+\overarc{CD'}=180</math>, <math>\overarc{CD'}=90</math>. Thus, <math>\angle COD'=90</math>. | Looking next to circle arcs, we know that <math>\angle AEC=\frac{\overarc{AC}+\overarc{BD}}{2}=45</math>, so <math>\overarc{AC}+\overarc{BD}=90</math>. Since <math>\overarc{BD'}=\overarc{BD}</math>, and <math>\overarc{AC}+\overarc{BD'}+\overarc{CD'}=180</math>, <math>\overarc{CD'}=90</math>. Thus, <math>\angle COD'=90</math>. | ||
− | Since <math>OC=OD'=5\sqrt{2}</math>, by the Pythagorean Theorem, the desired <math>CD'^2= \boxed{\textbf{(E) } 100}</math>. | + | Since <math>OC=OD'=5\sqrt{2}</math>, by the Pythagorean Theorem, the desired <math>CD'^2= \boxed{\textbf{(E)}\ 100}</math>. |
~sofas103 | ~sofas103 | ||
− | ==Video | + | == Solution 5 (Basically Solution 2 With Motivation) == |
+ | |||
+ | Basically, by PoP, you have that <cmath>CE \times DE = (10\sqrt{2}-2\sqrt{5})(2\sqrt{5}) = 20\sqrt{10} - 20.</cmath> Therefore, as <math>CE^2 + DE^2 = (CD)^2 - 2(CE \times DE),</math> basically, once you find <math>CD^2,</math> the problem is done. Now, this is an IMPORTANT concept: If you have a circle which you know the radius of and you want to find the length of a chord of that circle, drop an altitude from the center of the circle to the chord to find distance between the center of the circle and the chord. | ||
+ | |||
+ | In this case, let <math>M</math> be the midpoint of chord <math>CD.</math> Notice that now we can use our <math>45^\circ{}</math> angle, since <math>OME</math> is a <math>45^\circ{}-45^\circ{}-90^\circ{}</math> triangle so that <math>ME = x</math> and <math>OE = x\sqrt{2}.</math> However, we have that <math>OE = 5\sqrt{2}-2\sqrt{5},</math> so that <math>x = 5 - \sqrt{10}.</math> Now, notice that <math>x^2 = 35 - 10\sqrt{10},</math> so that <cmath>CM^2 = 50 - x^2 = 50 - (35 - 10\sqrt{10}) = 15 + 10 \sqrt{10}</cmath> and <cmath>CD^2 = 60 + 4\sqrt{10}.</cmath> Therefore, <cmath>CE^2 + DE^2 = CD^2 - 2(CE \times DE) = (60+4\sqrt{10}) - 2(2\sqrt{10} - 20) = (60+4\sqrt{10}) + (40-4\sqrt{10}) = \boxed{\textbf{(E)}\ 100}.</cmath> | ||
+ | |||
+ | This may not be the “shortest solution”, but in my opinion is very well motivated and doesn’t require much creativity. [Not requiring much creativity, it also saves more time than you’d think. ;)] | ||
+ | |||
+ | ~ Professor-Mom | ||
+ | |||
+ | == Solution 6 (Double Power of a Point) == | ||
+ | |||
+ | <asy> | ||
+ | /* Made by sofas103; edited by MRENTHUSIASM */ | ||
+ | size(250); | ||
+ | |||
+ | pair O, A, B, C, D, E, P, Q; | ||
+ | O = origin; | ||
+ | A = (-5*sqrt(2),0); | ||
+ | B = (5*sqrt(2),0); | ||
+ | E = (5*sqrt(2)-2*sqrt(5),0); | ||
+ | P = (5*sqrt(2), -4.5); | ||
+ | Q=(5*sqrt(2)-2*sqrt(5)-0.5, 0); | ||
+ | path p; | ||
+ | p = Circle(O,5*sqrt(2)); | ||
+ | C = intersectionpoint(p,E--E+10*dir(135)); | ||
+ | D = intersectionpoint(p,E--E+10*dir(-45)); | ||
+ | draw(p); | ||
+ | dot("$A$",A,1.5*dir(A),linewidth(4)); | ||
+ | dot("$B$",B,1.5*dir(B),linewidth(4)); | ||
+ | dot("$E$",E,1.5*dir(180+135/2),linewidth(4)); | ||
+ | dot("$C$",C,1.5*dir(C),linewidth(4)); | ||
+ | dot("$D$",D,1.5*dir(D),linewidth(4)); | ||
+ | dot("$P$",P,1.5*dir(P),linewidth(4)); | ||
+ | label("$45^{\circ}$",Q,NW); | ||
+ | draw(A--B^^C--P--B); | ||
+ | draw((5*sqrt(2), 0)--(5*sqrt(2)-0.4, 0)--(5*sqrt(2)-0.4, -0.4)--(5*sqrt(2), -0.4)); | ||
+ | </asy> | ||
+ | For ease of notation, let <math>x = DE</math> and <math>y=EC</math>. | ||
+ | Extend <math>\overline{CD}</math> to point <math>P</math> until <math>\overline{BP}</math> is perpendicular to <math>AB</math>. It's given that <math>\angle AEC = 45^{\circ}</math>, so, by vertical angles, we have <math>\angle BEP = \angle EPB = 45^{\circ}</math>. | ||
+ | |||
+ | Since <math>PEB</math> is a <math>45-45-90</math> right triangle, we have <math>BE = BP = 2\sqrt{5}</math> and <math>PE=2\sqrt{10}</math>. Hence, <math>PD = 2\sqrt{10}-x.</math> | ||
+ | |||
+ | By Power of a Point, we have | ||
+ | |||
+ | <cmath>PB^2 = PD\cdot PC</cmath> | ||
+ | <cmath>20 = \left(2\sqrt{10}-x\right)\left(y + 2\sqrt{10}\right).</cmath> | ||
+ | |||
+ | Isolating the variables after expanding gives <math>x-y = 2\sqrt{10}-10.</math> | ||
+ | |||
+ | Using Power of a Point again, we have | ||
+ | |||
+ | <cmath>DE\cdot EC = BE\cdot EA</cmath> | ||
+ | <cmath>xy = 2\sqrt{5}\left(10\sqrt{2}-2\sqrt{5} \right)</cmath> | ||
+ | <cmath>xy = 20\sqrt{10}-20.</cmath> | ||
+ | To get <math>x^2 + y^2</math>, we can perform the operation <math>(x-y)^2 + 2xy</math>. Plugging these values in, | ||
+ | |||
+ | <cmath>(x-y)^2 + 2xy = \left(2\sqrt{10}-10\right)^2 + 2\left(20\sqrt{10}-20\right)</cmath> | ||
+ | <cmath> = 40 + 100 - 40\sqrt{10} + 40\sqrt{10} - 40 = \boxed{\textbf{(E)}\ 100}</cmath> | ||
+ | |||
+ | -Benedict T (countmath1) | ||
+ | |||
+ | == Solution 7 (Cheating) == | ||
+ | |||
+ | Perhaps not reliable in general, but very useful as a last resort. | ||
+ | The choice of the radius <math>5\sqrt2</math> is strange, and is probably motivated by a nice answer in the end, so we only consider integer options. Notice that a 5 also appears in the condition <math>BE=2\sqrt5</math>, therefore it will likely be present in the answer as well; the only integer containing a factor of 5 amongst the choices is 100, thus the answer is <math>\boxed{\textbf{(E)}\ 100}</math> | ||
+ | |||
+ | ~Maths357 | ||
+ | |||
+ | == Solution 7 (Coordinate Geometry + Vieta) == | ||
+ | |||
+ | set the origin at E (instead of circle center O) will reduce calculation. | ||
+ | line CD is y= -x , | ||
+ | let OE = a , circle radius r=<math>5\sqrt{2}</math>, | ||
+ | circle <math>(x + a)^2 + y^2 = r^2</math> intersects line y = -x at 2 points <math> C(X_c, Y_c)</math> and <math> D(X_d, Y_d) </math> | ||
+ | <math>X_c = -Y_c </math> , <math>X_d = -Y_d </math> | ||
+ | |||
+ | substitute y with x , circle becomes <math>(x + a)^2 + y^2 = (x + a)^2 + (-x)^2 = 2x^2 + 2ax + a^2 - r^2 = 0 </math> | ||
+ | |||
+ | <math>X_c , X_d</math> will be 2 roots of above quadratic and we will apply vieta <math>X_c+ X_d , X_c \cdot X_d </math> below | ||
+ | |||
+ | <cmath> CE^2 + DE^2 = ( (-X_c) + Y_c)^2 + ( X_d + (- Y_d) )^2 \\ | ||
+ | = ( (-X_c) + (-X_c))^2 + ( X_d + (X_d) )^2 \\ | ||
+ | = 4 (X_c^2 + X_d^2) \\ | ||
+ | = 4 ((X_c + X_d)^2) - 2X_cX_d \\ | ||
+ | = 4 ( (\frac{-2a}{2 \cdot 2})^2 + \frac{a^2- r^2}{2} ) | ||
+ | |||
+ | </cmath> | ||
+ | |||
+ | plug in a = OE = OB - OE = <math> 5\sqrt{2} - 2 \sqrt{5} </math> and <math>r = 5\sqrt{2}</math> we get answer <math>\boxed{\textbf{(E) }100}</math>. | ||
+ | [[Image:2020_AMC_12B_P20.PNG|thumb|center|600px]] | ||
+ | ~[https://artofproblemsolving.com/wiki/index.php/User:Cyantist luckuso] | ||
+ | |||
+ | == Video Solution by OmegaLearn == | ||
+ | https://youtu.be/LpOegT0fKy8?t=228 | ||
+ | |||
+ | ~ pi_is_3.14 | ||
+ | |||
+ | ==Video Solution by On The Spot STEM== | ||
https://www.youtube.com/watch?v=h-hhRa93lK4 | https://www.youtube.com/watch?v=h-hhRa93lK4 | ||
− | + | ==Video Solution by TheBeautyOfMath== | |
− | |||
https://youtu.be/0xgTR3UEqbQ | https://youtu.be/0xgTR3UEqbQ | ||
− | |||
− | |||
==See Also== | ==See Also== |
Latest revision as of 11:51, 17 August 2024
Contents
- 1 Problem
- 2 Diagram
- 3 Solution 1 (Pythagorean Theorem)
- 4 Solution 2 (Power of a Point)
- 5 Solution 3 (Law of Cosines)
- 6 Solution 4 (Reflections)
- 7 Solution 5 (Basically Solution 2 With Motivation)
- 8 Solution 6 (Double Power of a Point)
- 9 Solution 7 (Cheating)
- 10 Solution 7 (Coordinate Geometry + Vieta)
- 11 Video Solution by OmegaLearn
- 12 Video Solution by On The Spot STEM
- 13 Video Solution by TheBeautyOfMath
- 14 See Also
Problem
Let be a diameter in a circle of radius
Let
be a chord in the circle that intersects
at a point
such that
and
What is
Diagram
~Shihan ~MRENTHUSIASM
Solution 1 (Pythagorean Theorem)
Let be the center of the circle, and
be the midpoint of
. Let
and
. This implies that
. Since
, we now want to find
. Since
is a right angle, by Pythagorean theorem
. Thus, our answer is
.
~JHawk0224
Solution 2 (Power of a Point)
Let be the center of the circle, and
be the midpoint of
. Draw triangle
, and median
. Because
,
is isosceles, so
is also an altitude of
.
, and because angle
is
degrees and triangle
is right,
. Because triangle
is right,
. Thus,
.
We are looking for +
which is also
.
Because ,
.
By Power of a Point, , so
.
Finally, .
Solution 3 (Law of Cosines)
Let be the center of the circle. Notice how
, where
is the radius of the circle. By applying the law of cosines on triangle
,
Similarly, by applying the law of cosines on triangle ,
By subtracting these two equations, we get We can rearrange it to get
Because both and
are both positive, we can safely divide both sides by
to obtain
. Because
,
Through power of a point, we can find out that , so
~Math_Wiz_3.14 (legibility changes by eagleye)
Solution 4 (Reflections)
Let
be the center of the circle. By reflecting
across the line
to produce
, we have that
. Since
,
. Since
, by the Pythagorean Theorem, our desired solution is just
.
Looking next to circle arcs, we know that
, so
. Since
, and
,
. Thus,
.
Since
, by the Pythagorean Theorem, the desired
.
~sofas103
Solution 5 (Basically Solution 2 With Motivation)
Basically, by PoP, you have that Therefore, as
basically, once you find
the problem is done. Now, this is an IMPORTANT concept: If you have a circle which you know the radius of and you want to find the length of a chord of that circle, drop an altitude from the center of the circle to the chord to find distance between the center of the circle and the chord.
In this case, let be the midpoint of chord
Notice that now we can use our
angle, since
is a
triangle so that
and
However, we have that
so that
Now, notice that
so that
and
Therefore,
This may not be the “shortest solution”, but in my opinion is very well motivated and doesn’t require much creativity. [Not requiring much creativity, it also saves more time than you’d think. ;)]
~ Professor-Mom
Solution 6 (Double Power of a Point)
For ease of notation, let
and
.
Extend
to point
until
is perpendicular to
. It's given that
, so, by vertical angles, we have
.
Since is a
right triangle, we have
and
. Hence,
By Power of a Point, we have
Isolating the variables after expanding gives
Using Power of a Point again, we have
To get
, we can perform the operation
. Plugging these values in,
-Benedict T (countmath1)
Solution 7 (Cheating)
Perhaps not reliable in general, but very useful as a last resort.
The choice of the radius is strange, and is probably motivated by a nice answer in the end, so we only consider integer options. Notice that a 5 also appears in the condition
, therefore it will likely be present in the answer as well; the only integer containing a factor of 5 amongst the choices is 100, thus the answer is
~Maths357
Solution 7 (Coordinate Geometry + Vieta)
set the origin at E (instead of circle center O) will reduce calculation.
line CD is y= -x ,
let OE = a , circle radius r=,
circle
intersects line y = -x at 2 points
and
,
substitute y with x , circle becomes
will be 2 roots of above quadratic and we will apply vieta
below
plug in a = OE = OB - OE = and
we get answer
.
Video Solution by OmegaLearn
https://youtu.be/LpOegT0fKy8?t=228
~ pi_is_3.14
Video Solution by On The Spot STEM
https://www.youtube.com/watch?v=h-hhRa93lK4
Video Solution by TheBeautyOfMath
See Also
2020 AMC 12B (Problems • Answer Key • Resources) | |
Preceded by Problem 11 |
Followed by Problem 13 |
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.