Difference between revisions of "2020 AMC 12B Problems/Problem 12"
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==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 | ||
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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)== | ||
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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> | 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> | + | 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) | ~Math_Wiz_3.14 (legibility changes by eagleye) | ||
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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 |
Revision as of 15:23, 3 July 2021
Contents
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
~MRENTHUSIASM (by Geometry Expressions)
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
Video Solutions
https://www.youtube.com/watch?v=h-hhRa93lK4
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.