Difference between revisions of "2011 AMC 12B Problems/Problem 20"
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<math>\textbf{(A)}\ 24 \qquad \textbf{(B)}\ 14\sqrt{3} \qquad \textbf{(C)}\ \frac{195}{8} \qquad \textbf{(D)}\ \frac{129\sqrt{7}}{14} \qquad \textbf{(E)}\ \frac{69\sqrt{2}}{4}</math> | <math>\textbf{(A)}\ 24 \qquad \textbf{(B)}\ 14\sqrt{3} \qquad \textbf{(C)}\ \frac{195}{8} \qquad \textbf{(D)}\ \frac{129\sqrt{7}}{14} \qquad \textbf{(E)}\ \frac{69\sqrt{2}}{4}</math> | ||
− | + | ==Solution 1 (Coordinates)== | |
− | |||
Let us also consider the circumcircle of <math>\triangle ADF</math>. | Let us also consider the circumcircle of <math>\triangle ADF</math>. | ||
Note that if we draw the perpendicular bisector of each side, we will have the circumcenter of <math>\triangle ABC</math> which is <math>P</math>, Also, since <math>m\angle ADP = m\angle AFP = 90^\circ</math>. <math>ADPF</math> is cyclic, similarly, <math>BDPE</math> and <math>CEPF</math> are also cyclic. With this, we know that the circumcircles of <math>\triangle ADF</math>, <math>\triangle BDE</math> and <math>\triangle CEF</math> all intersect at <math>P</math>, so <math>P</math> is <math>X</math>. | Note that if we draw the perpendicular bisector of each side, we will have the circumcenter of <math>\triangle ABC</math> which is <math>P</math>, Also, since <math>m\angle ADP = m\angle AFP = 90^\circ</math>. <math>ADPF</math> is cyclic, similarly, <math>BDPE</math> and <math>CEPF</math> are also cyclic. With this, we know that the circumcircles of <math>\triangle ADF</math>, <math>\triangle BDE</math> and <math>\triangle CEF</math> all intersect at <math>P</math>, so <math>P</math> is <math>X</math>. | ||
− | The question now becomes | + | The question now becomes calculating the sum of the distance from each vertex to the circumcenter. |
We can calculate the distances with coordinate geometry. (Note that <math>XA = XB = XC</math> because <math>X</math> is the circumcenter.) | We can calculate the distances with coordinate geometry. (Note that <math>XA = XB = XC</math> because <math>X</math> is the circumcenter.) | ||
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Let <math>A = (5,12)</math>, <math>B = (0,0)</math>, <math>C = (14, 0)</math>, <math>X= (x_0, y_0)</math> | Let <math>A = (5,12)</math>, <math>B = (0,0)</math>, <math>C = (14, 0)</math>, <math>X= (x_0, y_0)</math> | ||
− | Then <math>X</math> is on the line <math>x = 7</math> and also the line with slope <math>-\frac{5}{12}</math> | + | Then <math>X</math> is on the line <math>x = 7</math> and also the line with slope <math>-\frac{5}{12}</math> that passes through <math>(2.5, 6)</math> (realize this is due to the fact that <math>XD</math> is the perpendicular bisector of <math>AB</math>). |
<math>y_0 = 6-\frac{45}{24} = \frac{33}{8}</math> | <math>y_0 = 6-\frac{45}{24} = \frac{33}{8}</math> | ||
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So <math>X = (7, \frac{33}{8})</math> | So <math>X = (7, \frac{33}{8})</math> | ||
− | and <math>XA +XB+XC = 3XB = 3\sqrt{7^2 + \left(\frac{33}{8}\right)^2} = 3\times\frac{65}{8}=\frac{195}{8}</math> | + | and <math>XA +XB+XC = 3XB = 3\sqrt{7^2 + \left(\frac{33}{8}\right)^2} = 3\times\frac{65}{8}=\boxed{\frac{195}{8}}</math> |
− | + | Remark: the intersection of the three circles is called a Miquel point. | |
− | Consider an additional circumcircle on <math>\triangle ADF</math>. After drawing the diagram, it is noticed that each triangle has side values: <math>7</math>, <math>\frac{15}{2}</math>, <math>\frac{13}{2}</math>. Thus they are congruent, and their respective circumcircles are. | + | |
+ | ==Solution 2 (Algebra)== | ||
+ | Consider an additional circumcircle on <math>\triangle ADF</math>. After drawing the diagram, it is noticed that each triangle has side values: <math>7</math>, <math>\frac{15}{2}</math>, <math>\frac{13}{2}</math>. Thus they are congruent, and their respective circumcircles are. | ||
+ | |||
+ | |||
+ | Let <math>M</math> & <math>N</math> be <math>\triangle BDE</math> & <math>\triangle CEF</math>'s circumcircles' respective centers. Since <math>\triangle BDE</math> & <math>\triangle CEF</math> are congruent, the distance <math>M</math> & <math>N</math> each are from <math>\overline{BC}</math> are equal, so <math>\overline{MN} || \overline{BC}</math>. The angle between <math>\overline {MN}</math> & <math>\overline{EX}</math> is <math>90^{\circ}</math>, and since <math>\overline{MN} || \overline{BC}</math>, <math>\angle XEC</math> is also <math>90^{\circ}</math>. <math>\triangle XEC</math> is a right triangle inscribed in a circle, so <math>\overline{XC}</math> must be the diameter of <math>N</math>. Using the same logic & reasoning, we could deduce that <math>XA</math> & <math>XB</math> are also circumdiameters. | ||
+ | |||
+ | |||
+ | Since the circumcircles are congruent, circumdiameters <math>XA</math>, <math>XB</math>, and <math>XC</math> are congruent. Therefore, the solution can be found by calculating one of these circumdiameters and multiplying it by a factor of <math>3</math>. We can find the circumradius quite easily with the formula <math>\sqrt{(s)(s-a)(s-b)(s-c)} = \frac{abc}{4R}</math>, such that <math>s=\frac{a+b+c}{2}</math> and <math>R</math> is the circumradius. Since <math>s = \frac{21}{2}</math>: | ||
<cmath> \sqrt{(\frac{21}{2})(4)(3)(\frac{7}{2})} = \frac{\frac{15}{2}\cdot\frac{13}{2}\cdot 7}{4R} </cmath> | <cmath> \sqrt{(\frac{21}{2})(4)(3)(\frac{7}{2})} = \frac{\frac{15}{2}\cdot\frac{13}{2}\cdot 7}{4R} </cmath> | ||
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After a few algebraic manipulations: | After a few algebraic manipulations: | ||
− | <math>\Rightarrow R = \frac{65}{16} \Rightarrow | + | <math>\Rightarrow R=\frac{65}{16} \Rightarrow XA = XB = XC = \frac{65}{8} \Rightarrow XA + XB + XC = \boxed{\frac{195}{8}}</math>. |
+ | |||
+ | ==Solution 3 (Homothety)== | ||
+ | Let <math>O</math> be the circumcenter of <math>\triangle ABC,</math> and <math>h_A</math> denote the length of the altitude from <math>A.</math> Note that a homothety centered at <math>B</math> with ratio <math>\frac{1}{2}</math> takes the circumcircle of <math>\triangle BAC</math> to the circumcircle of <math>\triangle BDE</math>. It also takes the point diametrically opposite <math>B</math> on the circumcircle of <math>\triangle BAC</math> to <math>O.</math> Therefore, <math>O</math> lies on the circumcircle of <math>\triangle BDE.</math> Similarly, it lies on the circumcircle of <math>\triangle CEF.</math> By Pythagorean triples, <math>h_A=12.</math> Finally, our answer is <cmath>3R=3\cdot \frac{abc}{4\{ABC\}}=3\cdot \frac{abc}{2ah_A}=3\cdot \frac{bc}{2h_A}=\boxed{\frac{195}{8}}.</cmath> | ||
+ | |||
+ | |||
+ | ==Solution 4 (basically Solution 1 but without coordinates)== | ||
+ | |||
+ | Since Solution 1 has already proven that the circumcenter of <math>\triangle ABC</math> coincides with <math>X</math>, we'll go from there. Note that the radius of the circumcenter of any given triangle is <math>\frac{a}{2\sin{A}}</math>, and since <math>b=15</math> and <math>\sin{B}=\frac{12}{13}</math>, it can be easily seen that <math>XA = XB = XC = \frac{65}{8}</math> and therefore our answer is <cmath>3\cdot \frac{65}{8}=\boxed{\frac{195}{8}}.</cmath> | ||
+ | |||
+ | ==Solution 5== | ||
+ | <center> | ||
+ | [[File:Screen Shot 2021-08-06 at 7.30.10 PM.png|300px]] | ||
+ | </center> | ||
+ | |||
+ | Since <math>ED</math> is a midline of <math>\triangle CAB,</math> we have that <math>\triangle CED \sim \triangle CAB</math> with a side length ratio of <math>1:2.</math> | ||
+ | |||
+ | Consider a homothety of scale factor <math>2</math> with on <math>\triangle CED</math> concerning point <math>C</math>. Note that this sends <math>(CEDX)</math> to <math>(ABCC')</math> with <math>CX=XC'.</math> By properties of homotheties, <math>C,X,</math> and <math>C'</math> are collinear. Similarly, we obtain that <math>BX=XB',</math> with all three points collinear. Let <math>O</math> denote the circumcenter of <math>\triangle ABC.</math> It is well-known that <math>OX \perp CC'</math> and analogously <math>OX \perp BB'.</math> However, there is only one perpendicular line to <math>OX</math> passing through <math>X,</math>, therefore, <math>O</math> coincides with <math>X.</math> | ||
+ | |||
+ | It follows that <math>AX=BX=CX=R,</math> where <math>R</math> is the circumradius of <math>\triangle ABC,</math> and this can be computed using the formula <cmath>R=\frac{abc}{4[ABC]},</cmath> from which we quickly obtain <cmath>R=\frac{65}{8} \implies AX+BX+CX=\boxed{\frac{195}{8}}.</cmath> | ||
+ | |||
+ | == Solution 6 (Trigonometry) == | ||
+ | |||
+ | [[File:2011AMC12B20.png|center|500px]] | ||
+ | |||
+ | <math>\angle BXE = \angle BDE</math>, <math>\angle CXE = \angle CFE</math>, as the angles are on the same circle. | ||
+ | |||
+ | <math>\triangle BDE \sim \triangle ABC</math>, <math>\triangle CFE \sim \triangle ABC</math> | ||
+ | |||
+ | <math>\angle BDE = \angle A</math>, <math>\angle CFE = \angle A</math> | ||
+ | |||
+ | <math>\angle BXE = \angle A</math>, <math>\angle CXE = \angle A</math> | ||
+ | |||
+ | Therefore <math>\angle BXE = \angle CXE</math>, and <math>XE</math> is the angle bisector of <math>\triangle XBC</math>. By the angle bisector theorem <math>\frac{XB}{XC} = \frac{BE}{CE} = 1</math>, <math>XB = XC</math>. In a similar fashion <math>XA = XB = XC = R</math>, where <math>R</math> is the circumcircle of <math>\triangle ABC</math>. | ||
+ | |||
+ | By the law of cosine, <math>\cos A = \frac{13^2 + 15^2 - 14^2}{2 \cdot 13 \cdot 15} = \frac{33}{65}</math>, <math>\sin A = \sqrt{1 - \left(\frac{33}{65}\right)^2} = \frac{56}{65}</math> | ||
+ | |||
+ | By the extended law of sines, <math>2R = \frac{BC}{\sin A} = \frac{14}{\frac{56}{65}} = \frac{65}{4}</math>, <math>R = \frac{65}{8}</math> | ||
+ | |||
+ | <math>XA + XB + XC = 3 R = \boxed{\textbf{(C) } \frac{195}{8}}</math> | ||
+ | |||
+ | ~[https://artofproblemsolving.com/wiki/index.php/User:Isabelchen isabelchen] | ||
+ | |||
+ | == Solution 7 (abwabwabwa)== | ||
+ | |||
+ | Claim, <math>X</math> is the circumcenter of triangle <math>\triangle{ABC}</math>. | ||
− | |||
− | |||
+ | Proof: Note that <math>\triangle{BDE}</math> and <math>\triangle{EFC}</math> are congruent. Consider the centers <math>O_1</math> and <math>O_2</math> of <math>\triangle{BDE}</math> and <math>\triangle{EFC}</math>, respectively. Let <math>B'</math> be the reflection of <math>B</math> over <math>O_1</math>, and let <math>C'</math> be the reflection of <math>C</math> over <math>O_2</math>. Since they form diameters, they must form right triangles <math>\triangle{BEB'}</math> and <math>\triangle{CEC'}</math>. However, because <math>\triangle{BDE} \cong \triangle{EFC}</math>, C' and B' are the same point. Thus, one point lies on both circumcircles, so this point is <math>X</math>. But then X lies on the perpendicular bisector of <math>BC</math>, and appyling this logic to all 3 sides, <math>X</math> must be the circumcenter. | ||
− | + | Memorizing that the circumradius of a <math>13, 14, 15</math> triangle is <math>\frac{65}{8}</math>, since <math>XA=XB=XC=\frac{65}{8}</math>, <math>XA+XB+XC = \boxed{\textbf{(C) }\frac{195}{8}}</math>. | |
− | + | -skibbysiggy | |
== See also == | == See also == | ||
{{AMC12 box|year=2011|num-b=19|num-a=21|ab=B}} | {{AMC12 box|year=2011|num-b=19|num-a=21|ab=B}} | ||
{{MAA Notice}} | {{MAA Notice}} |
Latest revision as of 16:28, 21 September 2024
Contents
[hide]Problem
Triangle has , and . The points , and are the midpoints of , and respectively. Let be the intersection of the circumcircles of and . What is ?
Solution 1 (Coordinates)
Let us also consider the circumcircle of .
Note that if we draw the perpendicular bisector of each side, we will have the circumcenter of which is , Also, since . is cyclic, similarly, and are also cyclic. With this, we know that the circumcircles of , and all intersect at , so is .
The question now becomes calculating the sum of the distance from each vertex to the circumcenter.
We can calculate the distances with coordinate geometry. (Note that because is the circumcenter.)
Let , , ,
Then is on the line and also the line with slope that passes through (realize this is due to the fact that is the perpendicular bisector of ).
So
and
Remark: the intersection of the three circles is called a Miquel point.
Solution 2 (Algebra)
Consider an additional circumcircle on . After drawing the diagram, it is noticed that each triangle has side values: , , . Thus they are congruent, and their respective circumcircles are.
Let & be & 's circumcircles' respective centers. Since & are congruent, the distance & each are from are equal, so . The angle between & is , and since , is also . is a right triangle inscribed in a circle, so must be the diameter of . Using the same logic & reasoning, we could deduce that & are also circumdiameters.
Since the circumcircles are congruent, circumdiameters , , and are congruent. Therefore, the solution can be found by calculating one of these circumdiameters and multiplying it by a factor of . We can find the circumradius quite easily with the formula , such that and is the circumradius. Since :
After a few algebraic manipulations:
.
Solution 3 (Homothety)
Let be the circumcenter of and denote the length of the altitude from Note that a homothety centered at with ratio takes the circumcircle of to the circumcircle of . It also takes the point diametrically opposite on the circumcircle of to Therefore, lies on the circumcircle of Similarly, it lies on the circumcircle of By Pythagorean triples, Finally, our answer is
Solution 4 (basically Solution 1 but without coordinates)
Since Solution 1 has already proven that the circumcenter of coincides with , we'll go from there. Note that the radius of the circumcenter of any given triangle is , and since and , it can be easily seen that and therefore our answer is
Solution 5
Since is a midline of we have that with a side length ratio of
Consider a homothety of scale factor with on concerning point . Note that this sends to with By properties of homotheties, and are collinear. Similarly, we obtain that with all three points collinear. Let denote the circumcenter of It is well-known that and analogously However, there is only one perpendicular line to passing through , therefore, coincides with
It follows that where is the circumradius of and this can be computed using the formula from which we quickly obtain
Solution 6 (Trigonometry)
, , as the angles are on the same circle.
,
,
,
Therefore , and is the angle bisector of . By the angle bisector theorem , . In a similar fashion , where is the circumcircle of .
By the law of cosine, ,
By the extended law of sines, ,
Solution 7 (abwabwabwa)
Claim, is the circumcenter of triangle .
Proof: Note that and are congruent. Consider the centers and of and , respectively. Let be the reflection of over , and let be the reflection of over . Since they form diameters, they must form right triangles and . However, because , C' and B' are the same point. Thus, one point lies on both circumcircles, so this point is . But then X lies on the perpendicular bisector of , and appyling this logic to all 3 sides, must be the circumcenter.
Memorizing that the circumradius of a triangle is , since , .
-skibbysiggy
See also
2011 AMC 12B (Problems • Answer Key • Resources) | |
Preceded by Problem 19 |
Followed by Problem 21 |
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.