Difference between revisions of "2014 AMC 12A Problems/Problem 20"
(→See Also) |
(→Solution 3) |
||
(35 intermediate revisions by 11 users not shown) | |||
Line 7: | Line 7: | ||
\textbf{(D) }14\qquad | \textbf{(D) }14\qquad | ||
\textbf{(E) }3\sqrt 3+9\qquad</math> | \textbf{(E) }3\sqrt 3+9\qquad</math> | ||
+ | [[Category: Introductory Geometry Problems]] | ||
− | ==Solution== | + | ==Solution 1== |
− | Let <math>C_1</math> be the reflection of <math>C</math> across <math>\overline{AB}</math>, and let <math>C_2</math> be the reflection of <math>C_1</math> across <math>\overline{AC}</math>. Then it is well-known that the quantity <math>BE+DE+CD</math> is minimized when it is equal to <math>C_2B</math>. (Proving this is a simple application of the triangle inequality; for an example of a simpler case, see | + | |
+ | |||
+ | Let <math>C_1</math> be the reflection of <math>C</math> across <math>\overline{AB}</math>, and let <math>C_2</math> be the reflection of <math>C_1</math> across <math>\overline{AC}</math>. Then it is well-known that the quantity <math>BE+DE+CD</math> is minimized when it is equal to <math>C_2B</math>. (Proving this is a simple application of the triangle inequality; for an example of a simpler case, see Heron's Shortest Path Problem.) As <math>A</math> lies on both <math>AB</math> and <math>AC</math>, we have <math>C_2A=C_1A=CA=6</math>. Furthermore, <math>\angle CAC_1=2\angle CAB=80^\circ</math> by the nature of the reflection, so <math>\angle C_2AB=\angle C_2AC+\angle CAB=80^\circ+40^\circ=120^\circ</math>. Therefore by the Law of Cosines <cmath>BC_2^2=6^2+10^2-2\cdot 6\cdot 10\cos 120^\circ=196\implies BC_2=\boxed{\textbf{(D) }14}.</cmath> | ||
+ | |||
+ | ==Solution 2== | ||
+ | In <math>\triangle BAC</math>, the three lines look like the Chinese character 又. Let <math>\triangle DEA</math>, <math>\triangle CDA</math>, and <math>\triangle BEA</math> have bases <math>DE</math>, <math>CD</math>, and <math>BE</math> respectively. Then, <math>\triangle DEA</math> has the same side <math>DA</math> as <math>\triangle CDA</math> and the same side <math>EA</math> as <math>\triangle BEA</math>. Connect all three triangles with <math>\triangle DEA</math> in the center and the two triangles sharing one of its sides. Then, <math>\pentagon BACDE</math> is formed with <math>BE+DE+CD</math> forming the base. | ||
+ | |||
+ | Intuitively, the pentagon's base is minimized when all three bottom sides are collinear. This is simply the original <math>\triangle BAC</math> except that <math>\angle BAC =120^\circ</math>. (In <math>\triangle DEA</math>, <math>\triangle CDA</math>, and <math>\triangle BEA</math>, <math>\angle A = 40^\circ</math>, and the three triangles connect at <math>A</math> to form the pentagon). Thus, <math>m\angle BAC = 40 \cdot 3</math>). | ||
+ | |||
+ | |||
+ | <math>BC</math> in this new triangle is then the minimum of <math>BE+DE+CD</math>. Applying law of cosines, <math>BC=\sqrt{6^2+10^2-2(6)(10)\cos (120^\circ)}=\sqrt{196}=14 \implies \boxed{\textbf{(D) }14}</math> | ||
+ | |||
+ | |||
+ | ~bjhhar | ||
+ | Would prime notation be clearer? | ||
+ | |||
+ | ==Solution 3== | ||
+ | <asy> | ||
+ | size(300); | ||
+ | defaultpen(linewidth(0.4)+fontsize(10)); | ||
+ | pen s = linewidth(0.8)+fontsize(8); | ||
+ | |||
+ | pair A,B,C,D,Ep,Bp,Cp; | ||
+ | A = (0,0); | ||
+ | B = 10*dir(-110);C = 6*dir(-70); | ||
+ | Bp = 10*dir(-30);Cp = 6*dir(-150); | ||
+ | D = IP(Cp--Bp, A--B); Ep = IP(Cp--Bp, A--C); | ||
+ | draw(A--B--C--A--Cp--Bp--A); | ||
+ | draw(Cp--B); | ||
+ | draw(C--Bp); | ||
+ | draw(C--D); | ||
+ | draw(B--Ep); | ||
+ | |||
+ | dot("$A$", A, N); | ||
+ | dot("$B$", B, SW); | ||
+ | dot("$C$", C, SE); | ||
+ | dot("$B'$", Bp, E); | ||
+ | dot("$C'$", Cp, W); | ||
+ | dot("$D$", D, dir(-70)); | ||
+ | dot("$E$", Ep, dir(60)); | ||
+ | |||
+ | MA("40^\circ",Cp,A,D, 1); | ||
+ | MA("40^\circ",D,A,Ep, 1); | ||
+ | MA("40^\circ",Ep,A,Bp, 1); | ||
+ | label("$6$", A--Cp); | ||
+ | label("$10$", Bp--A); | ||
+ | </asy> | ||
+ | |||
+ | (Diagram by shihan) | ||
+ | Reflect <math>C</math> across <math>AB</math> to <math>C'</math>. Similarly, reflect <math>B</math> across <math>AC</math> to <math>B'</math>. Clearly, <math>BE = B'E</math> and <math>CD = C'D</math>. Thus, the sum <math>BE + DE + CD = B'E + DE + C'D</math>. This value is minimized when <math>B'</math>, <math>C'</math>, <math>D</math> and <math>E</math> are collinear. To finish, we use the law of cosines on the triangle <math>AB'C'</math>: <math>B'C' = \sqrt{6^2 + 10^2 - 2(6)(10)\cos 120^{\circ}} = \boxed{\textbf{(D) } 14}.</math> | ||
+ | |||
+ | ==Remark== | ||
+ | |||
+ | This problem is similar to [https://en.wikipedia.org/wiki/Fagnano%27s_problem Fagnano's Problem]. | ||
+ | |||
+ | ~[https://artofproblemsolving.com/wiki/index.php/User:Isabelchen isabelchen] | ||
==See Also== | ==See Also== | ||
{{AMC12 box|year=2014|ab=A|num-b=19|num-a=21}} | {{AMC12 box|year=2014|ab=A|num-b=19|num-a=21}} | ||
{{MAA Notice}} | {{MAA Notice}} |
Latest revision as of 03:35, 22 January 2024
Problem
In , , , and . Points and lie on and respectively. What is the minimum possible value of ?
Solution 1
Let be the reflection of across , and let be the reflection of across . Then it is well-known that the quantity is minimized when it is equal to . (Proving this is a simple application of the triangle inequality; for an example of a simpler case, see Heron's Shortest Path Problem.) As lies on both and , we have . Furthermore, by the nature of the reflection, so . Therefore by the Law of Cosines
Solution 2
In , the three lines look like the Chinese character 又. Let , , and have bases , , and respectively. Then, has the same side as and the same side as . Connect all three triangles with in the center and the two triangles sharing one of its sides. Then, is formed with forming the base.
Intuitively, the pentagon's base is minimized when all three bottom sides are collinear. This is simply the original except that . (In , , and , , and the three triangles connect at to form the pentagon). Thus, ).
in this new triangle is then the minimum of . Applying law of cosines,
~bjhhar
Would prime notation be clearer?
Solution 3
(Diagram by shihan) Reflect across to . Similarly, reflect across to . Clearly, and . Thus, the sum . This value is minimized when , , and are collinear. To finish, we use the law of cosines on the triangle :
Remark
This problem is similar to Fagnano's Problem.
See Also
2014 AMC 12A (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.