Difference between revisions of "2014 AMC 12A Problems/Problem 20"
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− | 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{ | + | 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== | ==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. | 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 | + | 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{ | + | <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> |
Line 49: | Line 49: | ||
dot("$C'$", Cp, W); | dot("$C'$", Cp, W); | ||
dot("$D$", D, dir(-70)); | dot("$D$", D, dir(-70)); | ||
− | dot("$E$", Ep, dir( | + | dot("$E$", Ep, dir(60)); |
MA("40^\circ",Cp,A,D, 1); | MA("40^\circ",Cp,A,D, 1); | ||
Line 59: | Line 59: | ||
(Diagram by shihan) | (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} = 14</math> | + | 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 04:35, 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 |
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