Difference between revisions of "2011 AMC 12A Problems/Problem 25"
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== Solution == | == Solution == | ||
+ | 25) Answer: (D) 80 degree | ||
+ | |||
+ | Given: <math>BC = 1</math>, <math>\angle BAC = 60^{\circ}</math>, <math>\angle CBA \le 90^{\circ}</math>, <math>AC \ge BC</math> | ||
+ | |||
+ | <math>H</math>, <math>I</math>, <math>O</math> are orthocenter, incenter, and circumcenter. and <math>BOIHC</math> has maximum area. | ||
+ | |||
+ | Find <math>\angle CBA</math>. | ||
+ | |||
+ | <br /> | ||
+ | '''Solution:''' | ||
+ | |||
+ | 1) Let's draw a circle with center <math>O</math> (which will be the circumcircle of <math>\triangle ABC</math>. Since <math>\angle BAC = 60^{\circ}</math>, <math>\overline{BC}</math> is a chord that intercept an arc of <math>120 ^{\circ}</math> | ||
+ | |||
+ | 2) Draw any chord that can be <math>BC</math>, and lets define that as unit length. | ||
+ | |||
+ | 3) Draw the diameter <math>\perp</math> to <math>BC</math>. Let's call the interception of the diameter with <math>BC</math> <math>M</math> (because it is the midpoint) and interception with the circle <math>X</math>. | ||
+ | |||
+ | 4) Note that OMB and XMC is fixed, hence the area is a constant. Thus, <math>XOIHC</math> also achieved maximum area. | ||
+ | |||
+ | <br /> | ||
+ | \textbf{Lemma)} <math>m\angle BOC = m \angle BIC = m \angle BHC = 120^{\circ}</math> | ||
+ | |||
+ | For <math>m\angle BOC</math>, we fixed it to <math>120^{\circ}</math> when we drew the diagram. | ||
+ | |||
+ | Let <math>m\angle ABC = \beta</math>, <math>m\angle ACB = \gamma</math> | ||
+ | |||
+ | <br /> | ||
+ | Now, lets isolate the points <math>A</math>,<math>B</math>,<math>C</math>, and <math>I</math>. | ||
+ | |||
+ | <math>m\angle IBC = \frac{\beta}{2}</math>, <math>m\angle ICB = \frac{\gamma}{2}</math> | ||
+ | |||
+ | <math>m\angle BIC = 180^{\circ} - \frac{\beta}{2} - \frac{\gamma}{2} = 180^{\circ} - \frac{180^{\circ} - 120^{\circ}}{2} = 120 ^{\circ}</math> | ||
+ | |||
+ | <br /> | ||
+ | Now, lets isolate the points <math>A</math>,<math>B</math>,<math>C</math>, and <math>H</math>. | ||
+ | |||
+ | <math>m\angle HBC = \beta - 30^{\circ}</math>, <math>m\angle HCB = \gamma - 30^{\circ}</math> | ||
+ | |||
+ | <math>m\angle BHC = 180^{\circ} - \beta - \gamma + 60^{\circ} = 240^{\circ} - 120^{\circ} = 120^{\circ}</math> | ||
+ | |||
+ | <br /> | ||
+ | Lemma proven. The lemma yields that BOIHC is a cyclic pentagon. | ||
+ | |||
+ | Since we got that XOIHC also achieved maximum area, | ||
+ | |||
+ | Let <math>m\angle XOI = x_1</math>, <math>m\angle OIH = x_2</math>, <math>m\angle IHC = x_3</math>, and the radius is <math>R</math> (which will drop out.) | ||
+ | |||
+ | then area = <math>\frac{r^2}{2}(\sin x_1 + \sin x_2 + \sin x_3)</math>, where <math>x_1 + x_2 + x_3 = 60^\circ</math> | ||
+ | |||
+ | So we want to maximize <math>f(x_1, x_2) = \sin x_1 + \sin x_2 + \sin x_3</math>, Note that <math>x_3 = 60 ^\circ - x_1 - x_2</math>. | ||
+ | |||
+ | Let's do some multi-variable calculus. | ||
+ | |||
+ | <math>f_{x_1} = \cos x_1 - \cos (x_3)</math>, <math>f_{x_2} = \cos x_2 - \cos (x_3)</math> | ||
+ | |||
+ | If both partial is zero, then <math>x_1 = x_2 = x_3 = 20^\circ</math>, and it is very easy to show that <math>f(x_1, x_2)</math> is maximum here with second derivative test (left for the reader). | ||
+ | |||
+ | <br /> | ||
+ | Now, we need to verify that such situation exist and find the angle for this situation. | ||
+ | |||
+ | Let's extend <math>AI</math> to the direction of <math>X</math>, since <math>AI</math> is the angle bisector, <math>AI</math> should intersection the midpoint of the arc, which is <math>X</math>. Hence, if such case exist, <math>m\angle AXB = m \angle ACB = 40 ^\circ</math>, which yield that <math>m\angle CBA = 80 ^\circ</math>. | ||
+ | |||
+ | If the angle is <math>80 ^\circ</math>, it is clear that since <math>I</math> and <math>H</math> are on the second circle (follow from lemma). <math>I</math> will be at the right place. <math>H</math> can be easily verify too. | ||
+ | |||
+ | <br /> | ||
+ | Hence, the answer is <math>(D)</math>. | ||
+ | |||
== See also == | == See also == | ||
{{AMC12 box|year=2011|num-b=24|after=Last Problem|ab=A}} | {{AMC12 box|year=2011|num-b=24|after=Last Problem|ab=A}} |
Revision as of 00:21, 15 February 2011
Problem
Triangle has , , , and . Let , , and be the orthocenter, incenter, and circumcenter of , repsectively. Assume that the area of pentagon is the maximum possible. What is ?
Solution
25) Answer: (D) 80 degree
Given: , , ,
, , are orthocenter, incenter, and circumcenter. and has maximum area.
Find .
Solution:
1) Let's draw a circle with center (which will be the circumcircle of . Since , is a chord that intercept an arc of
2) Draw any chord that can be , and lets define that as unit length.
3) Draw the diameter to . Let's call the interception of the diameter with (because it is the midpoint) and interception with the circle .
4) Note that OMB and XMC is fixed, hence the area is a constant. Thus, also achieved maximum area.
\textbf{Lemma)}
For , we fixed it to when we drew the diagram.
Let ,
Now, lets isolate the points ,,, and .
,
Now, lets isolate the points ,,, and .
,
Lemma proven. The lemma yields that BOIHC is a cyclic pentagon.
Since we got that XOIHC also achieved maximum area,
Let , , , and the radius is (which will drop out.)
then area = , where
So we want to maximize , Note that .
Let's do some multi-variable calculus.
,
If both partial is zero, then , and it is very easy to show that is maximum here with second derivative test (left for the reader).
Now, we need to verify that such situation exist and find the angle for this situation.
Let's extend to the direction of , since is the angle bisector, should intersection the midpoint of the arc, which is . Hence, if such case exist, , which yield that .
If the angle is , it is clear that since and are on the second circle (follow from lemma). will be at the right place. can be easily verify too.
Hence, the answer is .
See also
2011 AMC 12A (Problems • Answer Key • Resources) | |
Preceded by Problem 24 |
Followed by Last Problem |
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 |