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Difference between revisions of "2011 AMC 12A Problems/Problem 25"

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== Solution ==
 
== Solution ==
  
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<br />
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1) Let the circumcircle of <math>\triangle ABC</math> have a center <math>O</math>. Since <math>\angle BAC = 60^{\circ}</math>, <math>\overline{BC}</math> is a chord that intercept an arc of <math>120 ^{\circ}</math>
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'''Solution:'''
 
  
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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>
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2) Define the length of <math>BC</math> as a unit length.
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2) Draw any chord that can be <math>BC</math>, and let's 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>.
 
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>.
  
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4) Note that OMB and XMC is fixed, hence the area is a constant. Thus, <math>XOIHC</math> also achieved maximum area.
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4) Since OMB and XMC are fixed, the area is a constant. Thus, <math>XOIHC</math> also achieved maximum area.
  
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<br />
 
 
'''Lemma:'''  
 
'''Lemma:'''  
  
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Let <math>m\angle ABC = \beta</math>, <math>m\angle ACB = \gamma</math>
 
Let <math>m\angle ABC = \beta</math>, <math>m\angle ACB = \gamma</math>
  
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<br />
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Now, let's isolate the points <math>A</math>,<math>B</math>,<math>C</math>, and <math>I</math>.
 
Now, let's isolate the points <math>A</math>,<math>B</math>,<math>C</math>, and <math>I</math>.
  
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<math>m\angle BIC = 180^{\circ} - \frac{\beta}{2} - \frac{\gamma}{2} = 180^{\circ} - ({180^{\circ} - 120^{\circ})= 120 ^{\circ}</math>
 
<math>m\angle BIC = 180^{\circ} - \frac{\beta}{2} - \frac{\gamma}{2} = 180^{\circ} - ({180^{\circ} - 120^{\circ})= 120 ^{\circ}</math>
  
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<br />
+
 
 
Now, lets isolate the points <math>A</math>,<math>B</math>,<math>C</math>, and <math>H</math>.
 
Now, lets isolate the points <math>A</math>,<math>B</math>,<math>C</math>, and <math>H</math>.
  
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<math>m\angle BHC = 180^{\circ} - \beta - \gamma + 60^{\circ} = 240^{\circ} - 120^{\circ} = 120^{\circ}</math>
 
<math>m\angle BHC = 180^{\circ} - \beta - \gamma + 60^{\circ} = 240^{\circ} - 120^{\circ} = 120^{\circ}</math>
  
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<br />
 
 
Lemma proven. The lemma yields that BOIHC is a cyclic pentagon.
 
Lemma proven. The lemma yields that BOIHC is a cyclic pentagon.
  
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Since we got that XOIHC also achieved maximum area,
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Since 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.)
 
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.)
  
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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>
+
Then the 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>.
 
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>.
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If the partial derivatives with respect to <math>x_1</math> and <math>x_2</math> are 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 the maximum with the second derivative test (left for the reader).
 
If the partial derivatives with respect to <math>x_1</math> and <math>x_2</math> are 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 the maximum with the second derivative test (left for the reader).
  
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<br />
 
 
Now, we need to verify that such a situation exists and find the angle for this situation.
 
Now, we need to verify that such a situation exists and find the angle for this situation.
  
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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 a case exists, <math>m\angle AXB = m \angle ACB = 40 ^\circ</math>, which yields <math>m\angle CBA = 80 ^\circ</math>.
+
Let's extend <math>AI</math> to the direction of <math>X</math>, since <math>AI</math> is the angle bisector, so that <math>AI</math> intersects the midpoint of the arc <math>X</math>. Hence, if such a case exists, <math>m\angle AXB = m \angle ACB = 40 ^\circ</math>, so <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 (follows from the lemma). <math>I</math> will be at the right place. <math>H</math> can be easily verified too.
 
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 (follows from the lemma). <math>I</math> will be at the right place. <math>H</math> can be easily verified too.
  
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<br />
 
 
Hence, the answer is <math>(D) 80</math>.
 
Hence, the answer is <math>(D) 80</math>.
  

Revision as of 16:09, 22 September 2013

Problem

Triangle $ABC$ has $\angle BAC = 60^{\circ}$, $\angle CBA \leq 90^{\circ}$, $BC=1$, and $AC \geq AB$. Let $H$, $I$, and $O$ be the orthocenter, incenter, and circumcenter of $\triangle ABC$, respectively. Assume that the area of pentagon $BCOIH$ is the maximum possible. What is $\angle CBA$?

$\textbf{(A)}\ 60^{\circ} \qquad \textbf{(B)}\ 72^{\circ} \qquad \textbf{(C)}\ 75^{\circ} \qquad \textbf{(D)}\ 80^{\circ} \qquad \textbf{(E)}\ 90^{\circ}$

Solution

1) Let the circumcircle of $\triangle ABC$ have a center $O$. Since $\angle BAC = 60^{\circ}$, $\overline{BC}$ is a chord that intercept an arc of $120 ^{\circ}$

2) Define the length of $BC$ as a unit length.

3) Draw the diameter $\perp$ to $BC$. Let's call the interception of the diameter with $BC$ $M$ (because it is the midpoint) and interception with the circle $X$.

4) Since OMB and XMC are fixed, the area is a constant. Thus, $XOIHC$ also achieved maximum area.

Lemma:

$m\angle BOC = m \angle BIC = m \angle BHC = 120^{\circ}$

For $m\angle BOC$, we fixed it to $120^{\circ}$ when we drew the diagram.

Let $m\angle ABC = \beta$, $m\angle ACB = \gamma$


Now, let's isolate the points $A$,$B$,$C$, and $I$.

$m\angle IBC = \frac{\beta}{2}$, $m\angle ICB = \frac{\gamma}{2}$

$m\angle BIC = 180^{\circ} - \frac{\beta}{2} - \frac{\gamma}{2} = 180^{\circ} - ({180^{\circ} - 120^{\circ})= 120 ^{\circ}$ (Error compiling LaTeX. Unknown error_msg)


Now, lets isolate the points $A$,$B$,$C$, and $H$.

$m\angle HBC = \beta - 30^{\circ}$, $m\angle HCB = \gamma - 30^{\circ}$

$m\angle BHC = 180^{\circ} - \beta - \gamma + 60^{\circ} = 240^{\circ} - 120^{\circ} = 120^{\circ}$

Lemma proven. The lemma yields that BOIHC is a cyclic pentagon.

Since XOIHC also achieved maximum area,

Let $m\angle XOI = x_1$, $m\angle OIH = x_2$, $m\angle IHC = x_3$, and the radius is $R$ (which will drop out.)

Then the area = $\frac{r^2}{2}(\sin x_1 + \sin x_2 + \sin x_3)$, where $x_1 + x_2 + x_3 = 60^\circ$

So we want to maximize $f(x_1, x_2) = \sin x_1 + \sin x_2 + \sin x_3$, Note that $x_3 = 60 ^\circ - x_1 - x_2$.

Let's do some multivariable calculus.

$f_{x_1} = \cos x_1 - \cos (x_3)$, $f_{x_2} = \cos x_2 - \cos (x_3)$

If the partial derivatives with respect to $x_1$ and $x_2$ are zero, then $x_1 = x_2 = x_3 = 20^\circ$, and it is very easy to show that $f(x_1, x_2)$ is the maximum with the second derivative test (left for the reader).

Now, we need to verify that such a situation exists and find the angle for this situation.

Let's extend $AI$ to the direction of $X$, since $AI$ is the angle bisector, so that $AI$ intersects the midpoint of the arc $X$. Hence, if such a case exists, $m\angle AXB = m \angle ACB = 40 ^\circ$, so $m\angle CBA = 80 ^\circ$.

If the angle is $80 ^\circ$, it is clear that since $I$ and $H$ are on the second circle (follows from the lemma). $I$ will be at the right place. $H$ can be easily verified too.

Hence, the answer is $(D) 80$.

See also

2011 AMC 12A (Problems • Answer Key • Resources)
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