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

Revision as of 20:53, 3 July 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

Solution:

1) Let's draw a circle with center $O$ (which will be the circumcircle of $\triangle ABC$ . Since $\angle BAC = 60^{\circ}$ , $\overline{BC}$ is a chord that intercept an arc of $120 ^{\circ}$

2) Draw any chord that can be $BC$ , and let's define that as 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) Note that OMB and XMC is fixed, hence 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 we got that 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 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, $AI$ should intersection the midpoint of the arc, which is $X$ . Hence, if such a case exists, $m\angle AXB = m \angle ACB = 40 ^\circ$ , which yields $m\angle CBA = 80 ^\circ$ .

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 $(D) 80$ .

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 == See also ==
 {{AMC12 box|year=2011|num-b=24|after=Last Problem|ab=A}}
+{{MAA Notice}}

2011 AMC 12A (Problems • Answer Key • Resources)
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Difference between revisions of "2011 AMC 12A Problems/Problem 25"

Revision as of 20:53, 3 July 2013

Problem

Solution

See also