2011 AMC 12A Problems/Problem 25

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 $\triangle ABC$ , repsectively. 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

25) Answer: (D) 80 degree

Given: $BC = 1$ , $\angle BAC = 60^{\circ}$ , $\angle CBA \le 90^{\circ}$ , $AC \ge BC$

$H$ , $I$ , $O$ are orthocenter, incenter, and circumcenter. and $BOIHC$ has maximum area.

Find $\angle CBA$ .

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 lets 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.

\textbf{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, lets isolate the points $A$ , $B$ , $C$ , and $I$ .

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

$m\angle BIC = 180^{\circ} - \frac{\beta}{2} - \frac{\gamma}{2} = 180^{\circ} - \frac{180^{\circ} - 120^{\circ}}{2} = 120 ^{\circ}$

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 multi-variable calculus.

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

If both partial is zero, then $x_1 = x_2 = x_3 = 20^\circ$ , and it is very easy to show that $f(x_1, x_2)$ 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 $AI$ 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 case exist, $m\angle AXB = m \angle ACB = 40 ^\circ$ , which yield that $m\angle CBA = 80 ^\circ$ .

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

Hence, the answer is $(D)$ .

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

Page

Toolbox

Search

2011 AMC 12A Problems/Problem 25

Problem

Solution

See also