Difference between revisions of "2011 AMC 12A Problems/Problem 25"
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== Problem == | == Problem == | ||
| + | Triangle <math>ABC</math> has <math>\angle BAC = 60^\{\circ}</math>, <math>\angle CBA \leq 90^\{\circ}</math>, <math>BC=1</math>, and <math>AC \geq AB</math>. Let <math>H</math>, <math>I</math>, and <math>O</math> be the orthocenter, incenter, and circumcenter of <math>\triangle ABC</math>, repsectively. Assume that the area of pentagon <math>BCOIH</math> is the maximum possible. What is <math>\angle CBA</math>? | ||
| + | |||
| + | <math> | ||
| + | \textbf{(A)}\ 60^{\circ} \qquad | ||
| + | \textbf{(B)}\ 72^{\circ} \qquad | ||
| + | \textbf{(C)}\ 75^{\circ} \qquad | ||
| + | \textbf{(D)}\ 80^{\circ} \qquad | ||
| + | \textbf{(E)}\ 90^{\circ} </math> | ||
| + | |||
== Solution == | == Solution == | ||
== 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 02:37, 10 February 2011
Problem
Triangle 
has $\angle BAC = 60^\{\circ}$ (Error compiling LaTeX. Unknown error_msg), $\angle CBA \leq 90^\{\circ}$ (Error compiling LaTeX. Unknown error_msg), 
, and 
. Let 
, 
, and 
be the orthocenter, incenter, and circumcenter of 
, repsectively. Assume that the area of pentagon 
is the maximum possible. What is 
?
Solution
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 | |
