Difference between revisions of "2006 AMC 12B Problems/Problem 23"
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== Problem == | == Problem == | ||
− | Isosceles <math>\triangle ABC</math> has a right angle at <math>C</math>. Point <math>P</math> is inside <math>\triangle ABC</math>, such that <math>PA=11</math>, <math>PB=7</math>, and <math>PC=6</math>. Legs <math>\overline{AC}</math> and <math>\overline{BC}</math> have length <math>s=\sqrt{a+b\sqrt{2} | + | Isosceles <math>\triangle ABC</math> has a right angle at <math>C</math>. Point <math>P</math> is inside <math>\triangle ABC</math>, such that <math>PA=11</math>, <math>PB=7</math>, and <math>PC=6</math>. Legs <math>\overline{AC}</math> and <math>\overline{BC}</math> have length <math>s=\sqrt{a+b\sqrt{2}}</math>, where <math>a</math> and <math>b</math> are positive integers. What is <math>a+b</math>? |
<asy> | <asy> |
Revision as of 16:22, 14 May 2015
Contents
[hide]Problem
Isosceles has a right angle at
. Point
is inside
, such that
,
, and
. Legs
and
have length
, where
and
are positive integers. What is
?
Solution
Using the Law of Cosines on
, we have:
Using the Law of Cosines on , we have:
Now we use .
Note that we know that we want the solution with since we know that
. Thus,
.
Solution 2
Rotate triangle 90 degrees counterclockwise about
so that the image of
rests on
. Now let the image of
be
. Note that
, meaning triangle
is right isosceles, and
. Then
. Now because
and
, we observe that
, by the Pythagorean Theorem on
. Now we have that
. So we take the cosine of the second equality, using that fact that
, to get
. Finally, we use the fact that
and use the Law of Cosines on triangle
to arrive at the value of
.
Or notice that since and
, we have
, and Law of Cosines on triangle
gives the value of
.
See also
2006 AMC 12B (Problems • Answer Key • Resources) | |
Preceded by Problem 22 |
Followed by Problem 24 |
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.