Difference between revisions of "2017 AMC 12B Problems/Problem 24"
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− | Let <math>CD=1</math>, <math>BC=x</math>, <math>AB=x^2</math>. Note that <math>AB/BC=x</math>. The Pythagorean theorem states that <math>BD=sqrt(x^2+1)</math>. Since <math>BCD~ABC~CEB</math>, the ratios of side lengths must be equal. Since <math>BC=x</math>, <math>CE=x^2/sqrt(x^2+1)</math> and <math>BE=x/sqrt(x^2+1)</math>. Let Point F be a point on <math>BC</math> such that <math>EF</math> is an altitude of triangle <math>CEB</math>. Note that <math>CEB~CFE~EFB</math>, so <math>BF</math> and <math>CF</math> can be calculated. Solving for these lengths gives <math>BF=x/(x^2+1)</math> and <math>CF=x^3/(x^2+1)</math>. | + | Let <math>CD=1</math>, <math>BC=x</math>, <math>AB=x^2</math>. Note that <math>AB/BC=x</math>. The Pythagorean theorem states that <math>BD=sqrt(x^2+1)</math>. Since <math>BCD~ABC~CEB</math>, the ratios of side lengths must be equal. Since <math>BC=x</math>, <math>CE=x^2/sqrt(x^2+1)</math> and <math>BE=x/sqrt(x^2+1)</math>. Let Point F be a point on <math>BC</math> such that <math>EF</math> is an altitude of triangle <math>CEB</math>. Note that <math>CEB~CFE~EFB</math>, so <math>BF</math> and <math>CF</math> can be calculated. Solving for these lengths gives <math>BF=x/(x^2+1)</math> and <math>CF=x^3/(x^2+1)</math>. Since <math>CF</math> and <math>BF</math> form altitudes of triangles <math>CED</math> and <math>BEA</math>, respectively, the areas of these triangles can be calculated. Additionally, the area of triangle <math>BEC</math> can be calculated, as it is a right triangle. Solving for each of these yields: |
+ | <math>Area(BEC)=Area(CED)=Area(BEA)=(x^3)/(2(x^2+1))</math> | ||
+ | <math>Area(ABCD)=Area(AED)+Area(DEC)+Area(CEB)+Area(BEA).</math> | ||
+ | <math>(AB+CD)(BC)/2= 17*Area(CEB)+ Area(CEB) + Area(CEB) + Area(CEB)</math> | ||
+ | <math>(x^3+x)/2=(20x^3)/(2(x^2+1))</math> | ||
+ | <math>(x)(x^2+1)=20x^3/(x^2+1)</math> | ||
+ | <math>(x^2+1)^2=20x^2</math> | ||
+ | <math>x^4-18x^2+1=0</math> | ||
+ | <math>x^2=9+4sqrt(5)=4+2(2sqrt(5))+5</math> (Minus yields a negative value) | ||
+ | <math>x=2+sqrt(5)</math> | ||
+ | Thus the answer is D: 2+sqrt(5) | ||
+ | |||
==See Also== | ==See Also== | ||
{{AMC12 box|year=2017|ab=B|num-b=23|num-a=25}} | {{AMC12 box|year=2017|ab=B|num-b=23|num-a=25}} | ||
{{MAA Notice}} | {{MAA Notice}} |
Revision as of 21:41, 16 February 2017
Problem
Quadrilateral has right angles at
and
, Triangle
~ Triangle
, and
. There is a point
in the interior of
such that Triangle
~ Triangle
and the area of Triangle
is
times the area of Triangle
. What is
Solution
Solution by TorrTar
Let ,
,
. Note that
. The Pythagorean theorem states that
. Since
, the ratios of side lengths must be equal. Since
,
and
. Let Point F be a point on
such that
is an altitude of triangle
. Note that
, so
and
can be calculated. Solving for these lengths gives
and
. Since
and
form altitudes of triangles
and
, respectively, the areas of these triangles can be calculated. Additionally, the area of triangle
can be calculated, as it is a right triangle. Solving for each of these yields:
(Minus yields a negative value)
Thus the answer is D: 2+sqrt(5)
See Also
2017 AMC 12B (Problems • Answer Key • Resources) | |
Preceded by Problem 23 |
Followed by Problem 25 |
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.