Difference between revisions of "2016 AMC 12B Problems/Problem 17"
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Then <math>PQ = \frac{8}{15} II' = </math> <cmath>\boxed{\textbf{(D)}\frac{8}{15}\sqrt{5}}</cmath> | Then <math>PQ = \frac{8}{15} II' = </math> <cmath>\boxed{\textbf{(D)}\frac{8}{15}\sqrt{5}}</cmath> | ||
− | ==Solution 3 ( | + | ==Solution 3 (FAST)== |
<math>PQ</math> lies on altitude <math>AH</math>, which we find to have a length of <math>3\sqrt{5}</math> by Heron's Formula and dividing twice the area by <math>BC</math>. From H we can construct a segment <math>HX</math> with <math>X</math> on <math>CE</math> such that <math>HX</math> is parallel to <math>EB</math>. A similar construction gives <math>Y</math> on <math>BD</math> such that <math>HY</math> is parallel to <math>DC</math>. We can hence generate a system of ratios that will allow us to find <math>PQ/AH</math>. Note that such a system will generate a rational number for the ratio <math>PQ/AH</math>. Thus, we choose the only answer that has a <math>\sqrt{5}</math> term in it, giving us <math>\boxed{\textbf{(D)}} </math>. | <math>PQ</math> lies on altitude <math>AH</math>, which we find to have a length of <math>3\sqrt{5}</math> by Heron's Formula and dividing twice the area by <math>BC</math>. From H we can construct a segment <math>HX</math> with <math>X</math> on <math>CE</math> such that <math>HX</math> is parallel to <math>EB</math>. A similar construction gives <math>Y</math> on <math>BD</math> such that <math>HY</math> is parallel to <math>DC</math>. We can hence generate a system of ratios that will allow us to find <math>PQ/AH</math>. Note that such a system will generate a rational number for the ratio <math>PQ/AH</math>. Thus, we choose the only answer that has a <math>\sqrt{5}</math> term in it, giving us <math>\boxed{\textbf{(D)}} </math>. | ||
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Let <math>h=AH</math> and <math>BH=x</math>. Then, <math>CH=8-x</math>. By the Pythagorean Theorem on right triangles <math>ABH</math> and <math>ACH</math>, we have <cmath>h^2+x^2=49</cmath> <cmath>x^2+(8-x)^2=81.</cmath> Subtracting the prior from the latter yields <math>-16x+64=32\implies x=2</math>. So, <math>BH=2</math>, <math>CH=6</math>, and <math>AH=3\sqrt{5}</math>. Continue with Solution 1. | Let <math>h=AH</math> and <math>BH=x</math>. Then, <math>CH=8-x</math>. By the Pythagorean Theorem on right triangles <math>ABH</math> and <math>ACH</math>, we have <cmath>h^2+x^2=49</cmath> <cmath>x^2+(8-x)^2=81.</cmath> Subtracting the prior from the latter yields <math>-16x+64=32\implies x=2</math>. So, <math>BH=2</math>, <math>CH=6</math>, and <math>AH=3\sqrt{5}</math>. Continue with Solution 1. | ||
+ | |||
+ | ==Video Solution== | ||
+ | https://www.youtube.com/watch?v=ccB-z4_OHqw | ||
+ | |||
+ | ==Video Solution by CanadaMath (Problem 11-20)== | ||
+ | Fast Forward to 26:14 for problem 17 | ||
+ | https://www.youtube.com/watch?v=4osvFatUv1o | ||
+ | |||
+ | ~THEMATHCANADIAN | ||
+ | |||
+ | |||
==See Also== | ==See Also== | ||
{{AMC12 box|year=2016|ab=B|num-b=16|num-a=18}} | {{AMC12 box|year=2016|ab=B|num-b=16|num-a=18}} | ||
{{MAA Notice}} | {{MAA Notice}} |
Latest revision as of 01:26, 10 November 2024
Contents
Problem
In shown in the figure,
,
,
, and
is an altitude. Points
and
lie on sides
and
, respectively, so that
and
are angle bisectors, intersecting
at
and
, respectively. What is
?
Solution 1
Get the area of the triangle by Heron's Formula:
Use the area to find the height
with known base
:
Apply the Angle Bisector Theorem on
and
, we get
and
, respectively.
To find
,
,
, and
, apply variables, such that
is
and
is
. Solving them out, you will get
,
,
, and
. Then, since
according to the Segment Addition Postulate, and thus manipulating, you get
=
Solution 2
Let the intersection of and
be the point
. Then let the foot of the altitude from
to
be
. Note that
is an inradius and that
, where
is the semiperimeter of the triangle.
Using Heron's Formula, we see that , so
.
Then since and
are parallel,
and
.
Thus, and
, so
.
By the Dual Principle, and
. With the same method as Solution 1,
and
.
Then
Solution 3 (FAST)
lies on altitude
, which we find to have a length of
by Heron's Formula and dividing twice the area by
. From H we can construct a segment
with
on
such that
is parallel to
. A similar construction gives
on
such that
is parallel to
. We can hence generate a system of ratios that will allow us to find
. Note that such a system will generate a rational number for the ratio
. Thus, we choose the only answer that has a
term in it, giving us
.
Solution 4
Let and
. Then,
. By the Pythagorean Theorem on right triangles
and
, we have
Subtracting the prior from the latter yields
. So,
,
, and
. Continue with Solution 1.
Video Solution
https://www.youtube.com/watch?v=ccB-z4_OHqw
Video Solution by CanadaMath (Problem 11-20)
Fast Forward to 26:14 for problem 17 https://www.youtube.com/watch?v=4osvFatUv1o
~THEMATHCANADIAN
See Also
2016 AMC 12B (Problems • Answer Key • Resources) | |
Preceded by Problem 16 |
Followed by Problem 18 |
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 |
These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions.