Difference between revisions of "2010 AMC 12A Problems/Problem 17"
(Created page with '== Problem == Equiangular hexagon <math>ABCDEF</math> has side lengths <math>AB=CD=EF=1</math> and <math>BC=DE=FA=r</math>. The area of <math>\triangle ACE</math> is <math>70\%</…') |
(→Solution 2) |
||
(34 intermediate revisions by 23 users not shown) | |||
Line 1: | Line 1: | ||
+ | {{duplicate|[[2010 AMC 12A Problems|2010 AMC 12A #17]] and [[2010 AMC 10A Problems|2010 AMC 10A #19]]}} | ||
+ | |||
== Problem == | == Problem == | ||
Equiangular hexagon <math>ABCDEF</math> has side lengths <math>AB=CD=EF=1</math> and <math>BC=DE=FA=r</math>. The area of <math>\triangle ACE</math> is <math>70\%</math> of the area of the hexagon. What is the sum of all possible values of <math>r</math>? | Equiangular hexagon <math>ABCDEF</math> has side lengths <math>AB=CD=EF=1</math> and <math>BC=DE=FA=r</math>. The area of <math>\triangle ACE</math> is <math>70\%</math> of the area of the hexagon. What is the sum of all possible values of <math>r</math>? | ||
Line 4: | Line 6: | ||
<math>\textbf{(A)}\ \frac{4\sqrt{3}}{3} \qquad \textbf{(B)} \frac{10}{3} \qquad \textbf{(C)}\ 4 \qquad \textbf{(D)}\ \frac{17}{4} \qquad \textbf{(E)}\ 6</math> | <math>\textbf{(A)}\ \frac{4\sqrt{3}}{3} \qquad \textbf{(B)} \frac{10}{3} \qquad \textbf{(C)}\ 4 \qquad \textbf{(D)}\ \frac{17}{4} \qquad \textbf{(E)}\ 6</math> | ||
− | == Solution == | + | == Solution 1== |
− | It is clear that <math>\triangle ACE</math> is an equilateral triangle. From the [[Law of Cosines]], we get that <math>AC^2 = r^2+1^2-2r\cos{\frac{2\pi}{3}} = r^2+r+1</math>. Therefore, the area of <math>\triangle ACE</math> is <math>\frac{\sqrt{3}}{4}(r^2+r+1)</math>. | + | It is clear that <math>\triangle ACE</math> is an equilateral triangle. From the [[Law of Cosines]] on <math>\triangle ABC</math>, we get that <math>AC^2 = r^2+1^2-2r\cos{\frac{2\pi}{3}} = r^2+r+1</math>. Therefore, the area of <math>\triangle ACE</math> is <math>\frac{\sqrt{3}}{4}(r^2+r+1)</math> by area of an equilateral triangle. |
+ | |||
+ | If we extend <math>BC</math>, <math>DE</math> and <math>FA</math> so that <math>FA</math> and <math>BC</math> meet at <math>X</math>, <math>BC</math> and <math>DE</math> meet at <math>Y</math>, and <math>DE</math> and <math>FA</math> meet at <math>Z</math>, we find that hexagon <math>ABCDEF</math> is formed by taking equilateral triangle <math>XYZ</math> of side length <math>r+2</math> and removing three equilateral triangles, <math>ABX</math>, <math>CDY</math> and <math>EFZ</math>, of side length <math>1</math>. The area of <math>ABCDEF</math> is therefore | ||
+ | |||
+ | <math>\frac{\sqrt{3}}{4}(r+2)^2-\frac{3\sqrt{3}}{4} = \frac{\sqrt{3}}{4}(r^2+4r+1)</math>. | ||
+ | |||
+ | |||
+ | Based on the initial conditions, | ||
+ | |||
+ | <cmath>\frac{\sqrt{3}}{4}(r^2+r+1) = \frac{7}{10}\left(\frac{\sqrt{3}}{4}\right)(r^2+4r+1)</cmath> | ||
+ | |||
+ | Simplifying this gives us <math>r^2-6r+1 = 0</math>. By [[Vieta's Formulas]] we know that the sum of the possible value of <math>r</math> is <math>\boxed{\textbf{(E)}\ 6}</math>. | ||
+ | |||
+ | ==Solution 2== | ||
+ | Step 1: Use [[Law of Cosines]] in the same manner as the previous solution to get <math>AC=\sqrt{r^2+r+1}</math>. | ||
+ | |||
+ | Step 2: <math>\triangle{ABC}</math>~<math>\triangle{CDE}</math>~<math>\triangle{EFA}</math> via SAS congruency. Using the formula <math>[ABC]=\frac{ab \sin C}{2}= \frac{r \sqrt{3}}{4}</math>. The area of the hexagon is equal to <math>[ACE] + 3[ABC]</math>. We are given that <math>70\%</math> of this area is equal to <math>[ACE]</math>; solving for <math>AC</math> in terms of <math>r</math> gives <math>AC=\sqrt{7r}</math>. | ||
+ | |||
+ | Step 3: <math>\sqrt{7r}=\sqrt{r^2+r+1} \implies r^2-6r+1=0</math> and by [[Vieta's Formulas]] , we get <math>\boxed{\textbf{E}}</math>. | ||
+ | |||
+ | Note: To verify that the quadratic <math>r^2-6r+1</math> has two positive roots, we can either solve for the roots directly or note that discriminant is positive, and there are no negative roots (because then <math>r^2, -6r, 1</math> would all be positive). | ||
+ | |||
+ | ==Solution 3== | ||
+ | Find the area of the triangle <math>ACE</math> as how it was done in solution 1. Find the sum of the areas of the congruent triangles <math>ABC, CDE, EFA</math> as how it was done in solution 2. The sum of these areas is the area of the hexagon, hence the areas of the congruent triangles <math>ABC, CDE, EFA</math> is <math>30\%</math> of the area of the hexagon. Hence <math>\frac{7}{3}</math> times the latter is equal to the triangle <math>ACE</math>. Hence <math>\frac{7}{3}\cdot\frac{3\sqrt{3}}{4}r=\frac{\sqrt{3}}{4}(r^2+r+1)</math>. We can simplify this to <math>7r=r^2+r+1\implies r^2-6r+1=0</math>. By Vieta's, we get the sum of all possible values of <math>r</math> is <math>-\frac{-6}{1}=6\text{ or } \boxed{\textbf{E}}</math>. | ||
+ | -vsamc | ||
+ | |||
+ | ===Proof Triangle ACE is Equilateral.=== | ||
+ | We know <math>\triangle{ABC}</math>, <math>\triangle{CDE}</math>, and <math>\triangle{EFA}</math> are congruent by SAS, so the side opposite the 120 degree angle is also the same (since the triangles are congruent). Thus <math>\triangle{ACE}</math> is equilateral. | ||
+ | Q.E.D. | ||
+ | ~mathboy282 | ||
+ | |||
+ | ==Video Solution by the Beauty of Math== | ||
+ | https://youtu.be/rsURe5Xh-j0?t=961 | ||
− | + | == See also == | |
+ | {{AMC10 box|year=2010|ab=A|num-b=18|num-a=20}} | ||
+ | {{AMC12 box|year=2010|num-b=16|num-a=18|ab=A}} | ||
− | + | [[Category:Introductory Geometry Problems]] | |
+ | {{MAA Notice}} |
Latest revision as of 10:13, 14 October 2022
- The following problem is from both the 2010 AMC 12A #17 and 2010 AMC 10A #19, so both problems redirect to this page.
Contents
Problem
Equiangular hexagon has side lengths
and
. The area of
is
of the area of the hexagon. What is the sum of all possible values of
?
Solution 1
It is clear that is an equilateral triangle. From the Law of Cosines on
, we get that
. Therefore, the area of
is
by area of an equilateral triangle.
If we extend ,
and
so that
and
meet at
,
and
meet at
, and
and
meet at
, we find that hexagon
is formed by taking equilateral triangle
of side length
and removing three equilateral triangles,
,
and
, of side length
. The area of
is therefore
.
Based on the initial conditions,
Simplifying this gives us . By Vieta's Formulas we know that the sum of the possible value of
is
.
Solution 2
Step 1: Use Law of Cosines in the same manner as the previous solution to get .
Step 2: ~
~
via SAS congruency. Using the formula
. The area of the hexagon is equal to
. We are given that
of this area is equal to
; solving for
in terms of
gives
.
Step 3: and by Vieta's Formulas , we get
.
Note: To verify that the quadratic has two positive roots, we can either solve for the roots directly or note that discriminant is positive, and there are no negative roots (because then
would all be positive).
Solution 3
Find the area of the triangle as how it was done in solution 1. Find the sum of the areas of the congruent triangles
as how it was done in solution 2. The sum of these areas is the area of the hexagon, hence the areas of the congruent triangles
is
of the area of the hexagon. Hence
times the latter is equal to the triangle
. Hence
. We can simplify this to
. By Vieta's, we get the sum of all possible values of
is
.
-vsamc
Proof Triangle ACE is Equilateral.
We know ,
, and
are congruent by SAS, so the side opposite the 120 degree angle is also the same (since the triangles are congruent). Thus
is equilateral.
Q.E.D.
~mathboy282
Video Solution by the Beauty of Math
https://youtu.be/rsURe5Xh-j0?t=961
See also
2010 AMC 10A (Problems • Answer Key • Resources) | ||
Preceded by Problem 18 |
Followed by Problem 20 | |
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 10 Problems and Solutions |
2010 AMC 12A (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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.