Difference between revisions of "2016 AMC 12B Problems/Problem 21"
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+ | == Problem == | ||
Let <math>ABCD</math> be a unit square. Let <math>Q_1</math> be the midpoint of <math>\overline{CD}</math>. For <math>i=1,2,\dots,</math> let <math>P_i</math> be the intersection of <math>\overline{AQ_i}</math> and <math>\overline{BD}</math>, and let <math>Q_{i+1}</math> be the foot of the perpendicular from <math>P_i</math> to <math>\overline{CD}</math>. What is | Let <math>ABCD</math> be a unit square. Let <math>Q_1</math> be the midpoint of <math>\overline{CD}</math>. For <math>i=1,2,\dots,</math> let <math>P_i</math> be the intersection of <math>\overline{AQ_i}</math> and <math>\overline{BD}</math>, and let <math>Q_{i+1}</math> be the foot of the perpendicular from <math>P_i</math> to <math>\overline{CD}</math>. What is | ||
<cmath>\sum_{i=1}^{\infty} \text{Area of } \triangle DQ_i P_i \, ?</cmath> | <cmath>\sum_{i=1}^{\infty} \text{Area of } \triangle DQ_i P_i \, ?</cmath> | ||
Line 8: | Line 9: | ||
\textbf{(E)}\ 1</math> | \textbf{(E)}\ 1</math> | ||
− | ==Solution== | + | == Solutions == |
+ | === Solution 1 === | ||
+ | (By Qwertazertl) | ||
+ | |||
+ | We are tasked with finding the sum of the areas of every <math>\triangle DQ_i^{}P_i^{}</math> where <math>i</math> is a positive integer. We can start by finding the area of the first triangle, <math>\triangle DQ_1^{}P_1^{}</math>. This is equal to <math>\frac{1}{2}</math> ⋅ <math>DQ_1^{}</math> ⋅ <math>P_1^{}Q_2^{}</math>. Notice that since triangle <math>\triangle DQ_1^{}P_1^{}</math> is similar to triangle <math>\triangle ABP_1^{}</math> in a 1 : 2 ratio, <math>P_1^{}Q_2^{}</math> must equal <math>\frac{1}{3}</math> (since we are dealing with a unit square whose side lengths are 1). <math>DQ_1^{}</math> is of course equal to <math>\frac{1}{2}</math> as it is the mid-point of CD. Thus, the area of the first triangle is <math>[DQ_1P_1]=\frac{1}{2}</math> ⋅ <math>\frac{1}{2}</math> ⋅ <math>\frac{1}{3}</math>. | ||
+ | |||
+ | The second triangle has a base <math>DQ_2^{}</math> equal to that of <math>P_1^{}Q_2^{}</math> (see that <math>\triangle DQ_2^{}P_1^{}</math> ~ <math>\triangle DCB</math>) and using the same similar triangle logic as with the first triangle, we find the area to be <math>[DQ_2P_2]=\frac{1}{2}</math> ⋅ <math>\frac{1}{3}</math> ⋅ <math>\frac{1}{4}</math>. If we continue and test the next few triangles, we will find that the sum of all <math>\triangle DQ_i^{}P_i^{}</math> is equal to | ||
+ | <cmath>\frac{1}{2} \sum\limits_{n=2}^\infty \frac{1}{n(n+1)}</cmath> | ||
+ | or | ||
+ | <cmath>\frac{1}{2} \sum\limits_{n=2}^\infty \left(\frac{1}{n} - \frac{1}{n+1}\right)</cmath> | ||
+ | |||
+ | This is known as a telescoping series because we can see that every term after the first <math>\frac{1}{n}</math> is going to cancel out. Thus, the summation is equal to <math>\frac{1}{2}</math> and after multiplying by the half out in front, we find that the answer is <math>\boxed{\textbf{(B) }\frac{1}{4}}</math>. | ||
+ | |||
+ | === Solution 2 === | ||
+ | (By mastermind.hk16) | ||
+ | |||
+ | Note that <math>AD \|\ P_iQ_{i+1}\ \forall i \in \mathbb{N}</math>. So <math>\triangle ADQ_i \sim \triangle P_{i}Q_{i+1}Q_{i} \ \forall i \in \mathbb{N}</math> | ||
+ | |||
+ | Hence <math>\frac{Q_iQ_{i+1}}{DQ_{i}}=\frac{P_{i}Q_{i+1}}{AD} \ \ \Longrightarrow DQ_i \cdot P_iQ_{i+1}=Q_iQ_{i+1}</math> | ||
+ | |||
+ | We compute <math>\frac{1}{2} \sum_{i=1}^{\infty}DQ_i \cdot P_iQ_{i+1}= \frac{1}{2} \sum_{i=1}^{\infty}Q_iQ_{i+1}=\frac{1}{2} \cdot DQ_1 =\frac{1}{4}</math> | ||
+ | because <math>Q_i \rightarrow D</math> as <math>i \rightarrow \infty</math>. | ||
+ | |||
+ | |||
+ | === Solution 3 === | ||
+ | (By user0003) | ||
+ | |||
+ | We plot the figure on a coordinate plane with <math>D=(0,0)</math> and <math>A</math> in the positive y-direction from the origin. If <math>Q_i=(k, 0)</math> for some <math>k \neq 0</math>, then the line <math>AQ_i</math> can be represented as <math>y=-\frac{x}{k}+1</math>. The intersection of this and <math>BD</math>, which is the line <math>y=x</math>, is | ||
+ | |||
+ | <cmath>P_i = \left(\frac{k}{k+1}, \frac{k}{k+1}\right)</cmath>. | ||
+ | |||
+ | As <math>Q_{i+1}</math> is the projection of <math>P_i</math> onto the x-axis, it lies at <math>\left(\frac{k}{k+1}, 0\right)</math>. We have thus established that moving from <math>Q_i</math> to <math>Q_{i+1}</math> is equivalent to the transformation <math>x \rightarrow \frac{x}{x+1}</math> on the x-coordinate. The closed form of of the x-coordinate of <math>Q_i</math> can be deduced to be <math>\frac{1}{1+i}</math>, which can be determined empirically and proven via induction on the initial case <math>Q_1 = \left(\frac{1}{2}, 0\right)</math>. Now | ||
+ | |||
+ | <cmath>[\Delta DQ_iP_i] = \frac{1}{2}(DQ_i)(Q_{i+1}P_i) = \frac{1}{2}(DQ_i)(DQ_{i+1}),</cmath> | ||
+ | |||
+ | suggesting that <math>[\Delta DQ_iP_i]</math> is equivalent to <math>\frac{1}{2(i+1)(i+2)}</math>. The sum of this from <math>i=1</math> to <math>\infty</math> is a classic telescoping sequence as in Solution 1 and is equal to <math>\boxed{\textbf{(B) }\frac{1}{4}}</math>. | ||
+ | |||
+ | === Solution 4 Diagram and Detailed Steps=== | ||
+ | |||
+ | [[Image:2016_AMC_12B_Problem_21.png|thumb|center|800px| ]] | ||
+ | |||
+ | |||
+ | Midpoint <math>Q_1</math> of <math>\overline{CD}</math>: <math>Q_1 = \left(\frac{1}{2}, 0\right)</math> | ||
+ | |||
+ | Equation of <math>\overline{AQ_1}</math>: Slope <math>m = \frac{0 - 1}{\frac{1}{2} - 0} = -2</math> Equation: <math>y = -2x + 1</math> | ||
+ | |||
+ | Line <math>\overline{BD}</math>: - Equation: <math>y = x</math> | ||
+ | |||
+ | Intersection <math>P_1</math> of <math>\overline{AQ_1}</math> and <math>\overline{BD}</math>: | ||
+ | - Solve:<math>-2x + 1 = x \implies 1 = 3x \implies x = \frac{1}{3}</math> Therefore, <math>P_1 = \left(\frac{1}{3}, \frac{1}{3}\right)</math> | ||
+ | |||
+ | Now, using the pattern for subsequent points <math>P_k</math> and <math>Q_k</math>: | ||
+ | |||
+ | General <math>Q_k</math> - For <math>k \geq 1</math>, <math>Q_k = \left(\frac{1}{k+1}, 0\right)</math> | ||
+ | |||
+ | Equation of <math>\overline{AQ_k}</math> Slope <math>m = -(k+1)</math> Equation: <math>y = -(k+1)x + 1</math> | ||
+ | |||
+ | Intersection <math>P_k</math> of <math>\overline{AQ_k}</math> and <math>\overline{BD}</math>: | ||
+ | - Line <math>\overline{BD}</math>: <math>y = x</math> Solve:<math> -(k+1)x + 1 = x \implies 1 = (k+2)x \implies x = \frac{1}{k+2} </math> | ||
+ | - Therefore, <math>P_k = \left(\frac{1}{k+2}, \frac{1}{k+2}\right)</math> | ||
+ | |||
+ | <math>Q_{k+1}</math> is the foot of the perpendicular from <math>P_k</math> to <math>\overline{CD}</math>, so <math> Q_{k+1} = \left(\frac{1}{k+2}, 0\right)</math> | ||
+ | |||
+ | Area of <math>\triangle DQ_kP_k</math> = <math>\frac{1}{2} \times DQ_k \times \text{Height}\ (y\ of\ P_{k+2})= \frac{1}{2} \times \frac{1}{k+1} \times \frac{1}{k+2} = \frac{1}{2} ( \frac{1}{k+1} - \frac{1}{k+2} ) </math> | ||
+ | |||
+ | This recursive process confirms the telescoping series: | ||
+ | |||
+ | <cmath> | ||
+ | \sum_{k=1}^{\infty} \frac{1}{2(k+1)(k+2)} = \frac{1}{2} \left( \left( \frac{1}{2} - \frac{1}{3} \right) + \left( \frac{1}{3} - \frac{1}{4} \right) + \left( \frac{1}{4} - \frac{1}{5} \right) + \cdots \right) | ||
+ | </cmath> | ||
+ | |||
+ | Most terms cancel, and we are left with: <math>\frac{1}{2} \cdot \frac{1}{2} = \boxed{\textbf{(B) }\frac{1}{4}}</math>. | ||
+ | |||
+ | ~[https://artofproblemsolving.com/wiki/index.php/User:Cyantist luckuso] | ||
==See Also== | ==See Also== | ||
{{AMC12 box|year=2016|ab=B|num-b=20|num-a=22}} | {{AMC12 box|year=2016|ab=B|num-b=20|num-a=22}} | ||
{{MAA Notice}} | {{MAA Notice}} |
Latest revision as of 11:34, 21 July 2024
Contents
Problem
Let be a unit square. Let
be the midpoint of
. For
let
be the intersection of
and
, and let
be the foot of the perpendicular from
to
. What is
Solutions
Solution 1
(By Qwertazertl)
We are tasked with finding the sum of the areas of every where
is a positive integer. We can start by finding the area of the first triangle,
. This is equal to
⋅
⋅
. Notice that since triangle
is similar to triangle
in a 1 : 2 ratio,
must equal
(since we are dealing with a unit square whose side lengths are 1).
is of course equal to
as it is the mid-point of CD. Thus, the area of the first triangle is
⋅
⋅
.
The second triangle has a base equal to that of
(see that
~
) and using the same similar triangle logic as with the first triangle, we find the area to be
⋅
⋅
. If we continue and test the next few triangles, we will find that the sum of all
is equal to
or
This is known as a telescoping series because we can see that every term after the first is going to cancel out. Thus, the summation is equal to
and after multiplying by the half out in front, we find that the answer is
.
Solution 2
(By mastermind.hk16)
Note that . So
Hence
We compute
because
as
.
Solution 3
(By user0003)
We plot the figure on a coordinate plane with and
in the positive y-direction from the origin. If
for some
, then the line
can be represented as
. The intersection of this and
, which is the line
, is
.
As is the projection of
onto the x-axis, it lies at
. We have thus established that moving from
to
is equivalent to the transformation
on the x-coordinate. The closed form of of the x-coordinate of
can be deduced to be
, which can be determined empirically and proven via induction on the initial case
. Now
suggesting that is equivalent to
. The sum of this from
to
is a classic telescoping sequence as in Solution 1 and is equal to
.
Solution 4 Diagram and Detailed Steps
Midpoint of
:
Equation of : Slope
Equation:
Line : - Equation:
Intersection of
and
:
- Solve:Therefore,
![]()
Now, using the pattern for subsequent points and
:
General - For
,
Equation of Slope
Equation:
Intersection of
and
:
- Line:
Solve:
- Therefore,
![]()
is the foot of the perpendicular from
to
, so
Area of =
This recursive process confirms the telescoping series:
Most terms cancel, and we are left with: .
See Also
2016 AMC 12B (Problems • Answer Key • Resources) | |
Preceded by Problem 20 |
Followed by Problem 22 |
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.