Difference between revisions of "2018 AIME II Problems/Problem 7"

(Adding problem section)
(Solution 2)
 
(16 intermediate revisions by 8 users not shown)
Line 1: Line 1:
 
+
==Problem 7==
==Problem==
 
== Problem ==
 
 
 
 
Triangle <math>ABC</math> has side lengths <math>AB = 9</math>, <math>BC =</math> <math>5\sqrt{3}</math>, and <math>AC = 12</math>. Points <math>A = P_{0}, P_{1}, P_{2}, ... , P_{2450} = B</math> are on segment <math>\overline{AB}</math> with <math>P_{k}</math> between <math>P_{k-1}</math> and <math>P_{k+1}</math> for <math>k = 1, 2, ..., 2449</math>, and points <math>A = Q_{0}, Q_{1}, Q_{2}, ... , Q_{2450} = C</math> are on segment <math>\overline{AC}</math> with <math>Q_{k}</math> between <math>Q_{k-1}</math> and <math>Q_{k+1}</math> for <math>k = 1, 2, ..., 2449</math>. Furthermore, each segment <math>\overline{P_{k}Q_{k}}</math>, <math>k = 1, 2, ..., 2449</math>, is parallel to <math>\overline{BC}</math>. The segments cut the triangle into <math>2450</math> regions, consisting of <math>2449</math> trapezoids and <math>1</math> triangle. Each of the <math>2450</math> regions has the same area. Find the number of segments <math>\overline{P_{k}Q_{k}}</math>, <math>k = 1, 2, ..., 2450</math>, that have rational length.
 
Triangle <math>ABC</math> has side lengths <math>AB = 9</math>, <math>BC =</math> <math>5\sqrt{3}</math>, and <math>AC = 12</math>. Points <math>A = P_{0}, P_{1}, P_{2}, ... , P_{2450} = B</math> are on segment <math>\overline{AB}</math> with <math>P_{k}</math> between <math>P_{k-1}</math> and <math>P_{k+1}</math> for <math>k = 1, 2, ..., 2449</math>, and points <math>A = Q_{0}, Q_{1}, Q_{2}, ... , Q_{2450} = C</math> are on segment <math>\overline{AC}</math> with <math>Q_{k}</math> between <math>Q_{k-1}</math> and <math>Q_{k+1}</math> for <math>k = 1, 2, ..., 2449</math>. Furthermore, each segment <math>\overline{P_{k}Q_{k}}</math>, <math>k = 1, 2, ..., 2449</math>, is parallel to <math>\overline{BC}</math>. The segments cut the triangle into <math>2450</math> regions, consisting of <math>2449</math> trapezoids and <math>1</math> triangle. Each of the <math>2450</math> regions has the same area. Find the number of segments <math>\overline{P_{k}Q_{k}}</math>, <math>k = 1, 2, ..., 2450</math>, that have rational length.
  
 
== Solution 1 ==
 
== Solution 1 ==
For each <math>k</math> between <math>2</math> and <math>2450</math>, the area of the trapezoid with <math>\overline{P_kQ_k}</math> as its bottom base is the difference between the areas of two triangles, both similar to <math>\triangle{ABC}</math>. Let <math>d_k</math> be the length of segment <math>\overline{P_kQ_k}</math>. The area of the trapezoid with bases <math>\overline{P_{k-1}Q_{k-1}}</math> and <math>P_kQ_k</math> is <math>(\frac{d_k}{5\sqrt{3}})^2 - (\frac{d_{k-1}}{5\sqrt{3}})^2 = \frac{d_k^2-d_{k-1}^2}{75}</math> times the area of <math>\triangle{ABC}</math>. (This logic also applies to the topmost triangle if we notice that <math>d_0 = 0</math>.) However, we also know that the area of each shape is <math>\frac{1}{2450}</math> times the area of <math>\triangle{ABC}</math>. We then have <math>\frac{d_k^2-d_{k-1}^2}{75} = \frac{1}{2450}</math>. Simplifying, <math>d_k^2-d_{k-1}^2 = \frac{3}{98}</math>. However, we know that <math>d_0^2 = 0</math>, so <math>d_1^2 = \frac{3}{98}</math>, and in general, <math>d_k^2 = \frac{3k}{98}</math> and <math>d_k = \frac{\sqrt{\frac{3k}{2}}}{7}</math>. The smallest <math>k</math> that gives a rational <math>d_k</math> is <math>6</math>, so <math>d_k</math> is rational if and only if <math>k = 6n^2</math> for some integer <math>n</math>.The largest <math>n</math> such that <math>6n^2</math> is less than <math>2450</math> is <math>20</math>, so <math>k</math> has <math>\boxed{020}</math> possible values.
+
For each <math>k</math> between <math>2</math> and <math>2450</math>, the area of the trapezoid with <math>\overline{P_kQ_k}</math> as its bottom base is the difference between the areas of two triangles, both similar to <math>\triangle{ABC}</math>. Let <math>d_k</math> be the length of segment <math>\overline{P_kQ_k}</math>. The area of the trapezoid with bases <math>\overline{P_{k-1}Q_{k-1}}</math> and <math>P_kQ_k</math> is <math>\left(\frac{d_k}{5\sqrt{3}}\right)^2 - \left(\frac{d_{k-1}}{5\sqrt{3}}\right)^2 = \frac{d_k^2-d_{k-1}^2}{75}</math> times the area of <math>\triangle{ABC}</math>. (This logic also applies to the topmost triangle if we notice that <math>d_0 = 0</math>.) However, we also know that the area of each shape is <math>\frac{1}{2450}</math> times the area of <math>\triangle{ABC}</math>. We then have <math>\frac{d_k^2-d_{k-1}^2}{75} = \frac{1}{2450}</math>. Simplifying, <math>d_k^2-d_{k-1}^2 = \frac{3}{98}</math>. However, we know that <math>d_0^2 = 0</math>, so <math>d_1^2 = \frac{3}{98}</math>, and in general, <math>d_k^2 = \frac{3k}{98}</math> and <math>d_k = \frac{\sqrt{\frac{3k}{2}}}{7}</math>. The smallest <math>k</math> that gives a rational <math>d_k</math> is <math>6</math>, so <math>d_k</math> is rational if and only if <math>k = 6n^2</math> for some integer <math>n</math>.The largest <math>n</math> such that <math>6n^2</math> is less than <math>2450</math> is <math>20</math>, so <math>k</math> has <math>\boxed{020}</math> possible values.
  
 
Solution by zeroman
 
Solution by zeroman
  
 
==Solution 2==
 
==Solution 2==
We have that there are <math>2449</math> trapezoids and <math>1</math> triangle of equal area, with that one triangle being <math>AP_1Q_1</math>. Notice, if we "stack" the trapezoids on top of <math>\bigtriangleup AP_1Q_1</math> the way they already are, we'd create a similar triangle, all of which are similar to <math>\bigtriangleup ABC</math>, and since the trapezoids and <math>\bigtriangleup AP_1Q_1</math> have equal area, each of these similar triangles, <math>AP_kQ_k</math> have area <math>\frac{k}{2450}\left[ ABC\right]</math>, and so <math>\frac{\left[ AP_kQ_k\right]}{\left[ABC\right]}=\frac{k}{2450}</math>. We want the ratio of the side lengths <math>P_kQ_k:BC</math>. Since area is a 2-dimensional unit of measurement, and side lengths are 1-dimensional, the ratio is simply the square root of the areas, or <math>\frac{P_kQ_k}{BC}=\sqrt{\frac{k}{2450}} \implies P_kQ_k=BC\cdot \sqrt{\frac{k}{2450}}=5\sqrt{3}\cdot\sqrt{\frac{k}{2450}}=\frac{1}{7}\cdot \sqrt{\frac{3k}{2}}=\frac{3}{7}\sqrt{\frac{k}{6}} \implies k=6n^2<2450 \implies 0<n\leq 20</math>, so there are <math>\boxed{020}</math> solutions.  
+
We have that there are <math>2449</math> trapezoids and <math>1</math> triangle of equal area, with that one triangle being <math>AP_1Q_1</math>. Notice, if we "stack" the trapezoids on top of <math>\bigtriangleup AP_1Q_1</math> the way they already are, we'd create a similar triangle, all of which are similar to <math>\bigtriangleup ABC</math>, and since the trapezoids and <math>\bigtriangleup AP_1Q_1</math> have equal area, each of these similar triangles, <math>AP_kQ_k</math> have area <math>\frac{k}{2450}\left[ ABC\right]</math>, and so <math>\frac{\left[ AP_kQ_k\right]}{\left[ABC\right]}=\frac{k}{2450}</math>. We want the ratio of the side lengths <math>P_kQ_k:BC</math>. Since area is a 2-dimensional unit of measurement, and side lengths are 1-dimensional, the ratio is simply the square root of the areas, or <cmath>\frac{P_kQ_k}{BC}=\sqrt{\frac{k}{2450}}</cmath> <cmath>\implies P_kQ_k=BC\cdot \sqrt{\frac{k}{2450}}=5\sqrt{3}\cdot\sqrt{\frac{k}{2450}}=\frac{1}{7}\cdot \sqrt{\frac{3k}{2}}=\frac{3}{7}\sqrt{\frac{k}{6}}</cmath> <cmath>\implies k=6n^2<2450 </cmath> <cmath>\implies 0<n\leq 20</cmath> so there are <math>\boxed{020}</math> solutions.  
 +
 
 +
~Solution by ktong
  
Solution by ktong
+
~Beautified by jdong2006
  
 
==Solution 3==
 
==Solution 3==
  
Let <math>T_1</math> stand for <math>AP_1Q_1</math>, and <math>T_k = AP_kQ_k</math>. All triangles <math>T</math> are similar by AAA. Let the area of <math>T_1</math> be <math>x</math>. The next trapezoid will also have an area of <math>x</math>, as given. Therefore, <math>T_k</math> has an area of <math>\sqrt{k}x</math>. The ratio of the areas is equal to the square of the scale factor for any plane figure and its image. Therefore, <math>AP_k=AP_1\cdot \sqrt{k}</math>, and the same if <math>Q</math> is substituted for <math>P</math> throughout. We want the side <math>P_kQ_k</math> to be rational. Setting up proportions: <cmath>5\sqrt{3} : \sqrt{2450}=35\sqrt{2}</cmath> <cmath>\sqrt{6} : 14</cmath> which shows that <math>x=\frac{\sqrt{6}}{14}</math>. In order for <math>\sqrt{k}x</math> to be rational, <math>\sqrt{k}</math> must be some rational multiple of <math>\sqrt{6}</math>. This is achieved at <math>k=\sqrt{6}, 2\sqrt{6}, ... , 20\sqrt{6}</math>. We end there as <math>21\sqrt{6}=\sqrt{2646}</math>. There are 20 numbers from 1 to 20, so there are <math>\boxed{020}</math> solutions.
+
Let <math>T_1</math> stand for <math>AP_1Q_1</math>, and <math>T_k = AP_kQ_k</math>. All triangles <math>T</math> are similar by AA. Let the area of <math>T_1</math> be <math>x</math>. The next trapezoid will also have an area of <math>x</math>, as given. Therefore, <math>T_k</math> has an area of <math>kx</math>. The ratio of the areas is equal to the square of the scale factor for any plane figure and its image. Therefore, <math>P_k Q_k=P_1 Q_1\cdot \sqrt{k}</math>, and the same if <math>Q</math> is substituted for <math>P</math> throughout. We want the side <math>P_k Q_k</math> to be rational. Setting up proportions: <cmath>5\sqrt{3} : \sqrt{2450}=35\sqrt{2}</cmath> <cmath>\sqrt{6} : 14</cmath> which shows that <math>P_1 Q_1=\frac{\sqrt{6}}{14}</math>. In order for <math>\sqrt{k} P_1 Q_1</math> to be rational, <math>\sqrt{k}</math> must be some rational multiple of <math>\sqrt{6}</math>. This is achieved at <math>\sqrt{k}=\sqrt{6}, 2\sqrt{6}, \ldots, 20\sqrt{6}</math>. We end there as <math>21\sqrt{6}=\sqrt{2646}</math>. There are 20 numbers from 1 to 20, so there are <math>\boxed{020}</math> solutions.
 +
 
 +
Solution by a1b2
  
Solution by [[User:a1b2|a1b2]]
+
==See Also==
 
 
{{AIME box|year=2018|n=II|num-b=6|num-a=8}}
 
{{AIME box|year=2018|n=II|num-b=6|num-a=8}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Latest revision as of 00:12, 3 September 2021

Problem 7

Triangle $ABC$ has side lengths $AB = 9$, $BC =$ $5\sqrt{3}$, and $AC = 12$. Points $A = P_{0}, P_{1}, P_{2}, ... , P_{2450} = B$ are on segment $\overline{AB}$ with $P_{k}$ between $P_{k-1}$ and $P_{k+1}$ for $k = 1, 2, ..., 2449$, and points $A = Q_{0}, Q_{1}, Q_{2}, ... , Q_{2450} = C$ are on segment $\overline{AC}$ with $Q_{k}$ between $Q_{k-1}$ and $Q_{k+1}$ for $k = 1, 2, ..., 2449$. Furthermore, each segment $\overline{P_{k}Q_{k}}$, $k = 1, 2, ..., 2449$, is parallel to $\overline{BC}$. The segments cut the triangle into $2450$ regions, consisting of $2449$ trapezoids and $1$ triangle. Each of the $2450$ regions has the same area. Find the number of segments $\overline{P_{k}Q_{k}}$, $k = 1, 2, ..., 2450$, that have rational length.

Solution 1

For each $k$ between $2$ and $2450$, the area of the trapezoid with $\overline{P_kQ_k}$ as its bottom base is the difference between the areas of two triangles, both similar to $\triangle{ABC}$. Let $d_k$ be the length of segment $\overline{P_kQ_k}$. The area of the trapezoid with bases $\overline{P_{k-1}Q_{k-1}}$ and $P_kQ_k$ is $\left(\frac{d_k}{5\sqrt{3}}\right)^2 - \left(\frac{d_{k-1}}{5\sqrt{3}}\right)^2 = \frac{d_k^2-d_{k-1}^2}{75}$ times the area of $\triangle{ABC}$. (This logic also applies to the topmost triangle if we notice that $d_0 = 0$.) However, we also know that the area of each shape is $\frac{1}{2450}$ times the area of $\triangle{ABC}$. We then have $\frac{d_k^2-d_{k-1}^2}{75} = \frac{1}{2450}$. Simplifying, $d_k^2-d_{k-1}^2 = \frac{3}{98}$. However, we know that $d_0^2 = 0$, so $d_1^2 = \frac{3}{98}$, and in general, $d_k^2 = \frac{3k}{98}$ and $d_k = \frac{\sqrt{\frac{3k}{2}}}{7}$. The smallest $k$ that gives a rational $d_k$ is $6$, so $d_k$ is rational if and only if $k = 6n^2$ for some integer $n$.The largest $n$ such that $6n^2$ is less than $2450$ is $20$, so $k$ has $\boxed{020}$ possible values.

Solution by zeroman

Solution 2

We have that there are $2449$ trapezoids and $1$ triangle of equal area, with that one triangle being $AP_1Q_1$. Notice, if we "stack" the trapezoids on top of $\bigtriangleup AP_1Q_1$ the way they already are, we'd create a similar triangle, all of which are similar to $\bigtriangleup ABC$, and since the trapezoids and $\bigtriangleup AP_1Q_1$ have equal area, each of these similar triangles, $AP_kQ_k$ have area $\frac{k}{2450}\left[ ABC\right]$, and so $\frac{\left[ AP_kQ_k\right]}{\left[ABC\right]}=\frac{k}{2450}$. We want the ratio of the side lengths $P_kQ_k:BC$. Since area is a 2-dimensional unit of measurement, and side lengths are 1-dimensional, the ratio is simply the square root of the areas, or \[\frac{P_kQ_k}{BC}=\sqrt{\frac{k}{2450}}\] \[\implies P_kQ_k=BC\cdot \sqrt{\frac{k}{2450}}=5\sqrt{3}\cdot\sqrt{\frac{k}{2450}}=\frac{1}{7}\cdot \sqrt{\frac{3k}{2}}=\frac{3}{7}\sqrt{\frac{k}{6}}\] \[\implies k=6n^2<2450\] \[\implies 0<n\leq 20\] so there are $\boxed{020}$ solutions.

~Solution by ktong

~Beautified by jdong2006

Solution 3

Let $T_1$ stand for $AP_1Q_1$, and $T_k = AP_kQ_k$. All triangles $T$ are similar by AA. Let the area of $T_1$ be $x$. The next trapezoid will also have an area of $x$, as given. Therefore, $T_k$ has an area of $kx$. The ratio of the areas is equal to the square of the scale factor for any plane figure and its image. Therefore, $P_k Q_k=P_1 Q_1\cdot \sqrt{k}$, and the same if $Q$ is substituted for $P$ throughout. We want the side $P_k Q_k$ to be rational. Setting up proportions: \[5\sqrt{3} : \sqrt{2450}=35\sqrt{2}\] \[\sqrt{6} : 14\] which shows that $P_1 Q_1=\frac{\sqrt{6}}{14}$. In order for $\sqrt{k} P_1 Q_1$ to be rational, $\sqrt{k}$ must be some rational multiple of $\sqrt{6}$. This is achieved at $\sqrt{k}=\sqrt{6}, 2\sqrt{6}, \ldots, 20\sqrt{6}$. We end there as $21\sqrt{6}=\sqrt{2646}$. There are 20 numbers from 1 to 20, so there are $\boxed{020}$ solutions.

Solution by a1b2

See Also

2018 AIME II (ProblemsAnswer KeyResources)
Preceded by
Problem 6
Followed by
Problem 8
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15
All AIME Problems and Solutions

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png

Invalid username
Login to AoPS