Difference between revisions of "2018 AIME II Problems/Problem 7"
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− | + | ==Problem 7== | |
− | ==Problem== | ||
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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 | + | 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 | ||
− | + | ~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 | + | 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. |
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+ | Solution by 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 03:26, 23 December 2023
Problem 7
Triangle has side lengths
,
, and
. Points
are on segment
with
between
and
for
, and points
are on segment
with
between
and
for
. Furthermore, each segment
,
, is parallel to
. The segments cut the triangle into
regions, consisting of
trapezoids and
triangle. Each of the
regions has the same area. Find the number of segments
,
, that have rational length.
Solution 1
For each between
and
, the area of the trapezoid with
as its bottom base is the difference between the areas of two triangles, both similar to
. Let
be the length of segment
. The area of the trapezoid with bases
and
is
times the area of
. (This logic also applies to the topmost triangle if we notice that
.) However, we also know that the area of each shape is
times the area of
. We then have
. Simplifying,
. However, we know that
, so
, and in general,
and
. The smallest
that gives a rational
is
, so
is rational if and only if
for some integer
.The largest
such that
is less than
is
, so
has
possible values.
Solution by zeroman
Solution 2
We have that there are trapezoids and
triangle of equal area, with that one triangle being
. Notice, if we "stack" the trapezoids on top of
the way they already are, we'd create a similar triangle, all of which are similar to
, and since the trapezoids and
have equal area, each of these similar triangles
have area
, and so
. We want the ratio of the side lengths
. 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
so there are
solutions.
~Solution by ktong
~Beautified by jdong2006
Solution 3
Let stand for
, and
. All triangles
are similar by AA. Let the area of
be
. The next trapezoid will also have an area of
, as given. Therefore,
has an area of
. The ratio of the areas is equal to the square of the scale factor for any plane figure and its image. Therefore,
, and the same if
is substituted for
throughout. We want the side
to be rational. Setting up proportions:
which shows that
. In order for
to be rational,
must be some rational multiple of
. This is achieved at
. We end there as
. There are 20 numbers from 1 to 20, so there are
solutions.
Solution by a1b2
See Also
2018 AIME II (Problems • Answer Key • Resources) | ||
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 |
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