Difference between revisions of "2018 AIME II Problems/Problem 7"
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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. | ||
Revision as of 15:33, 9 August 2018
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
Solution 3
Let stand for
, and
. All triangles
are similar by AAA. 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
2018 AIME II (Problems • Answer Key • Resources) | ||
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