Difference between revisions of "2015 AIME II Problems/Problem 7"
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==Problem== | ==Problem== | ||
− | Triangle <math>ABC</math> has side lengths <math>AB = 12</math>, <math>BC = 25</math>, and <math>CA = 17</math>. Rectangle <math>PQRS</math> has vertex <math>P</math> on <math>\overline{AB}</math>, vertex <math>Q</math> on <math>\overline{AC}</math>, and vertices <math>R</math> and <math>S</math> on <math>\overline{BC}</math>. In terms of the side length <math>PQ = | + | Triangle <math>ABC</math> has side lengths <math>AB = 12</math>, <math>BC = 25</math>, and <math>CA = 17</math>. Rectangle <math>PQRS</math> has vertex <math>P</math> on <math>\overline{AB}</math>, vertex <math>Q</math> on <math>\overline{AC}</math>, and vertices <math>R</math> and <math>S</math> on <math>\overline{BC}</math>. In terms of the side length <math>PQ = \omega</math>, the area of <math>PQRS</math> can be expressed as the quadratic polynomial <cmath>Area(PQRS) = \alpha \omega - \beta \omega^2.</cmath> |
− | |||
− | Area( | ||
Then the coefficient <math>\beta = \frac{m}{n}</math>, where <math>m</math> and <math>n</math> are relatively prime positive integers. Find <math>m+n</math>. | Then the coefficient <math>\beta = \frac{m}{n}</math>, where <math>m</math> and <math>n</math> are relatively prime positive integers. Find <math>m+n</math>. | ||
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<cmath>\alpha\omega - \beta\omega^2 = 25\alpha - 625\beta = 0</cmath> | <cmath>\alpha\omega - \beta\omega^2 = 25\alpha - 625\beta = 0</cmath> | ||
− | and <math>\alpha = 25\beta</math>. If <math>\omega = \frac{25}{2}</math>, we can reflect <math>APQ</math> over PQ, <math>PBS</math> over <math>PS</math>, and <math>QCR</math> over <math>QR</math> to completely cover rectangle <math>PQRS</math>, so the area of <math>PQRS</math> is half the area of the triangle. Using Heron's formula, since <math>s = \frac{12 + 17 + 25}{2} = 27</math>, | + | and <math>\alpha = 25\beta</math>. If <math>\omega = \frac{25}{2}</math>, we can reflect <math>APQ</math> over <math>PQ</math>, <math>PBS</math> over <math>PS</math>, and <math>QCR</math> over <math>QR</math> to completely cover rectangle <math>PQRS</math>, so the area of <math>PQRS</math> is half the area of the triangle. Using Heron's formula, since <math>s = \frac{12 + 17 + 25}{2} = 27</math>, |
<cmath> [ABC] = \sqrt{27 \cdot 15 \cdot 10 \cdot 2} = 90</cmath> | <cmath> [ABC] = \sqrt{27 \cdot 15 \cdot 10 \cdot 2} = 90</cmath> | ||
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label("$B$",B,SW); | label("$B$",B,SW); | ||
label("$C$",C,SE); | label("$C$",C,SE); | ||
− | label("$E$",E, | + | label("$E$",E,SE); |
label("$F$",F,NE); | label("$F$",F,NE); | ||
label("$P$",P,NW); | label("$P$",P,NW); | ||
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label("$S$",S,SW); | label("$S$",S,SW); | ||
draw(rightanglemark(B,E,A,12)); | draw(rightanglemark(B,E,A,12)); | ||
+ | dot(E); | ||
+ | dot(F); | ||
</asy> | </asy> | ||
Similar triangles can also solve the problem. | Similar triangles can also solve the problem. | ||
− | First, solve for the area of the triangle. <math>[ABC] = 90</math>. This can be done by Heron's Formula or placing an <math>8-15-17</math> right triangle on <math> | + | First, solve for the area of the triangle. <math>[ABC] = 90</math>. This can be done by Heron's Formula or placing an <math>8-15-17</math> right triangle on <math>AC</math> and solving. (The <math>8</math> side would be collinear with line <math>AB</math>) |
After finding the area, solve for the altitude to <math>BC</math>. Let <math>E</math> be the intersection of the altitude from <math>A</math> and side <math>BC</math>. Then <math>AE = \frac{36}{5}</math>. | After finding the area, solve for the altitude to <math>BC</math>. Let <math>E</math> be the intersection of the altitude from <math>A</math> and side <math>BC</math>. Then <math>AE = \frac{36}{5}</math>. | ||
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This means that <math>\beta = \frac{36}{125} \Rightarrow (m,n) = (36,125) \Rightarrow m+n = \boxed{161}</math>. | This means that <math>\beta = \frac{36}{125} \Rightarrow (m,n) = (36,125) \Rightarrow m+n = \boxed{161}</math>. | ||
− | |||
− | |||
==Solution 3== | ==Solution 3== | ||
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Now, draw a parallel to <math>AB</math> from <math>Q</math>, intersecting <math>BC</math> at <math>T</math>. Then <math>BT = w</math> in parallelogram <math>QPBT</math>, and so <math>CT = 25 - w</math>. Clearly, <math>CQT</math> and <math>CAB</math> are similar triangles, and so their altitudes have lengths proportional to their corresponding base sides, and so | Now, draw a parallel to <math>AB</math> from <math>Q</math>, intersecting <math>BC</math> at <math>T</math>. Then <math>BT = w</math> in parallelogram <math>QPBT</math>, and so <math>CT = 25 - w</math>. Clearly, <math>CQT</math> and <math>CAB</math> are similar triangles, and so their altitudes have lengths proportional to their corresponding base sides, and so | ||
<cmath>\frac{QR}{\frac{36}{5}} = \frac{25 - w}{25}.</cmath> | <cmath>\frac{QR}{\frac{36}{5}} = \frac{25 - w}{25}.</cmath> | ||
− | Solving gives <math>[PQRS] = \dfrac{36}{5} \cdot \dfrac{25 - w}{25} = \dfrac{36w}{5} - \dfrac{36w^2}{125}</math>, so the answer is <math>36 + 125 = 161</math>. | + | Solving gives <math>[PQRS] = \dfrac{36}{5} \cdot \dfrac{25 - w}{25} \cdot w = \dfrac{36w}{5} - \dfrac{36w^2}{125}</math>, so the answer is <math>36 + 125 = 161</math>. |
+ | |||
+ | ==Solution 4== | ||
+ | Using the diagram from Solution 2 above, label <math>AF</math> to be <math>h</math>. Through Heron's formula, the area of <math>\triangle ABC</math> turns out to be <math>90</math>, so using <math>AE</math> as the height and <math>BC</math> as the base yields <math>AE=\frac{36}{5}</math>. Now, through the use of similarity between <math>\triangle APQ</math> and <math>\triangle ABC</math>, you find <math>\frac{w}{25}=\frac{h}{36/5}</math>. Thus, <math>h=\frac{36w}{125}</math>. To find the height of the rectangle, subtract <math>h</math> from <math>\frac{36}{5}</math> to get <math>\left(\frac{36}{5}-\frac{36w}{125}\right)</math>, and multiply this by the other given side <math>w</math> to get <math>\frac{36w}{5}-\frac{36w^2}{125}</math> for the area of the rectangle. Finally, <math>36+125=\boxed{161}</math>. | ||
+ | |||
+ | ==Solution 5== | ||
+ | Using the diagram as shown in Solution 2, let <math>AE=h</math> and <math>AP=L</math> | ||
+ | Now, by Heron's formula, we find that the <math>[ABC]=90</math>. Hence, <math>h=\frac{36}{5}</math> | ||
+ | |||
+ | Now, we see that <math>\sin{B}=\frac{PS}{12-L}\implies PS=\sin{B}(12-L)</math> | ||
+ | We easily find that <math>\sin{B}=\frac{3}{5}</math> | ||
+ | |||
+ | Hence, <math>PS=\frac{3}{5}(12-L)</math> | ||
+ | |||
+ | Now, we see that <math>[PQRS]=\frac{3}{5}(12-L)(w)</math> | ||
+ | |||
+ | Now, it is obvious that we want to find <math>L</math> in terms of <math>W</math>. | ||
+ | |||
+ | Looking at the diagram, we see that because <math>PQRS</math> is a rectangle, <math>\triangle{APQ}\sim{\triangle{ABC}}</math> | ||
+ | |||
+ | Hence.. we can now set up similar triangles. | ||
+ | |||
+ | We have that <math>\frac{AP}{AB}=\frac{PQ}{BC}\implies \frac{L}{12}=\frac{W}{25}\implies 25L=12W\implies L=\frac{12W}{25}</math>. | ||
+ | |||
+ | Plugging back in.. | ||
+ | |||
+ | |||
+ | <math>[PQRS]=\frac{3w}{5}(12-(\frac{12W}{25}))\implies \frac{3w}{5}(\frac{300-12W}{25})\implies \frac{900W-36W^2}{125}</math> | ||
+ | |||
+ | Simplifying, we get <math>\frac{36W}{5}-\frac{36W^2}{125}</math> | ||
+ | |||
+ | Hence, <math>125+36=\boxed{161}</math> | ||
+ | |||
+ | ==Solution 6== | ||
+ | Proceed as in solution 1. When <math>\omega</math> is equal to zero, <math>\alpha - \beta\omega=\alpha</math> is equal to the altitude. This means that <math>25\beta</math> is equal to <math>\frac{36}{5}</math>, so <math>\beta = \frac{36}{125}</math>, yielding <math>\boxed{161}</math>. | ||
+ | |||
+ | |||
+ | |||
+ | ==Video Solution== | ||
+ | https://www.youtube.com/watch?v=9re2qLzOKWk&t=554s | ||
+ | |||
+ | ~MathProblemSolvingSkills.com | ||
+ | |||
− | |||
==See also== | ==See also== | ||
{{AIME box|year=2015|n=II|num-b=6|num-a=8}} | {{AIME box|year=2015|n=II|num-b=6|num-a=8}} | ||
+ | [[Category: Intermediate Geometry Problems]] | ||
{{MAA Notice}} | {{MAA Notice}} |
Latest revision as of 14:08, 14 February 2023
Contents
Problem
Triangle has side lengths
,
, and
. Rectangle
has vertex
on
, vertex
on
, and vertices
and
on
. In terms of the side length
, the area of
can be expressed as the quadratic polynomial
Then the coefficient , where
and
are relatively prime positive integers. Find
.
Solution 1
If , the area of rectangle
is
, so
and . If
, we can reflect
over
,
over
, and
over
to completely cover rectangle
, so the area of
is half the area of the triangle. Using Heron's formula, since
,
so
and
so the answer is .
Solution 2
Similar triangles can also solve the problem.
First, solve for the area of the triangle. . This can be done by Heron's Formula or placing an
right triangle on
and solving. (The
side would be collinear with line
)
After finding the area, solve for the altitude to . Let
be the intersection of the altitude from
and side
. Then
.
Solving for
using the Pythagorean Formula, we get
. We then know that
.
Now consider the rectangle . Since
is collinear with
and parallel to
,
is parallel to
meaning
is similar to
.
Let be the intersection between
and
. By the similar triangles, we know that
. Since
. We can solve for
and
in terms of
. We get that
and
.
Let's work with . We know that
is parallel to
so
is similar to
. We can set up the proportion:
. Solving for
,
.
We can solve for then since we know that
and
.
Therefore, .
This means that .
Solution 3
Heron's Formula gives so the altitude from
to
has length
Now, draw a parallel to from
, intersecting
at
. Then
in parallelogram
, and so
. Clearly,
and
are similar triangles, and so their altitudes have lengths proportional to their corresponding base sides, and so
Solving gives
, so the answer is
.
Solution 4
Using the diagram from Solution 2 above, label to be
. Through Heron's formula, the area of
turns out to be
, so using
as the height and
as the base yields
. Now, through the use of similarity between
and
, you find
. Thus,
. To find the height of the rectangle, subtract
from
to get
, and multiply this by the other given side
to get
for the area of the rectangle. Finally,
.
Solution 5
Using the diagram as shown in Solution 2, let and
Now, by Heron's formula, we find that the
. Hence,
Now, we see that
We easily find that
Hence,
Now, we see that
Now, it is obvious that we want to find in terms of
.
Looking at the diagram, we see that because is a rectangle,
Hence.. we can now set up similar triangles.
We have that .
Plugging back in..
Simplifying, we get
Hence,
Solution 6
Proceed as in solution 1. When is equal to zero,
is equal to the altitude. This means that
is equal to
, so
, yielding
.
Video Solution
https://www.youtube.com/watch?v=9re2qLzOKWk&t=554s
~MathProblemSolvingSkills.com
See also
2015 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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.