Difference between revisions of "2002 AMC 8 Problems/Problem 20"
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The area of triangle <math>XYZ</math> is <math>8</math> square inches. You can turn the triangle into a rectangle by drawing a triangle on the left side with the hypotenuse of it being <math>YAX</math> and another triangle on the right side, with its hypotenuse being <math>XBZ</math>. After drawing the square, you can cut it into <math>8</math> squares. The area of the rectangle is <math>16</math> square inches because the triangle on the left is half of <math>XYCZ</math> and there's another triangle on the other side, equaling them to <math>8</math> square inches. We will be focusing on the left side of the rectangle of <math>XC</math> because it includes the shaded region. It's split into <math>4</math> equal squares. The area of this triangle is <math>8</math> square inches because the total area of the rectangle is <math>16</math> square inches and <math>16</math> divided by <math>2</math> is <math>8</math>. There are <math>4</math> sections, so you would do <math>8</math> divided by <math>4</math> in order to find the area of one square. That means that the area of the top right square is <math>2</math> inches and because it's not needed, we will subtract <math>2</math> from <math>8</math> to get rid of it. If you turn around the paper, you notice that the square with half of the shaded region, and the square above it is the shaded region, except flipped and turned. Therefore, the remaining of the area of it which is <math>6</math> should then be divided by <math>2</math> in order to find the shaded region since the shaded region is equal to the other square and half a square. | The area of triangle <math>XYZ</math> is <math>8</math> square inches. You can turn the triangle into a rectangle by drawing a triangle on the left side with the hypotenuse of it being <math>YAX</math> and another triangle on the right side, with its hypotenuse being <math>XBZ</math>. After drawing the square, you can cut it into <math>8</math> squares. The area of the rectangle is <math>16</math> square inches because the triangle on the left is half of <math>XYCZ</math> and there's another triangle on the other side, equaling them to <math>8</math> square inches. We will be focusing on the left side of the rectangle of <math>XC</math> because it includes the shaded region. It's split into <math>4</math> equal squares. The area of this triangle is <math>8</math> square inches because the total area of the rectangle is <math>16</math> square inches and <math>16</math> divided by <math>2</math> is <math>8</math>. There are <math>4</math> sections, so you would do <math>8</math> divided by <math>4</math> in order to find the area of one square. That means that the area of the top right square is <math>2</math> inches and because it's not needed, we will subtract <math>2</math> from <math>8</math> to get rid of it. If you turn around the paper, you notice that the square with half of the shaded region, and the square above it is the shaded region, except flipped and turned. Therefore, the remaining of the area of it which is <math>6</math> should then be divided by <math>2</math> in order to find the shaded region since the shaded region is equal to the other square and half a square. | ||
<math>6</math> divided by <math>2</math> is <math>3</math>, so the answer is <math>3</math>. | <math>6</math> divided by <math>2</math> is <math>3</math>, so the answer is <math>3</math>. | ||
+ | ==Solution 5 (Dividing into Congruent Triangles)== | ||
+ | <asy> | ||
+ | /* AMC8 2002 #20 Problem */ | ||
+ | draw((0,0)--(10,0)--(5,4)--cycle); | ||
+ | draw((2.5,2)--(7.5,2)); | ||
+ | draw((5,4)--(5,0)); | ||
+ | fill((0,0)--(2.5,2)--(5,2)--(5,0)--cycle, mediumgrey); | ||
+ | label(scale(0.8)*"$X$", (5,4), N); | ||
+ | label(scale(0.8)*"$Y$", (0,0), W); | ||
+ | label(scale(0.8)*"$Z$", (10,0), E); | ||
+ | label(scale(0.8)*"$A$", (2.5,2.2), W); | ||
+ | label(scale(0.8)*"$B$", (7.5,2.2), E); | ||
+ | label(scale(0.8)*"$C$", (5,0), S); | ||
+ | fill((0,-.8)--(1,-.8)--(1,-.95)--cycle, white); | ||
+ | /* Edit for solution */ | ||
+ | /* A--C */ | ||
+ | draw((2.5,2)--(5,0)); | ||
+ | /* A--[half of YC] */ | ||
+ | draw((2.5,2)--(2.5,0)); | ||
+ | label(scale(0.8)*"$D$",(2.5,0),S); | ||
+ | /* B--C */ | ||
+ | draw((7.5,2)--(5,0)); | ||
+ | /* B--[half of CZ] */ | ||
+ | draw((7.5,2)--(7.5,0)); | ||
+ | label(scale(0.8)*"$E$",(7.5,0),S); | ||
+ | /* drawing right angles */ | ||
+ | draw((2.5,0.5)--(2,0.5)); | ||
+ | draw((2,0.5)--(2,0)); | ||
+ | |||
+ | draw((2.5,0.5)--(3,0.5)); | ||
+ | draw((3,0.5)--(3,0)); | ||
+ | |||
+ | draw((7.5,0.5)--(7,0.5)); | ||
+ | draw((7,0.5)--(7,0)); | ||
+ | |||
+ | draw((7.5,0.5)--(8,0.5)); | ||
+ | draw((8,0.5)--(8,0)); | ||
+ | |||
+ | draw((5.5,2.5)--(4.5,2.5)--(4.5,1.5)--(5.5,1.5)--cycle); | ||
+ | |||
+ | label(scale(0.8)*"$F$", (5,2), NE); | ||
+ | |||
+ | label(scale(1)*"All credits for original unedited asymptote for the problem go to whoever made the asymptote in the 'Problem' section.", (5,-1), S); | ||
+ | </asy> | ||
+ | |||
+ | We draw the triangles as shown above. | ||
+ | |||
+ | Note that <math>\overline{AD}</math> is perpendicular to <math>\overline{YC}</math>; <math>\overline{BE}</math> is perpendicular to <math>\overline{CZ}</math>. | ||
+ | |||
+ | <math>\overline{AB}</math> is perpendicular to <math>\overline{YZ}</math>, so because of corresponding angles theorem, <math>\angle YAD=\angle AXF</math>. We are also told <math>\overline{YA}=\overline{AX}</math>. Both <math>\bigtriangleup \text{XAF}</math> and <math>\bigtriangleup \text{AYD}</math> have a <math>90\degree</math> angle. Because of AAS Congruency (Angle-Angle-Side; <math>90\degree</math>, <math>\angle YAD</math>, and side <math>\overline{YA}</math>), <math>\bigtriangleup \text{XAF}</math> is congruent to <math>\bigtriangleup \text{AYD}</math>. Because of symmetry, the same goes for <math>\bigtriangleup \text{XFB}</math> and <math>\bigtriangleup \text{BEZ}</math>. | ||
==Video Solution== | ==Video Solution== | ||
− | |||
https://www.youtube.com/watch?v=zwy5U5IQi88 ~David | https://www.youtube.com/watch?v=zwy5U5IQi88 ~David | ||
Revision as of 19:42, 28 December 2023
Contents
Problem
The area of triangle is 8 square inches. Points
and
are midpoints of congruent segments
and
. Altitude
bisects
. The area (in square inches) of the shaded region is
Solution 1
The shaded region is a right trapezoid. Assume WLOG that . Then because the area of
is equal to 8, the height of the triangle
. Because the line
is a midsegment, the top base of the trapezoid is
. Also,
divides
in two, so the height of the trapezoid is
. The bottom base is
. The area of the shaded region is
.
Solution 2
Since and
are the midpoints of
and
, respectively,
.
Draw segments
and
.
Drawing an altitude in an isoceles triangle splits the triangle into 2 congruent triangles and we also know that
.
is the line that connects the midpoints of two sides of a triangle together, which means that
is parallel to and half in length of
. Then
.
Since
is parallel to
, and
is the transversal,
Similarly,
Then, by SAS,
.
Since corresponding parts of congruent triangles are congruent,
.
Since ACY and BCZ are now isosceles triangles,
and
Using the fact that
is parallel to
,
and
.
Now
.
Draw an altitude through each of them such that each triangle is split into two congruent right triangles. Now there are a total of 8 congruent small triangles, each with area 1. The shaded area has three of these triangles, so it has area 3.
Basically the proof is to show . If you just look at the diagram you can easily see that the triangles are congruent and you would solve this a lot faster. Anyways, since those triangles are congruent, you can split each in half to find eight congruent triangles with area 1, and since the shaded region has three of these triangles, its area is
.
Solution 3
We know the area of triangle is
square inches. The area of a triangle can also be represented as
or in this problem
. By solving, we have
With SAS congruence, triangles and
are congruent. Hence, triangle
. (Let's say point
is the intersection between line segments
and
.) We can find the area of the trapezoid
by subtracting the area of triangle
from
.
We find the area of triangle by the
formula-
.
is
of
from solution 1. The area of
is
.
Therefore, the area of the shaded area- trapezoid has area
.
- sarah07
Solution 4 (Dummed down)
The area of triangle is
square inches. You can turn the triangle into a rectangle by drawing a triangle on the left side with the hypotenuse of it being
and another triangle on the right side, with its hypotenuse being
. After drawing the square, you can cut it into
squares. The area of the rectangle is
square inches because the triangle on the left is half of
and there's another triangle on the other side, equaling them to
square inches. We will be focusing on the left side of the rectangle of
because it includes the shaded region. It's split into
equal squares. The area of this triangle is
square inches because the total area of the rectangle is
square inches and
divided by
is
. There are
sections, so you would do
divided by
in order to find the area of one square. That means that the area of the top right square is
inches and because it's not needed, we will subtract
from
to get rid of it. If you turn around the paper, you notice that the square with half of the shaded region, and the square above it is the shaded region, except flipped and turned. Therefore, the remaining of the area of it which is
should then be divided by
in order to find the shaded region since the shaded region is equal to the other square and half a square.
divided by
is
, so the answer is
.
Solution 5 (Dividing into Congruent Triangles)
We draw the triangles as shown above.
Note that is perpendicular to
;
is perpendicular to
.
is perpendicular to
, so because of corresponding angles theorem,
. We are also told
. Both
and
have a $90\degree$ (Error compiling LaTeX. Unknown error_msg) angle. Because of AAS Congruency (Angle-Angle-Side; $90\degree$ (Error compiling LaTeX. Unknown error_msg),
, and side
),
is congruent to
. Because of symmetry, the same goes for
and
.
Video Solution
https://www.youtube.com/watch?v=zwy5U5IQi88 ~David
See Also
2002 AMC 8 (Problems • Answer Key • Resources) | ||
Preceded by Problem 19 |
Followed by Problem 21 | |
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 AJHSME/AMC 8 Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.