Difference between revisions of "2020 AMC 10A Problems/Problem 12"
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− | == Problem == | + | == Problem 12 == |
− | + | Triangle <math>AMC</math> is isosceles with <math>AM = AC</math>. Medians <math>\overline{MV}</math> and <math>\overline{CU}</math> are perpendicular to each other, and <math>MV=CU=12</math>. What is the area of <math>\triangle AMC?</math> | |
− | Triangle <math>AMC</math> is | ||
<asy> | <asy> | ||
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<cmath>72=\frac 34\cdot [AMC]</cmath> | <cmath>72=\frac 34\cdot [AMC]</cmath> | ||
<cmath>[AMC]=96\rightarrow \boxed{\textbf{(C)}}.</cmath> | <cmath>[AMC]=96\rightarrow \boxed{\textbf{(C)}}.</cmath> | ||
+ | |||
+ | Note: We know that since AU is the median AU divided by AM is 1/2. Since triangle AUV is similar to triangle AMC, and by a factor of 1/2, then every sidelength of AMC must also be scaled by a factor of 1/2 to get AUV. The base and height are all scaled by 1/2, so then the ratio is 1/2 x 1/2. (can someone convert this into latex) | ||
==Solution 2 (Trapezoid)== | ==Solution 2 (Trapezoid)== | ||
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</asy> | </asy> | ||
− | We know that <math>\triangle AUV \sim \triangle AMC</math>, and since the ratios of its sides are <math>\frac{1}{2}</math>, the ratio of of their areas is <math>(\frac{1}{2})^2=\frac{1}{4}</math>. | + | We know that <math>\triangle AUV \sim \triangle AMC</math>, and since the ratios of its sides are <math>\frac{1}{2}</math>, the ratio of of their areas is <math>\left(\frac{1}{2}\right)^2=\frac{1}{4}</math>. |
If <math>\triangle AUV</math> is <math>\frac{1}{4}</math> the area of <math>\triangle AMC</math>, then trapezoid <math>MUVC</math> is <math>\frac{3}{4}</math> the area of <math>\triangle AMC</math>. | If <math>\triangle AUV</math> is <math>\frac{1}{4}</math> the area of <math>\triangle AMC</math>, then trapezoid <math>MUVC</math> is <math>\frac{3}{4}</math> the area of <math>\triangle AMC</math>. | ||
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</asy> | </asy> | ||
− | Let <math>AB</math> be the height. Since medians divide each other into a <math>2:1</math> ratio, and the medians have length 12, we have <math>PC=MP=8</math> and <math>UP= | + | Let <math>AB</math> be the height. Since medians divide each other into a <math>2:1</math> ratio, and the medians have length 12, we have <math>PC=MP=8</math> and <math>UP=VP=4</math>. From right triangle <math>\triangle{MUP}</math>, <cmath>MU^2=MP^2+UP^2=8^2+4^2=80,</cmath> so <math>MU=\sqrt{80}=4\sqrt{5}</math>. Since <math>CU</math> is a median, <math>AM=8\sqrt{5}</math>. From right triangle <math>\triangle{MPC}</math>, <cmath>MC^2=MP^2+PC^2=8^2+8^2=128,</cmath> which implies <math>MC=\sqrt{128}=8\sqrt{2}</math>. By symmetry <math>MB=\dfrac{8\sqrt{2}}{2}=4\sqrt{2}</math>. |
Applying the Pythagorean Theorem to right triangle <math>\triangle{MAB}</math> gives <math>AB^2=AM^2-MB^2=8\sqrt{5}^2-4\sqrt{2}^2=288</math>, so <math>AB=\sqrt{288}=12\sqrt{2}</math>. Then the area of <math>\triangle{AMC}</math> is <cmath>\dfrac{AB \cdot MC}{2}=\dfrac{8\sqrt{2} \cdot 12\sqrt{2}}{2}=\dfrac{96 \cdot 2}{2}=\boxed{\textbf{(C) }96}</cmath> | Applying the Pythagorean Theorem to right triangle <math>\triangle{MAB}</math> gives <math>AB^2=AM^2-MB^2=8\sqrt{5}^2-4\sqrt{2}^2=288</math>, so <math>AB=\sqrt{288}=12\sqrt{2}</math>. Then the area of <math>\triangle{AMC}</math> is <cmath>\dfrac{AB \cdot MC}{2}=\dfrac{8\sqrt{2} \cdot 12\sqrt{2}}{2}=\dfrac{96 \cdot 2}{2}=\boxed{\textbf{(C) }96}</cmath> | ||
==Solution 6 (Drawing)== | ==Solution 6 (Drawing)== | ||
− | + | By similarity, the area of <math>AUV</math> is equal to <math>\frac{1}{4}</math>. | |
− | + | ||
− | + | The area of <math>UVCM</math> is equal to 72. | |
+ | |||
+ | Assuming the total area of the triangle is S, the equation will be : <math>\frac{3}{4}</math>S = 72. | ||
− | ==Solution 7== | + | S = <math>\boxed{\textbf{(C) }96}</math> |
+ | |||
+ | ==Solution 7(fastest)== | ||
Given a triangle with perpendicular medians with lengths <math>x</math> and <math>y</math>, the area will be <math>\frac{2xy}{3}=\boxed{\textbf{(C) }96}</math>. | Given a triangle with perpendicular medians with lengths <math>x</math> and <math>y</math>, the area will be <math>\frac{2xy}{3}=\boxed{\textbf{(C) }96}</math>. | ||
− | ==Solution 8 | + | ==Solution 8 == |
Connect the line segment <math>UV</math> and it's easy to see quadrilateral <math>UVMC</math> has an area of the product of its diagonals divided by <math>2</math> which is <math>72</math>. Now, solving for triangle <math>AUV</math> could be an option, but the drawing shows the area of <math>AUV</math> will be less than the quadrilateral meaning the the area of <math>AMC</math> is less than <math>72*2</math> but greater than <math>72</math>, leaving only one possible answer choice, <math>\boxed{\textbf{(C) } 96}</math>. | Connect the line segment <math>UV</math> and it's easy to see quadrilateral <math>UVMC</math> has an area of the product of its diagonals divided by <math>2</math> which is <math>72</math>. Now, solving for triangle <math>AUV</math> could be an option, but the drawing shows the area of <math>AUV</math> will be less than the quadrilateral meaning the the area of <math>AMC</math> is less than <math>72*2</math> but greater than <math>72</math>, leaving only one possible answer choice, <math>\boxed{\textbf{(C) } 96}</math>. | ||
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</asy> | </asy> | ||
− | Connect <math>AP</math>, and let <math>B</math> be the point where <math>AP</math> intersects <math>MC</math>. <math>MB=CB</math> because all medians of a triangle intersect at one point, which in this case is <math>P</math>. <math>MP:PV=2:1</math> because the point at which all medians intersect divides the medians into segments of ratio <math>2:1</math>, so <math>MP=8</math> and similarly <math>CP=8</math>. We apply the Pythagorean Theorem to triangle <math>MPC</math> and get <math>MC=\sqrt{128}=8\sqrt{2}</math>. The area of triangle <math>MPC</math> is <math>\dfrac{MP\cdot CP}{2}=32</math>, and that must equal to <math>\dfrac{MC\cdot BP}{2}</math>, so <math>BP=\dfrac{8}{\sqrt{2}}=4\cdot\sqrt{2}</math>. <math>BP=\dfrac{1}{3}BA</math>, so <math>BA=12\sqrt{2}</math>. The area of triangle <math>AMC</math> is equal to <math>\dfrac{MC\cdot BA}{2}=\dfrac{8 \cdot \sqrt{2} \cdot 12 \cdot \sqrt{2}}{2}=\boxed{\textbf{(C)} 96}</math>. | + | Connect <math>AP</math>, and let <math>B</math> be the point where <math>AP</math> intersects <math>MC</math>. <math>MB=CB</math> because all medians of a triangle intersect at one point, which in this case is <math>P</math>. <math>MP:PV=2:1</math> because the point at which all medians intersect divides the medians into segments of ratio <math>2:1</math>, so <math>MP=8</math> and similarly <math>CP=8</math>. We apply the Pythagorean Theorem to triangle <math>MPC</math> and get <math>MC=\sqrt{128}=8\sqrt{2}</math>. The area of triangle <math>MPC</math> is <math>\dfrac{MP\cdot CP}{2}=32</math>, and that must equal to <math>\dfrac{MC\cdot BP}{2}</math>, so <math>BP=\dfrac{8}{\sqrt{2}}=4\cdot\sqrt{2}</math>. <math>BP=\dfrac{1}{3}BA</math>, so <math>BA=12\sqrt{2}</math>. The area of triangle <math>AMC</math> is equal to <math>\dfrac{MC\cdot BA}{2}=\dfrac{8 \cdot \sqrt{2} \cdot 12 \cdot \sqrt{2}}{2}=\boxed{\textbf{(C)}\ 96}</math>. |
-SmileKat32 | -SmileKat32 | ||
− | ==Video Solution== | + | == Solution 10 (High IQ) == |
+ | <math>[\square MUVC] = 72</math>. Let intersection of line <math>AP</math> and base <math>MC</math> be <math>B</math> <cmath> [AUV]=[MUB]=[UVB]=[BVC] \implies \left[\frac{\triangle AMC}{4}\right] = \left[\frac{\square MUVC}{3}\right] \implies [AMC] = \boxed{\textbf{(C)}\ 96}</cmath> | ||
+ | |||
+ | ~herobrine-india | ||
+ | |||
+ | ==Solution 11 (Kite)== | ||
+ | <asy> | ||
+ | draw((-4,0)--(4,0)--(0,12)--cycle); | ||
+ | draw((-2,6)--(4,0)); | ||
+ | draw((2,6)--(-4,0)); | ||
+ | draw((-2,6)--(2,6)); | ||
+ | label("M", (-4,0), W); | ||
+ | label("C", (4,0), E); | ||
+ | label("A", (0, 12), N); | ||
+ | label("V", (2, 6), NE); | ||
+ | label("U", (-2, 6), NW); | ||
+ | label("P", (0, 3.6), S); | ||
+ | </asy> | ||
+ | |||
+ | Since <math>\overline{MV}</math> and <math>\overline{CU}</math> intersect at a right angle, this means <math>MUVC</math> is a kite. Hence, the area of this kite is <math>\frac{12 \cdot 12}{2} = 72</math>. | ||
+ | |||
+ | Also, notice that <math>\triangle AUV \sim \triangle AMC</math> by AAA Similarity. Since the ratios of its sides is <math>\frac{1}{2}</math>, the ratios of the area is <math>\left(\frac{1}{2}\right)^2=\frac{1}{4}</math>. Therefore, | ||
+ | |||
+ | <cmath>[AMC] = [MUVC] + \frac{1}{4} \cdot [AMC]</cmath> | ||
+ | |||
+ | Simplifying gives us <math>\frac{3}{4} \cdot [AMC] = 72</math>, so <math>[AMC] = 72 \cdot \frac{4}{3} = \boxed{\textbf{(C)}\ 96}</math> | ||
+ | |||
+ | ~MrThinker | ||
+ | |||
+ | ==Solution 12 (Educated guessing)== | ||
+ | Horizontally translate line <math>\overline{UC}</math> until point <math>U</math> is at point <math>V</math>, with <math>C</math> subsequently at <math>C'</math>, and then connect up <math>C'</math> and <math>C</math> to create <math>\triangle MVC'</math>, which is a right triangle. | ||
+ | |||
+ | <asy> | ||
+ | draw((-4,0)--(4,0)--(0,12)--cycle); | ||
+ | draw((-2,6)--(4,0)); | ||
+ | draw((2,6)--(-4,0)); | ||
+ | draw((-2,6)--(2,6)); | ||
+ | draw((2,6)--(8,0)); | ||
+ | draw((4,0)--(8,0)); | ||
+ | label("M", (-4,0), W); | ||
+ | label("C", (4,0), S); | ||
+ | label("A", (0, 12), N); | ||
+ | label("V", (2, 6), NE); | ||
+ | label("U", (-2, 6), NW); | ||
+ | label("P", (0, 3.6), S); | ||
+ | label("C'", (8,0), E); | ||
+ | </asy> | ||
+ | |||
+ | Notice that <math>\triangle MVC'</math> = <math>12 \cdot 12 \cdot \frac{1}{2} = 72</math>, and <math>\triangle MVC'</math> = <math>\triangle MVC + \triangle MUV</math> (since the latter has the same base and height as the sub-triangle <math>\triangle CVC'</math> inside <math>\triangle MVC'</math>). | ||
+ | |||
+ | From this, we can deduce that <math>\textbf{(B)}</math> cannot be true, since an incomplete part of <math>\triangle AMC</math> is equal to it. We can also deduce that <math>\textbf{(D)}</math> also cannot be true, since the unknown triangle <math>\triangle AUV = \triangle MUV</math>, and <math>\triangle MUV = \triangle CVC' < \triangle MVC'</math>. Therefore, the answer must be between <math>72</math> and <math>144</math>, leaving <math>\boxed{\textbf{(C)}\ 96}</math> as the only possible correct answer. | ||
+ | |||
+ | ~DifrractSliver | ||
+ | |||
+ | ==Video Solution 1== | ||
Education, The Study of Everything | Education, The Study of Everything | ||
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https://youtu.be/0TslJ3aDXac | https://youtu.be/0TslJ3aDXac | ||
− | + | ==Video Solution 2== | |
https://youtu.be/ZGwAasE32Y4 | https://youtu.be/ZGwAasE32Y4 | ||
~IceMatrix | ~IceMatrix | ||
+ | ==Video Solution 3== | ||
https://youtu.be/7ZvKOYuwSnE | https://youtu.be/7ZvKOYuwSnE | ||
~savannahsolver | ~savannahsolver | ||
+ | |||
+ | == Video Solution 4 by OmegaLearn == | ||
+ | https://youtu.be/4_x1sgcQCp4?t=2067 | ||
+ | |||
+ | ~ pi_is_3.14 | ||
==See Also== | ==See Also== |
Revision as of 08:56, 17 May 2024
Contents
- 1 Problem 12
- 2 Solution 1
- 3 Solution 2 (Trapezoid)
- 4 Solution 3 (Medians)
- 5 Solution 4 (Triangles)
- 6 Solution 5 (Only Pythagorean Theorem)
- 7 Solution 6 (Drawing)
- 8 Solution 7(fastest)
- 9 Solution 8
- 10 Solution 9
- 11 Solution 10 (High IQ)
- 12 Solution 11 (Kite)
- 13 Solution 12 (Educated guessing)
- 14 Video Solution 1
- 15 Video Solution 2
- 16 Video Solution 3
- 17 Video Solution 4 by OmegaLearn
- 18 See Also
Problem 12
Triangle is isosceles with
. Medians
and
are perpendicular to each other, and
. What is the area of
Solution 1
Since quadrilateral has perpendicular diagonals, its area can be found as half of the product of the length of the diagonals. Also note that
has
the area of triangle
by similarity, so
Thus,
Note: We know that since AU is the median AU divided by AM is 1/2. Since triangle AUV is similar to triangle AMC, and by a factor of 1/2, then every sidelength of AMC must also be scaled by a factor of 1/2 to get AUV. The base and height are all scaled by 1/2, so then the ratio is 1/2 x 1/2. (can someone convert this into latex)
Solution 2 (Trapezoid)
We know that , and since the ratios of its sides are
, the ratio of of their areas is
.
If is
the area of
, then trapezoid
is
the area of
.
Let's call the intersection of and
. Let
. Then
. Since
,
and
are heights of triangles
and
, respectively. Both of these triangles have base
.
Area of
Area of
Adding these two gives us the area of trapezoid , which is
.
This is of the triangle, so the area of the triangle is
~quacker88, diagram by programjames1
Solution 3 (Medians)
Draw median .
Since we know that all medians of a triangle intersect at the centroid, we know that passes through point
. We also know that medians of a triangle divide each other into segments of ratio
. Knowing this, we can see that
, and since the two segments sum to
,
and
are
and
, respectively.
Finally knowing that the medians divide the triangle into sections of equal area, finding the area of
is enough.
.
The area of . Multiplying this by
gives us
~quacker88
Solution 4 (Triangles)
We know that
,
, so
.
As , we can see that
and
with a side ratio of
.
So ,
.
With that, we can see that , and the area of trapezoid
is 72.
As said in solution 1, .
-QuadraticFunctions, solution 1 by ???
Solution 5 (Only Pythagorean Theorem)
Let be the height. Since medians divide each other into a
ratio, and the medians have length 12, we have
and
. From right triangle
,
so
. Since
is a median,
. From right triangle
,
which implies
. By symmetry
.
Applying the Pythagorean Theorem to right triangle gives
, so
. Then the area of
is
Solution 6 (Drawing)
By similarity, the area of is equal to
.
The area of is equal to 72.
Assuming the total area of the triangle is S, the equation will be : S = 72.
S =
Solution 7(fastest)
Given a triangle with perpendicular medians with lengths and
, the area will be
.
Solution 8
Connect the line segment and it's easy to see quadrilateral
has an area of the product of its diagonals divided by
which is
. Now, solving for triangle
could be an option, but the drawing shows the area of
will be less than the quadrilateral meaning the the area of
is less than
but greater than
, leaving only one possible answer choice,
.
-Rohan S.
Solution 9
Connect , and let
be the point where
intersects
.
because all medians of a triangle intersect at one point, which in this case is
.
because the point at which all medians intersect divides the medians into segments of ratio
, so
and similarly
. We apply the Pythagorean Theorem to triangle
and get
. The area of triangle
is
, and that must equal to
, so
.
, so
. The area of triangle
is equal to
.
-SmileKat32
Solution 10 (High IQ)
. Let intersection of line
and base
be
~herobrine-india
Solution 11 (Kite)
Since and
intersect at a right angle, this means
is a kite. Hence, the area of this kite is
.
Also, notice that by AAA Similarity. Since the ratios of its sides is
, the ratios of the area is
. Therefore,
Simplifying gives us , so
~MrThinker
Solution 12 (Educated guessing)
Horizontally translate line until point
is at point
, with
subsequently at
, and then connect up
and
to create
, which is a right triangle.
Notice that =
, and
=
(since the latter has the same base and height as the sub-triangle
inside
).
From this, we can deduce that cannot be true, since an incomplete part of
is equal to it. We can also deduce that
also cannot be true, since the unknown triangle
, and
. Therefore, the answer must be between
and
, leaving
as the only possible correct answer.
~DifrractSliver
Video Solution 1
Education, The Study of Everything
Video Solution 2
~IceMatrix
Video Solution 3
~savannahsolver
Video Solution 4 by OmegaLearn
https://youtu.be/4_x1sgcQCp4?t=2067
~ pi_is_3.14
See Also
2020 AMC 10A (Problems • Answer Key • Resources) | ||
Preceded by Problem 11 |
Followed by Problem 13 | |
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 AMC 10 Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.