https://artofproblemsolving.com/wiki/api.php?action=feedcontributions&user=Suhasp06&feedformat=atomAoPS Wiki - User contributions [en]2024-03-29T14:14:37ZUser contributionsMediaWiki 1.31.1https://artofproblemsolving.com/wiki/index.php?title=2020_AMC_10A_Problems/Problem_12&diff=1432442020 AMC 10A Problems/Problem 122021-01-25T20:21:21Z<p>Suhasp06: /* Problem */</p>
<hr />
<div>== Problem 12 == <br />
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><br />
<br />
<asy><br />
draw((-4,0)--(4,0)--(0,12)--cycle);<br />
draw((-2,6)--(4,0));<br />
draw((2,6)--(-4,0));<br />
label("M", (-4,0), W);<br />
label("C", (4,0), E);<br />
label("A", (0, 12), N);<br />
label("V", (2, 6), NE);<br />
label("U", (-2, 6), NW);<br />
label("P", (0, 3.6), S);<br />
</asy><br />
<br />
<math>\textbf{(A) } 48 \qquad \textbf{(B) } 72 \qquad \textbf{(C) } 96 \qquad \textbf{(D) } 144 \qquad \textbf{(E) } 192</math><br />
<br />
== Solution 1 ==<br />
Since quadrilateral <math>UVCM</math> has perpendicular diagonals, its area can be found as half of the product of the length of the diagonals. Also note that <math>\triangle AUV</math> has <math>\frac 14</math> the area of triangle <math>AMC</math> by similarity, so <math>[UVCM]=\frac 34\cdot [AMC].</math> Thus,<br />
<cmath>\frac 12 \cdot 12\cdot 12=\frac 34 \cdot [AMC]</cmath><br />
<cmath>72=\frac 34\cdot [AMC]</cmath><br />
<cmath>[AMC]=96\rightarrow \boxed{\textbf{(C)}}.</cmath><br />
<br />
==Solution 2 (Trapezoid)==<br />
<asy><br />
draw((-4,0)--(4,0)--(0,12)--cycle);<br />
draw((-2,6)--(4,0));<br />
draw((2,6)--(-4,0));<br />
draw((-2,6)--(2,6));<br />
label("M", (-4,0), W);<br />
label("C", (4,0), E);<br />
label("A", (0, 12), N);<br />
label("V", (2, 6), NE);<br />
label("U", (-2, 6), NW);<br />
label("P", (0, 3.6), S);<br />
</asy><br />
<br />
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>. <br />
<br />
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>. <br />
<br />
Let's call the intersection of <math>\overline{UC}</math> and <math>\overline{MV}</math> <math>P</math>. Let <math>\overline{UP}=x</math>. Then <math>\overline{PC}=12-x</math>. Since <math>\overline{UC} <br />
\perp \overline{MV}</math>, <math>\overline{UP}</math> and <math>\overline{CP}</math> are heights of triangles <math>\triangle MUV</math> and <math>\triangle MCV</math>, respectively. Both of these triangles have base <math>12</math>. <br />
<br />
Area of <math>\triangle MUV = \frac{x\cdot12}{2}=6x</math><br />
<br />
Area of <math>\triangle MCV = \frac{(12-x)\cdot12}{2}=72-6x</math><br />
<br />
Adding these two gives us the area of trapezoid <math>MUVC</math>, which is <math>6x+(72-6x)=72</math>.<br />
<br />
This is <math>\frac{3}{4}</math> of the triangle, so the area of the triangle is <math>\frac{4}{3}\cdot{72}=\boxed{\textbf{(C) } 96}</math> ~quacker88, diagram by programjames1<br />
<br />
==Solution 3 (Medians)==<br />
Draw median <math>\overline{AB}</math>.<br />
<asy><br />
draw((-4,0)--(4,0)--(0,12)--cycle);<br />
draw((-2,6)--(4,0));<br />
draw((2,6)--(-4,0));<br />
draw((0,12)--(0,0));<br />
label("M", (-4,0), W);<br />
label("C", (4,0), E);<br />
label("A", (0, 12), N);<br />
label("V", (2, 6), NE);<br />
label("U", (-2, 6), NW);<br />
label("P", (0.5, 4), E);<br />
label("B", (0, 0), S);<br />
</asy><br />
<br />
Since we know that all medians of a triangle intersect at the centroid, we know that <math>\overline{AB}</math> passes through point <math>P</math>. We also know that medians of a triangle divide each other into segments of ratio <math>2:1</math>. Knowing this, we can see that <math>\overline{PC}:\overline{UP}=2:1</math>, and since the two segments sum to <math>12</math>, <math>\overline{PC}</math> and <math>\overline{UP}</math> are <math>8</math> and <math>4</math>, respectively.<br />
<br />
Finally knowing that the medians divide the triangle into <math>6</math> sections of equal area, finding the area of <math>\triangle PUM</math> is enough. <math>\overline{PC} = \overline{MP} = 8</math>.<br />
<br />
The area of <math>\triangle PUM = \frac{4\cdot8}{2}=16</math>. Multiplying this by <math>6</math> gives us <math>6\cdot16=\boxed{\textbf{(C) }96}</math> <br />
<br />
~quacker88<br />
<br />
==Solution 4 (Triangles)==<br />
<asy><br />
draw((-4,0)--(4,0)--(0,12)--cycle);<br />
draw((-2,6)--(4,0));<br />
draw((2,6)--(-4,0));<br />
draw((-2,6)--(2,6));<br />
label("M", (-4,0), W);<br />
label("C", (4,0), E);<br />
label("A", (0, 12), N);<br />
label("V", (2, 6), NE);<br />
label("U", (-2, 6), NW);<br />
label("P", (0, 3.6), S);<br />
</asy><br />
We know that <math>AU = UM</math>, <math>AV = VC</math>, so <math>UV = \frac{1}{2} MC</math>.<br />
<br />
As <math>\angle UPM = \angle VPC = 90</math>, we can see that <math>\triangle UPM \cong \triangle VPC</math> and <math>\triangle UVP \sim \triangle MPC</math> with a side ratio of <math>1 : 2</math>.<br />
<br />
So <math>UP = VP = 4</math>, <math>MP = PC = 8</math>.<br />
<br />
With that, we can see that <math>[\triangle UPM] = 16</math>, and the area of trapezoid <math>MUVC</math> is 72.<br />
<br />
As said in solution 1, <math>[\triangle AMC] = 72 / \frac{3}{4} = \boxed{\textbf{(C) } 96}</math>.<br />
<br />
-QuadraticFunctions, solution 1 by ???<br />
<br />
==Solution 5 (Only Pythagorean Theorem)==<br />
<asy><br />
draw((-4,0)--(4,0)--(0,12)--cycle);<br />
draw((-2,6)--(4,0));<br />
draw((2,6)--(-4,0));<br />
draw((0,12)--(0,0));<br />
label("M", (-4,0), W);<br />
label("C", (4,0), E);<br />
label("A", (0, 12), N);<br />
label("V", (2, 6), NE);<br />
label("U", (-2, 6), NW);<br />
label("P", (0.5, 4), E);<br />
label("B", (0, 0), S);<br />
<br />
</asy><br />
<br />
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=UV=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>. <br />
<br />
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><br />
<br />
==Solution 6 (Drawing)==<br />
(NOT recommended)<br />
Transfer the given diagram, which happens to be to scale, onto a piece of a graph paper. Counting the boxes should give a reliable result since the answer choices are relatively far apart.<br />
<br />
<br />
==Solution 7==<br />
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>.<br />
<br />
==Solution 8 (Fastest)==<br />
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>.<br />
<br />
-Rohan S.<br />
<br />
==Solution 9==<br />
<asy><br />
draw((-4,0)--(4,0)--(0,12)--cycle);<br />
draw((-2,6)--(4,0));<br />
draw((2,6)--(-4,0));<br />
draw((0,12)--(0,0));<br />
label("M", (-4,0), W);<br />
label("C", (4,0), E);<br />
label("A", (0, 12), N);<br />
label("V", (2, 6), NE);<br />
label("U", (-2, 6), NW);<br />
label("P", (0.5, 4), E);<br />
label("B", (0, 0), S);<br />
</asy><br />
<br />
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>.<br />
<br />
-SmileKat32<br />
<br />
==Video Solution 1==<br />
<br />
Education, The Study of Everything<br />
<br />
https://youtu.be/0TslJ3aDXac<br />
<br />
==Video Solution 2==<br />
https://youtu.be/ZGwAasE32Y4<br />
<br />
~IceMatrix<br />
<br />
==Video Solution 3==<br />
https://youtu.be/7ZvKOYuwSnE<br />
<br />
~savannahsolver<br />
<br />
== Video Solution 4 ==<br />
https://youtu.be/4_x1sgcQCp4?t=2067<br />
<br />
~ pi_is_3.14<br />
<br />
==See Also==<br />
<br />
{{AMC10 box|year=2020|ab=A|num-b=11|num-a=13}}<br />
{{MAA Notice}}</div>Suhasp06https://artofproblemsolving.com/wiki/index.php?title=2020_AMC_10A_Problems/Problem_7&diff=1432412020 AMC 10A Problems/Problem 72021-01-25T19:56:21Z<p>Suhasp06: /* Solution */</p>
<hr />
<div>{{duplicate|[[2020 AMC 12A Problems|2020 AMC 12A #5]] and [[2020 AMC 10A Problems|2020 AMC 10A #7]]}}<br />
<br />
==Problem==<br />
The <math>25</math> integers from <math>-10</math> to <math>14,</math> inclusive, can be arranged to form a <math>5</math>-by-<math>5</math> square in which the sum of the numbers in each row, the sum of the numbers in each column, and the sum of the numbers along each of the main diagonals are all the same. What is the value of this common sum?<br />
<br />
<math>\textbf{(A) }2 \qquad\textbf{(B) } 5\qquad\textbf{(C) } 10\qquad\textbf{(D) } 25\qquad\textbf{(E) } 50</math><br />
<br />
== Solution == <br />
<br />
Without loss of generality, consider the five rows in the square. Each row must have the same sum of numbers, meaning that the sum of all the numbers in the square divided by <math>5</math> is the total value per row. The sum of the <math>25</math> integers is <math>-10+9+...+14=11+12+13+14=50</math>, and the common sum is <math>\frac{50}{5}=\boxed{\text{(C) }10}</math>.<br />
<br />
==Solution 2==<br />
<br />
Take the sum of the middle 5 values of the set (they will turn out to be the mean of each row). We get <math>0 + 1 + 2 + 3 + 4 = \boxed{\textbf{(C) } 10}</math> as our answer.<br />
~Baolan<br />
<br />
<br />
==Solution 3==<br />
<br />
Taking the average of the first and last terms, <math>-10</math> and <math>14</math>, we have that the mean of the set is <math>2</math>. There are 5 values in each row, column or diagonal, so the value of the common sum is <math>5\cdot2</math>, or <math>\boxed{\textbf{(C) } 10}</math>. <br />
~Arctic_Bunny, edited by KINGLOGIC<br />
<br />
==Solution 4==<br />
<br />
Let us consider the horizontal rows. Since there are five of them, each with constant sum <math>x</math>, we can add up the 25 numbers in 5 rows for a sum of <math>5x</math>. Since the sum of the 25 numbers used is <math>-10-9-8-\cdots{}+12+13+14+15=11+12+13+14+15=50</math>, <math>5x=50</math> and <math>x=\boxed{\textbf{(C) }10}</math>. <br />
~cw357<br />
<br />
==Video Solution 1==<br />
<br />
Education, the Study of Everything<br />
<br />
https://youtu.be/Zf4HCY-y5Z4<br />
<br />
==Video Solution 2==<br />
https://youtu.be/JEjib74EmiY<br />
<br />
~IceMatrix<br />
<br />
==Video Solution 3==<br />
https://youtu.be/PHHBIiIlCY0<br />
<br />
~savannahsolver<br />
<br />
== Video Solution 4==<br />
https://youtu.be/mgEZOXgIZXs?t=1<br />
<br />
~ pi_is_3.14<br />
<br />
==See Also==<br />
<br />
{{AMC10 box|year=2020|ab=A|num-b=6|num-a=8}}<br />
{{AMC12 box|year=2020|ab=A|num-b=4|num-a=6}}<br />
{{MAA Notice}}</div>Suhasp06