https://artofproblemsolving.com/wiki/api.php?action=feedcontributions&user=Eugenis&feedformat=atomAoPS Wiki - User contributions [en]2024-03-29T16:00:09ZUser contributionsMediaWiki 1.31.1https://artofproblemsolving.com/wiki/index.php?title=2008_AMC_10A_Problems/Problem_23&diff=777862008 AMC 10A Problems/Problem 232016-03-22T15:33:19Z<p>Eugenis: /* Solution */</p>
<hr />
<div>==Problem==<br />
Two subsets of the set <math>S=\lbrace a,b,c,d,e\rbrace</math> are to be chosen so that their union is <math>S</math> and their intersection contains exactly two elements. In how many ways can this be done, assuming that the order in which the subsets are chosen does not matter?<br />
<br />
<math>\mathrm{(A)}\ 20\qquad\mathrm{(B)}\ 40\qquad\mathrm{(C)}\ 60\qquad\mathrm{(D)}\ 160\qquad\mathrm{(E)}\ 320</math><br />
<br />
==Solution==<br />
<br />
<br />
===Solution 1===<br />
First choose the two letters to be repeated in each set. <math>\dbinom{5}{2}=10</math>. Now we have three remaining elements that we wish to place into two separate subsets. There are <math>2^3 = 8 </math> ways to do so (Do you see why? It's because each of the three remaining letters can be placed either into the first or second subset. Both of those subsets contain the two chosen elements, so their intersection is the two chosen elements). Unfortunately, we have over-counted (Take for example <math>S_{1} = \{a,b,c,d \}</math> and <math>S_{2} = \{a,b,e \}</math>). Notice how <math>S_{1}</math> and <math>S_{2}</math> are interchangeable. A simple division by two will fix this problem. Thus we have:<br />
<br />
<math> \dfrac{10 \times 8}{2} = 40 \implies \boxed{\text{B}} </math><br />
<br />
<br />
Alternatively, after picking the two elements in both sets in <math>\dbinom{5}{2}=10</math> ways, we can use stars and bars to assign the remaining 3 elements to the sets. There are 3 stars, and 1 bar, so there are 4 total ways of assigning the elements. Then there are <math>10\cdot4=40</math> ways to create the sets.<br />
<br />
<br />
===Solution 2===<br />
Another way of looking at this problem is to break it down into cases.<br />
<br />
First, our two subsets can have 2 and 5 elements. The 5-element subset (aka the set) will contain the 2-element subset. There are <math>\dbinom{5}{2}=10</math> ways to choose the 2-element subset. Thus, there are <math>10</math> ways to create these sets.<br />
<br />
<br />
Second, the subsets can have 3 and 4 elements. <math>3+4=7</math> non-distinct elements. <math>7-5=2</math> elements in the intersection. There are <math>\dbinom{5}{3}=10</math> ways to choose the 3-element subset. For the 4-element subset, two of the elements must be the remaining elements (not in the 3-element subset). The other two have to be a subset of the 3-element subset. There are <math>\dbinom{3}{2}=3</math> ways to choose these two elements, which means there are 3 ways to choose the 4-element subset. Therefore, there are <math>10\cdot3=30</math> ways to choose these sets.<br />
<br />
<br />
This leads us to the answer:<br />
<br />
<center><br />
<math> 10+30=40 \implies \boxed{\textbf{(B) 40}} </math><br />
<br />
===Solution 3 (provided by Eugenis)===<br />
We label the subsets subset 1 and subset 2. Suppose the first subset has <math>k</math> elements where <math>k<5.</math> The second element has <math>5-k</math> elements which the first subset does not contain (in order for the union to be the whole set). Additionally, the second set has 2 elements in common with the first subset. Therefore the number of ways to choose these sets is <math>\binom{5}{k} \cdot \binom{k}{2}.</math> Computing for <math>k<5</math> we have <math>10+30+30+10=80.</math> Divide by 2 for order to get <math>40.</math><br />
<br />
==See also==<br />
{{AMC10 box|year=2008|ab=A|num-b=22|num-a=24}}<br />
{{MAA Notice}}</div>Eugenishttps://artofproblemsolving.com/wiki/index.php?title=2015_AMC_8_Problems/Problem_25&diff=731722015 AMC 8 Problems/Problem 252015-11-25T22:56:58Z<p>Eugenis: /* Solution 1 */</p>
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<div>One-inch squares are cut from the corners of this 5 inch square. What is the area in square inches of the largest square that can be fitted into the remaining space?<br />
<br />
<math> \mathrm{(A) \ } 9\qquad \mathrm{(B) \ } 12\frac{1}{2}\qquad \mathrm{(C) \ } 15\qquad \mathrm{(D) \ } 15\frac{1}{2}\qquad \mathrm{(E) \ } 17</math><br />
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<asy><br />
draw((0,0)--(0,5)--(5,5)--(5,0)--cycle);<br />
filldraw((0,4)--(1,4)--(1,5)--(0,5)--cycle, gray);<br />
filldraw((0,0)--(1,0)--(1,1)--(0,1)--cycle, gray);<br />
filldraw((4,0)--(4,1)--(5,1)--(5,0)--cycle, gray);<br />
filldraw((4,4)--(4,5)--(5,5)--(5,4)--cycle, gray);<br />
</asy><br />
<br />
===Solution 1===<br />
We draw a diagram as shown.<br />
<asy><br />
draw((0,0)--(0,5)--(5,5)--(5,0)--cycle);<br />
filldraw((0,4)--(1,4)--(1,5)--(0,5)--cycle, gray);<br />
filldraw((0,0)--(1,0)--(1,1)--(0,1)--cycle, gray);<br />
filldraw((4,0)--(4,1)--(5,1)--(5,0)--cycle, gray);<br />
filldraw((4,4)--(4,5)--(5,5)--(5,4)--cycle, gray);<br />
path arc = arc((2.5,4),1.5,0,90);<br />
pair P = intersectionpoint(arc,(0,5)--(5,5));<br />
pair Pp=rotate(90,(2.5,2.5))*P, Ppp = rotate(90,(2.5,2.5))*Pp, Pppp=rotate(90,(2.5,2.5))*Ppp;<br />
draw(P--Pp--Ppp--Pppp--cycle);<br />
</asy> <br />
Let us focus on the big triangles taking up the rest of the space. The triangles on top of the unit square between the inscribed square, are similiar to the 4 big triangles by <math>AA.</math> Let the height of a big triangle be <math>x</math> then <math>\tfrac{x}{x-1}=\tfrac{5-x}{1}</math>.<br />
<cmath>x=-x^2+6x-5</cmath><br />
<cmath>x^2-5x+5=0</cmath><br />
<cmath>x=\dfrac{5\pm \sqrt{(-5)^2-(4)(1)(5)}}{2}</cmath><br />
<cmath>x=\dfrac{5\pm \sqrt{5}}{2}</cmath><br />
Which means <math>x=\dfrac{5-\sqrt{5}}{2}</math><br />
This means the area of each triangle is <math>\dfrac{5-\sqrt{5}}{2}*(5-\dfrac{5-\sqrt{5}}{2})*\dfrac{1}{2}=\dfrac{5}{2}</math><br />
This the area of the square is <math>25-(4*\dfrac{5}{2})=\boxed{C,~15}</math><br />
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===Solution 2=== <br />
<br />
We draw a diagram as shown:<br />
<asy><br />
pair Q,R,S,T;<br />
Q=(1.381966,0);<br />
R=(5,1.381966);<br />
S=(3.618034,5);<br />
T=(0,3.618034);<br />
draw((0,0)--(0,5)--(5,5)--(5,0)--cycle);<br />
filldraw((0,4)--(1,4)--(1,5)--(0,5)--cycle, gray);<br />
filldraw((0,0)--(1,0)--(1,1)--(0,1)--cycle, gray);<br />
filldraw((4,0)--(4,1)--(5,1)--(5,0)--cycle, gray);<br />
filldraw((4,4)--(4,5)--(5,5)--(5,4)--cycle, gray);<br />
draw(Q--R--S--T--cycle);<br />
draw((1,1)--(4,1)--(4,4)--(1,4)--cycle,dashed);<br />
</asy><br />
<br />
We wish to find the area of the larger triangle. The area of the larger square is composed of the smaller square and the four triangles. The triangles have base <math>3</math> and height <math>1</math>, so the combined area of the four triangles is <math>4 \cdot \frac 32=6</math>. The area of the smaller square is <math>9</math>. We add these to see that the area of the large square is <math>9+6=\boxed{\mathrm{(C) \ } 15}</math>.<br />
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===Solution 3===<br />
<asy><br />
draw((0,0)--(0,5)--(5,5)--(5,0)--cycle);<br />
filldraw((0,4)--(1,4)--(1,5)--(0,5)--cycle, gray);<br />
filldraw((0,0)--(1,0)--(1,1)--(0,1)--cycle, gray);<br />
filldraw((4,0)--(4,1)--(5,1)--(5,0)--cycle, gray);<br />
filldraw((4,4)--(4,5)--(5,5)--(5,4)--cycle, gray);<br />
path arc = arc((2.5,4),1.5,0,90);<br />
pair P = intersectionpoint(arc,(0,5)--(5,5));<br />
pair Pp=rotate(90,(2.5,2.5))*P, Ppp = rotate(90,(2.5,2.5))*Pp, Pppp=rotate(90,(2.5,2.5))*Ppp;<br />
draw(P--Pp--Ppp--Pppp--cycle);<br />
</asy><br />
<br />
Let's find the area of the triangles and the unit squares: on each side, there are 2 triangles. They both have 1 leg of length 1, and let's label the other legs x for one of thr triangles and y for the other. Note that x+y=3.<br />
The area of each of the triangles is <math>\frac{x}{2}</math> and <math>\frac{y}{2}</math>, and there are 4 of each. So now we need to find <math>(4)\frac{x}{2} + (4)\frac{y}{2}</math>.<br />
<math>(4)\frac{x}{2} + (4)\frac{y}{2}</math><br />
<math>\Rightarrow~~4\left(\frac{x}{2}+ \frac{y}{2}\right)</math><br />
<math>\Rightarrow~~4\left(\frac{x+y}{2}\right)</math><br />
Remember that x+y=3, so substituting this in we find that the area of all of the triangles is <math>4\left(\frac{3}{2}\right) = 6</math>. <br />
The area of the 4 unit squares is 4, so the area of the square we need is <math>25- (4+6) = 15 \Rightarrow \boxed{(C)}</math><br />
<br />
==See Also==<br />
<br />
{{AMC8 box|year=2015|num-b=24|after=Last Problem}}<br />
{{MAA Notice}}</div>Eugenishttps://artofproblemsolving.com/wiki/index.php?title=2001_AMC_10_Problems/Problem_24&diff=714362001 AMC 10 Problems/Problem 242015-08-02T16:05:07Z<p>Eugenis: /* Solution */</p>
<hr />
<div>==Problem==<br />
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In trapezoid <math> ABCD </math>, <math> \overline{AB} </math> and <math> \overline{CD} </math> are perpendicular to <math> \overline{AD} </math>, with <math> AB+CD=BC </math>, <math> AB<CD </math>, and <math> AD=7 </math>. What is <math> AB\cdot CD </math>?<br />
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<math> \textbf{(A)}\ 12 \qquad \textbf{(B)}\ 12.25 \qquad \textbf{(C)}\ 12.5 \qquad \textbf{(D)}\ 12.75 \qquad \textbf{(E)}\ 13 </math><br />
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[[Category: Introductory Geometry Problems]]<br />
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==Solution==<br />
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If <math> AB=x </math> and <math> CD=y </math>,then <math> BC=x+y </math>. By the [[Pythagorean theorem]], we have <math> (x+y)^2=(y-x)^2+49 </math> Solving the equation, we get <math> 4xy=49 \implies xy = \boxed{\textbf{(B)}\ 12.25} </math>.<br />
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==See Also==<br />
<br />
{{AMC10 box|year=2001|num-b=23|num-a=25}}<br />
{{MAA Notice}}</div>Eugenishttps://artofproblemsolving.com/wiki/index.php?title=2013_USAJMO_Problems/Problem_5&diff=700292013 USAJMO Problems/Problem 52015-04-26T01:49:55Z<p>Eugenis: /* Solution 2 */</p>
<hr />
<div>==Problem==<br />
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Quadrilateral <math>XABY</math> is inscribed in the semicircle <math>\omega</math> with diameter <math>XY</math>. Segments <math>AY</math> and <math>BX</math> meet at <math>P</math>. Point <math>Z</math> is the foot of the perpendicular from <math>P</math> to line <math>XY</math>. Point <math>C</math> lies on <math>\omega</math> such that line <math>XC</math> is perpendicular to line <math>AZ</math>. Let <math>Q</math> be the intersection of segments <math>AY</math> and <math>XC</math>. Prove that <cmath>\dfrac{BY}{XP}+\dfrac{CY}{XQ}=\dfrac{AY}{AX}.</cmath><br />
<br />
==Solution 1==<br />
Let us use coordinates. Let O, the center of the circle, be (0,0). WLOG the radius of the circle is 1, so set Y (1,0) and X (-1,0). Also, for arbitrary constants <math>a</math> and <math>b</math> set A <math>(\cos a, \sin a)</math> and B <math>(\cos b, \sin b)</math>. Now, let's use our coordinate tools. It is easily derived that the equation of <math>BX</math> is <math>y = \frac{\sin b}{1 + \cos b}(x + 1) = v(x+1)</math> and the equation of <math>AY</math> is <math>y = \frac{\sin a}{1 - \cos a}(x - 1) = u(x-1)</math>, where <math>u</math> and <math>v</math> are defined appropriately. Thus, by equating the y's in the equation we find the intersection of these lines, <math>P</math>, is <math>\left(\frac{u-v}{u+v}\right), \frac{2uv}{u+v})</math>. Also, <math>Z\left(\frac{u-v}{u+v}\right), 0)</math>. It shall be left to the reader to find the slope of <math>AZ</math>, the coordinates of Q and C, and use the distance formula to verify that <math>\frac{BY}{XP} + \frac{CY}{XQ} = \frac{AY}{AX}</math>.<br />
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==Solution 2==<br />
First of all <br />
<br />
<cmath>\angle BXY = \angle PAZ =\angle AXQ =\angle AXC</cmath> since the quadrilateral <math>APZX</math> is cyclic, and triangle <math>AXQ</math> is rectangle, and <math>CX</math> is orthogonal to <math>AZ</math>. Now <br />
<br />
<cmath>\angle BXY =\angle BAY =\angle AXC</cmath> because <math>XABY</math> is cyclic and we have proved that <br />
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<cmath>\angle AXC = \angle BXY</cmath> so <math>BC</math> is parallel to <math>AY</math>, and <cmath>AC=BY, CY=AB</cmath> Now by Ptolomey's theorem on <math>APZX</math> we have <cmath>(AX)(PZ)+(AP)(XZ)=(AZ)(PX)</cmath> we see that triangles <math>PXZ</math> and <math>QXA</math> are similar since <cmath>\angle QAX= \angle PZX= 90</cmath> and <cmath>\angle AXC = \angle BXY</cmath> is already proven, so <cmath>(AX)(PZ)=(AQ)(XZ)</cmath> Substituting yields <cmath>(AQ)(XZ)+(AP)(XZ)=(AZ)(PX)</cmath> dividing by <math>(PX)(XZ)</math> We get <cmath>\frac {AQ+AP}{XP} = \frac {AZ}{XZ}</cmath> Now triangles <math>AYZ</math>, and <math>XYP</math> are similar so <cmath>\frac {AY}{AZ}= \frac {XY}{XP}</cmath> but also triangles <math>XPY</math> and <math>XZB</math> are similar and we get <cmath>\frac {XY}{XP}= \frac {XB}{XZ}</cmath> Comparing we have, <cmath>\frac {AY}{XB}= \frac {AZ}{XZ}</cmath> Substituting, <cmath>\frac {AQ+AP}{XP}= \frac {AY}{XB}</cmath> Dividing the new relation by <math>AX</math> and multiplying by <math>XB</math> we get <cmath>\frac{XB(AQ+AP)}{(XP)(AX)} = \frac {AY}{AX}</cmath> but <cmath>\frac {XB}{AX}= \frac {XY}{XQ}</cmath> since triangles <math>AXB</math> and <math>QXY</math> are similar, because <cmath>\angle AYX= \angle ABX</cmath> and <cmath>\angle AXB= \angle CXY</cmath> since <math>CY=AB</math> Substituting again we get <cmath>\frac {XY(AQ)+XY(AP)}{(XP)(XQ)} =\frac {AY}{AX}</cmath> Now since triangles <math>ACQ</math> and <math>XYQ</math> are similar we have <cmath>XY(AQ)=AC(XQ)</cmath> and by the similarity of <math>APB</math> and <math>XPY</math>, we get <cmath>AB(CP)=XY(AP)</cmath> so substituting, and separating terms we get <cmath>\frac{AC}{XP} + \frac{AB}{XQ} = \frac{AY}{AX}</cmath> In the beginning we prove that <math>AC=BY</math> and <math>AB=CY</math> so <cmath>\frac{BY}{XP} + \frac{CY}{XQ} = \frac{AY}{AX}</cmath> <br />
<math>\blacksquare</math><br />
<br />
<br />
<br />
{{MAA Notice}}</div>Eugenishttps://artofproblemsolving.com/wiki/index.php?title=2013_USAJMO_Problems/Problem_5&diff=700282013 USAJMO Problems/Problem 52015-04-26T01:48:54Z<p>Eugenis: /* Solution 1 */</p>
<hr />
<div>==Problem==<br />
<br />
Quadrilateral <math>XABY</math> is inscribed in the semicircle <math>\omega</math> with diameter <math>XY</math>. Segments <math>AY</math> and <math>BX</math> meet at <math>P</math>. Point <math>Z</math> is the foot of the perpendicular from <math>P</math> to line <math>XY</math>. Point <math>C</math> lies on <math>\omega</math> such that line <math>XC</math> is perpendicular to line <math>AZ</math>. Let <math>Q</math> be the intersection of segments <math>AY</math> and <math>XC</math>. Prove that <cmath>\dfrac{BY}{XP}+\dfrac{CY}{XQ}=\dfrac{AY}{AX}.</cmath><br />
<br />
==Solution 1==<br />
Let us use coordinates. Let O, the center of the circle, be (0,0). WLOG the radius of the circle is 1, so set Y (1,0) and X (-1,0). Also, for arbitrary constants <math>a</math> and <math>b</math> set A <math>(\cos a, \sin a)</math> and B <math>(\cos b, \sin b)</math>. Now, let's use our coordinate tools. It is easily derived that the equation of <math>BX</math> is <math>y = \frac{\sin b}{1 + \cos b}(x + 1) = v(x+1)</math> and the equation of <math>AY</math> is <math>y = \frac{\sin a}{1 - \cos a}(x - 1) = u(x-1)</math>, where <math>u</math> and <math>v</math> are defined appropriately. Thus, by equating the y's in the equation we find the intersection of these lines, <math>P</math>, is <math>\left(\frac{u-v}{u+v}\right), \frac{2uv}{u+v})</math>. Also, <math>Z\left(\frac{u-v}{u+v}\right), 0)</math>. It shall be left to the reader to find the slope of <math>AZ</math>, the coordinates of Q and C, and use the distance formula to verify that <math>\frac{BY}{XP} + \frac{CY}{XQ} = \frac{AY}{AX}</math>.<br />
<br />
==Solution 2==<br />
First <math> \angle BXY = \angle PAZ =\angle AXQ =\angle AXC</math>, since the quadrilateral <math>APZX</math> is cyclic, and triangle <math>AXQ</math> is rectangle, and <math>CX</math> is orthogonal to <math>AZ</math>. Now <math>\angle BXY =\angle BAY =\angle AXC</math> because <math>XABY</math> is cyclic and we have proved that <math>\angle AXC = \angle BXY</math>, so <math>BC</math> is parallel to <math>AY</math>, and <math>AC=BY</math>, <math>CY=AB</math>. Now by Ptolomey's theorem on <math>APZX</math>, we have <math>(AX)(PZ)+(AP)(XZ)=(AZ)(PX)</math>, we see that triangles <math>PXZ</math> and <math>QXA</math> are similar since <math>\angle QAX= \angle PZX= 90</math> and <math>\angle AXC = \angle BXY</math>, already proven, so <math>(AX)(PZ)=(AQ)(XZ)</math>, substituting we get <math>(AQ)(XZ)+(AP)(XZ)=(AZ)(PX)</math>, dividing by <math>(PX)(XZ)</math>, we get <math>\frac {AQ+AP}{XP} = \frac {AZ}{XZ}</math>. Now triangles <math>AYZ</math>, and <math>XYP</math> are similar so <math>\frac {AY}{AZ}= \frac {XY}{XP}</math>, but also triangles <math>XPY</math> and <math>XZB</math> are similar and we get <math>\frac {XY}{XP}= \frac {XB}{XZ}</math>, comparing we have, <math>\frac {AY}{XB}= \frac {AZ}{XZ}</math> substituting, <math>\frac {AQ+AP}{XP}= \frac {AY}{XB}</math>. Dividing the new relation by <math>AX</math> and multiplying by <math>XB</math> we get <math>\frac{XB(AQ+AP)}{(XP)(AX)} = \frac {AY}{AX}</math>, but <math>\frac {XB}{AX}= \frac {XY}{XQ}</math>, since triangles <math>AXB</math> and <math>QXY</math> are similar, because <math>\angle AYX= \angle ABX</math> and <math>\angle AXB= \angle CXY</math> since <math>CY=AB</math>. Substituting again we get <math>\frac {XY(AQ)+XY(AP)}{(XP)(XQ)} =\frac {AY}{AX}</math>. Now since triangles <math>ACQ</math> and <math>XYQ</math> are similar we have <math>XY(AQ)=AC(XQ)</math> and by the similarity of <math>APB</math> and <math>XPY</math>, we get <math>AB(CP)=XY(AP)</math> so substituting, and separating terms we get <math>\frac{AC}{XP} + \frac{AB}{XQ} = \frac{AY}{AX}</math>, but in the beginning we prove that <math>AC=BY</math> and <math>AB=CY</math> so <math>\frac{BY}{XP} + \frac{CY}{XQ} = \frac{AY}{AX}</math>, and we are done.<br />
<br />
<br />
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{{MAA Notice}}</div>Eugenis