Difference between revisions of "2013 AIME II Problems/Problem 13"
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==Problem 13== | ==Problem 13== | ||
In <math>\triangle ABC</math>, <math>AC = BC</math>, and point <math>D</math> is on <math>\overline{BC}</math> so that <math>CD = 3\cdot BD</math>. Let <math>E</math> be the midpoint of <math>\overline{AD}</math>. Given that <math>CE = \sqrt{7}</math> and <math>BE = 3</math>, the area of <math>\triangle ABC</math> can be expressed in the form <math>m\sqrt{n}</math>, where <math>m</math> and <math>n</math> are positive integers and <math>n</math> is not divisible by the square of any prime. Find <math>m+n</math>. | In <math>\triangle ABC</math>, <math>AC = BC</math>, and point <math>D</math> is on <math>\overline{BC}</math> so that <math>CD = 3\cdot BD</math>. Let <math>E</math> be the midpoint of <math>\overline{AD}</math>. Given that <math>CE = \sqrt{7}</math> and <math>BE = 3</math>, the area of <math>\triangle ABC</math> can be expressed in the form <math>m\sqrt{n}</math>, where <math>m</math> and <math>n</math> are positive integers and <math>n</math> is not divisible by the square of any prime. Find <math>m+n</math>. | ||
− | == | + | ==Solutions== |
− | |||
− | + | ===Stewart's Solid Start=== | |
+ | Draw a good diagram. This involves paper and ruler. We can set <math>AE=ED=m</math>. Set <math>BD=k</math>, therefore <math>CD=3k, AC=4k</math>. Thereafter, by Stewart's Theorem on <math>\triangle ACD</math> and cevian <math>CE</math>, we get <math>2m^2+14=25k^2</math>. Also apply Stewart's Theorem on <math>\triangle CEB</math> with cevian <math>DE</math>. After simplification, <math>2m^2=17-6k^2</math>. Therefore, <math>k=1, m=\frac{\sqrt{22}}{2}</math>. Finally, note that (using [] for area) <math>[CED]=[CAE]=3[EDB]=3[AEB]=\frac{3}{8}[ABC]</math>, because of base-ratios. Using Heron's Formula on <math>\triangle EDB</math>, as it is simplest, we see that <math>[ABC]=3\sqrt{7}</math>, so your answer is <math>10</math>. Every step was straightforward and by adopting the simplest steps, we solved the problem quickly. | ||
− | <math> | + | === Solution 1 === |
+ | After drawing the figure, we suppose <math>BD=a</math>, so that <math>CD=3a</math>, <math>AC=4a</math>, and <math>AE=ED=b</math>. | ||
− | <math> | + | Using Law of Cosines for <math>\triangle AEC</math> and <math>\triangle CED</math>,we get |
− | So, <math>(1)+(2)</math>, we get< | + | <cmath>b^2+7-2\sqrt{7}\cdot \cos(\angle CED)=9a^2\qquad (1)</cmath> |
+ | <cmath>b^2+7+2\sqrt{7}\cdot \cos(\angle CED)=16a^2\qquad (2)</cmath> | ||
+ | So, <math>(1)+(2)</math>, we get<cmath>2b^2+14=25a^2. \qquad (3)</cmath> | ||
− | Using | + | Using Law of Cosines in <math>\triangle ACD</math>, we get |
− | < | + | <cmath>b^2+9a^2-2\cdot 2b\cdot 3a\cdot \cos(\angle ADC)=16a^2</cmath> |
− | So, < | + | So, <cmath>\cos(\angle ADC)=\frac{7a^2-4b^2}{12ab}.\qquad (4)</cmath> |
− | Using | + | Using Law of Cosines in <math>\triangle EDC</math> and <math>\triangle EDB</math>, we get |
− | < | + | <cmath>b^2+9a^2-2\cdot 3a\cdot b\cdot \cos(\angle ADC)=7\qquad (5)</cmath> |
− | < | + | <cmath>b^2+a^2+2\cdot a\cdot b\cdot \cos(\angle ADC)=9.\qquad (6)</cmath> |
− | <math>(5)+(6)</math>, and according to <math>(4)</math>, we can get < | + | <math>(5)+(6)</math>, and according to <math>(4)</math>, we can get |
+ | <cmath>37a^2+2b^2=48. \qquad (7)</cmath> | ||
− | Using <math>(3)</math> and <math>(7)</math>, we can solve <math>a=1</math> and <math>b=\frac{\sqrt{22}}{2}</math> | + | Using <math>(3)</math> and <math>(7)</math>, we can solve <math>a=1</math> and <math>b=\frac{\sqrt{22}}{2}</math>. |
− | Finally, we use | + | Finally, we use Law of Cosines for <math>\triangle ADB</math>, |
− | < | + | <cmath>4(\frac{\sqrt{22}}{2})^2+1+2\cdot2(\frac{\sqrt{22}}{2})\cdot \cos(ADC)=AB^2</cmath> |
− | then <math>AB=2\sqrt{7}</math> | + | then <math>AB=2\sqrt{7}</math>, so the height of this <math>\triangle ABC</math> is <math>\sqrt{4^2-(\sqrt{7})^2}=3</math>. |
− | so the height of this <math>\ | + | Then the area of <math>\triangle ABC</math> is <math>3\sqrt{7}</math>, so the answer is <math>\boxed{010}</math>. |
+ | |||
+ | === Solution 2 === | ||
+ | Let <math>X</math> be the foot of the altitude from <math>C</math> with other points labelled as shown below. | ||
+ | <asy> | ||
+ | size(200); | ||
+ | pair A=(0,0),B=(2*sqrt(7),0),C=(sqrt(7),3),D=(3*B+C)/4,L=C/5,M=3*B/7; | ||
+ | draw(A--B--C--cycle);draw(A--D^^B--L^^C--M); | ||
+ | label("$A$",A,SW);label("$B$",B,SE);label("$C$",C,N);label("$D$",D,NE);label("$L$",L,NW);label("$M$",M,S); | ||
+ | pair X=foot(C,A,B), Y=foot(L,A,B); | ||
+ | pair EE=D/2; | ||
+ | label("$X$",X,S);label("$E$",EE,NW);label("$Y$",Y,S); | ||
+ | draw(C--X^^L--Y,dotted); | ||
+ | draw(rightanglemark(B,X,C)^^rightanglemark(B,Y,L)); | ||
+ | </asy> | ||
+ | Now we proceed using [[mass points]]. To balance along the segment <math>BC</math>, we assign <math>B</math> a mass of <math>3</math> and <math>C</math> a mass of <math>1</math>. Therefore, <math>D</math> has a mass of <math>4</math>. As <math>E</math> is the midpoint of <math>AD</math>, we must assign <math>A</math> a mass of <math>4</math> as well. This gives <math>L</math> a mass of <math>5</math> and <math>M</math> a mass of <math>7</math>. | ||
+ | |||
+ | Now let <math>AB=b</math> be the base of the triangle, and let <math>CX=h</math> be the height. Then as <math>AM:MB=3:4</math>, and as <math>AX=\frac{b}{2}</math>, we know that <cmath>MX=\frac{b}{2}-\frac{3b}{7}=\frac{b}{14}.</cmath> Also, as <math>CE:EM=7:1</math>, we know that <math>EM=\frac{1}{\sqrt{7}}</math>. Therefore, by the Pythagorean Theorem on <math>\triangle {XCM}</math>, we know that <cmath>\frac{b^2}{196}+h^2=\left(\sqrt{7}+\frac{1}{\sqrt{7}}\right)^2=\frac{64}{7}.</cmath> | ||
+ | |||
+ | Also, as <math>LE:BE=5:3</math>, we know that <math>BL=\frac{8}{5}\cdot 3=\frac{24}{5}</math>. Furthermore, as <math>\triangle YLA\sim \triangle XCA</math>, and as <math>AL:LC=1:4</math>, we know that <math>LY=\frac{h}{5}</math> and <math>AY=\frac{b}{10}</math>, so <math>YB=\frac{9b}{10}</math>. Therefore, by the Pythagorean Theorem on <math>\triangle BLY</math>, we get <cmath>\frac{81b^2}{100}+\frac{h^2}{25}=\frac{576}{25}.</cmath> | ||
+ | Solving this system of equations yields <math>b=2\sqrt{7}</math> and <math>h=3</math>. Therefore, the area of the triangle is <math>3\sqrt{7}</math>, giving us an answer of <math>\boxed{010}</math>. | ||
+ | |||
+ | === Solution 3 === | ||
+ | Let the coordinates of A, B and C be (-a, 0), (a, 0) and (0, h) respectively. | ||
+ | Then <math>D = (\frac{3a}{4}, \frac{h}{4})</math> and <math>E = (-\frac{a}{8},\frac{h}{8}).</math> | ||
+ | <math>EC^2 = 7</math> implies <math>a^2 + 49h^2 = 448</math>; <math>EB^2 = 9</math> implies <math>81a^2 + h^2 = 576.</math> | ||
+ | Solve this system of equations simultaneously, <math>a=\sqrt{7}</math> and <math>h=3</math>. | ||
+ | Area of the triangle is ah = <math>3\sqrt{7}</math>, giving us an answer of <math>\boxed{010}</math>. | ||
+ | |||
+ | ===Solution 4=== | ||
+ | <asy> | ||
+ | size(200); | ||
+ | pair A=(0,0),B=(2*sqrt(7),0),C=(sqrt(7),3),D=(3*B+C)/4,L=C/5,M=3*B/7; | ||
+ | draw(A--B--C--cycle);draw(A--D^^B--L^^C--M); | ||
+ | label("$A$",A,SW);label("$B$",B,SE);label("$C$",C,N);label("$D$",D,NE);label("$L$",L,NW);label("$M$",M,S); | ||
+ | pair EE=D/2; | ||
+ | label("$\sqrt{7}$", C--EE, W); label("$x$", D--B, E); label("$3x$", C--D, E); label("$l$", EE--D, N); label("$3$", EE--B, N); | ||
+ | label("$E$",EE,NW); | ||
+ | </asy> | ||
+ | (Thanks to writer of Solution 2) | ||
+ | |||
+ | Let <math>BD = x</math>. Then <math>CD = 3x</math> and <math>AC = 4x</math>. Also, let <math>AE = ED = l</math>. Using Stewart's Theorem on <math>\bigtriangleup CEB</math> gives us the equation <math>(x)(3x)(4x) + (4x)(l^2) = 27x + 7x</math> or, after simplifying, <math>4l^2 = 34 - 12x^2</math>. We use Stewart's again on <math>\bigtriangleup CAD</math>: <math>(l)(l)(2l) + 7(2l) = (16x^2)(l) + (9x^2)(l)</math>, which becomes <math>2l^2 = 25x^2 - 14</math>. Substituting <math>2l^2 = 17 - 6x^2</math>, we see that <math>31x^2 = 31</math>, or <math>x = 1</math>. Then <math>l^2 = \frac{11}{2}</math>. | ||
+ | |||
+ | We now use Law of Cosines on <math>\bigtriangleup CAD</math>. <math>(2l)^2 = (4x)^2 + (3x)^2 - 2(4x)(3x)\cos C</math>. Plugging in for <math>x</math> and <math>l</math>, <math>22 = 16 + 9 - 2(4)(3)\cos C</math>, so <math>\cos C = \frac{1}{8}</math>. Using the Pythagorean trig identity <math>\sin^2 + \cos^2 = 1</math>, <math>\sin^2 C = 1 - \frac{1}{64}</math>, so <math>\sin C = \frac{3\sqrt{7}}{8}</math>. | ||
+ | |||
+ | <math>[ABC] = \frac{1}{2} AC \cdot BC \sin C = (\frac{1}{2})(4)(4)(\frac{3\sqrt{7}}{8}) = 3\sqrt{7}</math>, and our answer is <math>3 + 7 = \boxed{010}</math>. | ||
+ | |||
+ | ===Solution 5 (Barycentric Coordinates)=== | ||
+ | Let ABC be the reference triangle, with <math>A=(1,0,0)</math>, <math>B=(0,1,0)</math>, and <math>C=(0,0,1)</math>. We can easily calculate <math>D=(0,\frac{3}{4},\frac{1}{4})</math> and subsequently <math>E=(\frac{1}{2},\frac{3}{8},\frac{1}{8})</math>. Using distance formula on <math>\overline{EC}=(\frac{1}{2},\frac{3}{8},-\frac{7}{8})</math> and <math>\overline{EB}=(\frac{1}{2},-\frac{5}{8},\frac{1}{8})</math> gives | ||
− | Then the | + | <cmath> |
+ | \begin{align*} | ||
+ | \begin{cases} | ||
+ | 7&=|EC|^2=-a^2 \cdot \frac{3}{8} \cdot (-\frac{7}{8})-b^2 \cdot \frac{1}{2} \cdot (-\frac{7}{8})-c^2 \cdot \frac{1}{2} \cdot \frac{3}{8} \\ | ||
+ | 9&=|EB|^2=-a^2 \cdot (-\frac{5}{8}) \cdot \frac{1}{8}-b^2 \cdot \frac{1}{2} \cdot \frac{1}{8}-c^2 \cdot \frac{1}{2} \cdot (-\frac{5}{8}) \\ | ||
+ | \end{cases} | ||
+ | \end{align*} | ||
+ | </cmath> | ||
+ | |||
+ | But we know that <math>a=b</math>, so we can substitute and now we have two equations and two variables. So we can clear the denominators and prepare to cancel a variable: | ||
+ | |||
+ | <cmath> | ||
+ | \begin{align*} | ||
+ | \begin{cases} | ||
+ | 7\cdot 64&=3\cdot 7\cdot a^2+b^2\cdot 4\cdot 7-c^2\cdot 4\cdot 3\\ | ||
+ | 9\cdot 64&=5a^2-4b^2+4\cdot 5\cdot c^2 \\ | ||
+ | \end{cases} | ||
+ | \end{align*} | ||
+ | </cmath> | ||
+ | |||
+ | <cmath> | ||
+ | \begin{align*} | ||
+ | \begin{cases} | ||
+ | 7\cdot 64&=49a^2-12c^2 \\ | ||
+ | 9\cdot 64&=a^2+20c^2 \\ | ||
+ | \end{cases} | ||
+ | \end{align*} | ||
+ | </cmath> | ||
+ | |||
+ | <cmath> | ||
+ | \begin{align*} | ||
+ | \begin{cases} | ||
+ | 5\cdot 7\cdot 64&=245a^2-60c^2 \\ | ||
+ | 3\cdot 9\cdot 64&=3a^2+60c^2 \\ | ||
+ | \end{cases} | ||
+ | \end{align*} | ||
+ | </cmath> | ||
+ | |||
+ | Then we add the equations to get | ||
+ | |||
+ | <cmath> | ||
+ | \begin{align*} | ||
+ | 62\cdot 64&=248a^2 \\ | ||
+ | a^2 &=16 \\ | ||
+ | a &=4 \\ | ||
+ | \end{align*} | ||
+ | </cmath> | ||
+ | |||
+ | Then plugging gives <math>b=4</math> and <math>c=2\sqrt{7}</math>. Then the height from <math>C</math> is <math>3</math>, and the area is <math>3\sqrt{7}</math> and our answer is <math>\boxed{010}</math>. | ||
==See Also== | ==See Also== | ||
{{AIME box|year=2013|n=II|num-b=12|num-a=14}} | {{AIME box|year=2013|n=II|num-b=12|num-a=14}} | ||
{{MAA Notice}} | {{MAA Notice}} |
Revision as of 10:19, 20 December 2020
Contents
Problem 13
In , , and point is on so that . Let be the midpoint of . Given that and , the area of can be expressed in the form , where and are positive integers and is not divisible by the square of any prime. Find .
Solutions
Stewart's Solid Start
Draw a good diagram. This involves paper and ruler. We can set . Set , therefore . Thereafter, by Stewart's Theorem on and cevian , we get . Also apply Stewart's Theorem on with cevian . After simplification, . Therefore, . Finally, note that (using [] for area) , because of base-ratios. Using Heron's Formula on , as it is simplest, we see that , so your answer is . Every step was straightforward and by adopting the simplest steps, we solved the problem quickly.
Solution 1
After drawing the figure, we suppose , so that , , and .
Using Law of Cosines for and ,we get
So, , we get
Using Law of Cosines in , we get
So,
Using Law of Cosines in and , we get
, and according to , we can get
Using and , we can solve and .
Finally, we use Law of Cosines for ,
then , so the height of this is .
Then the area of is , so the answer is .
Solution 2
Let be the foot of the altitude from with other points labelled as shown below. Now we proceed using mass points. To balance along the segment , we assign a mass of and a mass of . Therefore, has a mass of . As is the midpoint of , we must assign a mass of as well. This gives a mass of and a mass of .
Now let be the base of the triangle, and let be the height. Then as , and as , we know that Also, as , we know that . Therefore, by the Pythagorean Theorem on , we know that
Also, as , we know that . Furthermore, as , and as , we know that and , so . Therefore, by the Pythagorean Theorem on , we get Solving this system of equations yields and . Therefore, the area of the triangle is , giving us an answer of .
Solution 3
Let the coordinates of A, B and C be (-a, 0), (a, 0) and (0, h) respectively. Then and implies ; implies Solve this system of equations simultaneously, and . Area of the triangle is ah = , giving us an answer of .
Solution 4
(Thanks to writer of Solution 2)
Let . Then and . Also, let . Using Stewart's Theorem on gives us the equation or, after simplifying, . We use Stewart's again on : , which becomes . Substituting , we see that , or . Then .
We now use Law of Cosines on . . Plugging in for and , , so . Using the Pythagorean trig identity , , so .
, and our answer is .
Solution 5 (Barycentric Coordinates)
Let ABC be the reference triangle, with , , and . We can easily calculate and subsequently . Using distance formula on and gives
But we know that , so we can substitute and now we have two equations and two variables. So we can clear the denominators and prepare to cancel a variable:
Then we add the equations to get
Then plugging gives and . Then the height from is , and the area is and our answer is .
See Also
2013 AIME II (Problems • Answer Key • Resources) | ||
Preceded by Problem 12 |
Followed by Problem 14 | |
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.