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Difference between revisions of "2018 AIME I Problems/Problem 4"

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~bluebacon008
 
~bluebacon008
  
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==Solution 2 (Algebra w/ Law of Cosines)==
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==Solution 2 (Easy Similar Triangles)==
 +
We start by adding a few points to the diagram. Call <math>F</math> the midpoint of <math>AE</math>, and <math>G</math> the midpoint of <math>BC</math>. We also have <math>\angle BAC = \theta</math>. Since triangle <math>ADE</math> is isosceles, <math>\angle AED = \theta</math>, and <math>\angle ADF = \angle EDF = 90 - \theta</math>. Since <math>\angle DEA = \theta</math>, <math>\angle DEC = 180 - \theta</math> and <math>\angle EDC = \angle ECD = \frac{\theta}{2}</math>. Since <math>FDC</math> is a right triangle, <math>\angle FDC = 180 - 90 - \frac{\theta}{2} = \frac{180-m}{2}</math>.
 +
 
 +
Since <math>\angle BAG = \frac{\theta}{2}</math> and <math>\angle ABG = \frac{180-m}{2}</math>, triangles <math>ABG</math> and <math>CDF</math> are similar by Angle-Angle similarity. Using similar triangle ratios, we have <math>\frac{AG}{BG} = \frac{CF}{DF}</math>. <math>AG = 8</math> and <math>BG = 6</math> because there are <math>2</math> <math>6-8-10</math> triangles in the problem. Call <math>AD = x</math>. Then <math>CE = x</math> and <math>AE = 10-x</math>. Thus <math>CF = x + \frac{10-x}{2}</math>. Our ratio now becomes <math>\frac{8}{6} = \frac{x+ \frac{10-x}{2}{DF}</math>. Solving for <math>DF</math> gives us <math>DF = \frac{30+3x}{8}</math>. Since <math>DF</math> is the height of the triangle <math>ADE</math>, <math>FE^2 + DF^2 = x^2</math>, or <math>DF = \sqrt{x^2 - (\frac{10-x}{2})^2}</math>. Solving the equation <math>\frac{30+3x}{8} = \sqrt{x^2 - (\frac{10-x}{2})^2}</math> gives us <math>x = \frac{250}{39}</math>, so our answer is <math>250+39 = \boxed{289}</math>.
 +
 
 +
==Solution 3 (Algebra w/ Law of Cosines)==
 
As in the diagram, let <math>DE = x</math>. Consider point <math>G</math> on the diagram shown above. Our goal is to be able to perform Pythagorean Theorem on <math>DG, GC</math>, and <math>DC</math>. Let <math>GE = 10-x</math>. Therefore, it is trivial to see that <math>GC^2 = \Big(x + \frac{10-x}{2}\Big)^2</math> (leave everything squared so that it cancels nicely at the end). By Pythagorean Theorem on Triangle <math>DGE</math>, we know that <math>DG^2 = x^2 - \Big(\frac{10-x}{2}\Big)^2</math>. Finally, we apply Law of Cosines on Triangle <math>DBC</math>. We know that <math>\cos(\angle DBC) = \frac{3}{5}</math>. Therefore, we get that <math>DC^2 = (10-x)^2 + 12^2 - 2(12)(10-x)\frac{3}{5}</math>. We can now do our final calculation:
 
As in the diagram, let <math>DE = x</math>. Consider point <math>G</math> on the diagram shown above. Our goal is to be able to perform Pythagorean Theorem on <math>DG, GC</math>, and <math>DC</math>. Let <math>GE = 10-x</math>. Therefore, it is trivial to see that <math>GC^2 = \Big(x + \frac{10-x}{2}\Big)^2</math> (leave everything squared so that it cancels nicely at the end). By Pythagorean Theorem on Triangle <math>DGE</math>, we know that <math>DG^2 = x^2 - \Big(\frac{10-x}{2}\Big)^2</math>. Finally, we apply Law of Cosines on Triangle <math>DBC</math>. We know that <math>\cos(\angle DBC) = \frac{3}{5}</math>. Therefore, we get that <math>DC^2 = (10-x)^2 + 12^2 - 2(12)(10-x)\frac{3}{5}</math>. We can now do our final calculation:
 
<cmath>
 
<cmath>
Line 52: Line 57:
  
  
−
==Solution 3 (Coordinates)==
+
==Solution 4 (Coordinates)==
 
Let <math>B = (0, 0)</math>, <math>C = (12, 0)</math>, and <math>A = (6, 8)</math>. Then, let <math>x</math> be in the interval <math>0<x<2</math> and parametrically define <math>D</math> and <math>E</math> as <math>(6-3x, 8-4x)</math> and <math>(12-3x, 4x)</math> respectively. Note that <math>AD = 5x</math>, so <math>DE = 5x</math>. This means that
 
Let <math>B = (0, 0)</math>, <math>C = (12, 0)</math>, and <math>A = (6, 8)</math>. Then, let <math>x</math> be in the interval <math>0<x<2</math> and parametrically define <math>D</math> and <math>E</math> as <math>(6-3x, 8-4x)</math> and <math>(12-3x, 4x)</math> respectively. Note that <math>AD = 5x</math>, so <math>DE = 5x</math>. This means that
 
<cmath>\begin{align*}
 
<cmath>\begin{align*}
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However, since <math>2</math> is extraneous by definition, <math>x=\dfrac{50}{39}\implies AD = \dfrac{250}{39}\implies\boxed{289}</math> ~ mathwiz0803
 
However, since <math>2</math> is extraneous by definition, <math>x=\dfrac{50}{39}\implies AD = \dfrac{250}{39}\implies\boxed{289}</math> ~ mathwiz0803
  
−
==Solution 4 (Law of Cosines)==
+
==Solution 5 (Law of Cosines)==
 
As shown in the diagram, let <math>x</math> denote <math>\overline{AD}</math>. Let us denote the foot of the altitude of <math>A</math> to <math>\overline{BC}</math> as <math>F</math>. Note that <math>\overline{AE}</math> can be expressed as <math>10-x</math> and <math>\triangle{ABF}</math> is a <math>6-8-10</math> triangle . Therefore, <math>\sin(\angle{BAF})=\frac{3}{5}</math> and <math>\cos(\angle{BAF})=\frac{4}{5}</math>. Before we can proceed with the Law of Cosines, we must determine <math>\cos(\angle{BAC})=\cos(2\cdot \angle{BAF})=\cos^2(\angle{BAF})-\sin^2(\angle{BAF})=\frac{7}{25}</math>. Using LOC, we can write the following statement:
 
As shown in the diagram, let <math>x</math> denote <math>\overline{AD}</math>. Let us denote the foot of the altitude of <math>A</math> to <math>\overline{BC}</math> as <math>F</math>. Note that <math>\overline{AE}</math> can be expressed as <math>10-x</math> and <math>\triangle{ABF}</math> is a <math>6-8-10</math> triangle . Therefore, <math>\sin(\angle{BAF})=\frac{3}{5}</math> and <math>\cos(\angle{BAF})=\frac{4}{5}</math>. Before we can proceed with the Law of Cosines, we must determine <math>\cos(\angle{BAC})=\cos(2\cdot \angle{BAF})=\cos^2(\angle{BAF})-\sin^2(\angle{BAF})=\frac{7}{25}</math>. Using LOC, we can write the following statement:
 
<cmath>(\overline{DE})^2=(\overline{AD})^2+\overline{AE}^2-2(\overline{AD})(\overline{AE})\cos(\angle{BAC})\implies</cmath>
 
<cmath>(\overline{DE})^2=(\overline{AD})^2+\overline{AE}^2-2(\overline{AD})(\overline{AE})\cos(\angle{BAC})\implies</cmath>
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Thus, the desired answer is <math>\boxed{289}</math> ~ blitzkrieg21
 
Thus, the desired answer is <math>\boxed{289}</math> ~ blitzkrieg21
  
−
==Solution 5==
+
==Solution 6==
 
In isosceles triangle, draw the altitude from <math>D</math> onto <math>\overline{AD}</math>. Let the point of intersection be <math>X</math>. Clearly, <math>AE=10-AD</math>, and hence <math>AX=\frac{10-AD}{2}</math>.
 
In isosceles triangle, draw the altitude from <math>D</math> onto <math>\overline{AD}</math>. Let the point of intersection be <math>X</math>. Clearly, <math>AE=10-AD</math>, and hence <math>AX=\frac{10-AD}{2}</math>.
  
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~novus677
 
~novus677
  
−
==Solution 6 (Fastest via Law of Cosines)==
+
==Solution 7 (Fastest via Law of Cosines)==
 
We can have 2 Law of Cosines applied on <math>\angle A</math> (one from <math>\triangle ADE</math> and one from <math>\triangle ABC</math>),
 
We can have 2 Law of Cosines applied on <math>\angle A</math> (one from <math>\triangle ADE</math> and one from <math>\triangle ABC</math>),
  
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'''-RootThreeOverTwo'''
 
'''-RootThreeOverTwo'''
  
−
==Solution 6 (Easiest way- Coordinates without bash)==
+
==Solution 8 (Easiest way- Coordinates without bash)==
 
Let <math>B=(0, 0)</math>, and <math>C=(12, 0)</math>. From there, we know that <math>A=(6, 8)</math>, so line <math>AB</math> is <math>y=\frac{4}{3}x</math>. Hence, <math>D=(a, \frac{4}{3}a)</math> for some <math>a</math>, and <math>BD=\frac{5}{3}a</math> so <math>AD=10-\frac{5}{3}a</math>. Now, notice that by symmetry, <math>E=(6+a, 8-\frac{4}{3}a)</math>, so <math>ED^2=6^2+(8-\frac{8}{3}a)^2</math>. Because <math>AD=ED</math>, we now have <math>(10-\frac{5}{3})^2=6^2+(8-\frac{8}{3}a)^2</math>, which simplifies to <math>\frac{64}{9}a^2-\frac{128}{3}a+100=\frac{25}{9}a^2-\frac{100}{3}a+100</math>, so <math>\frac{39}{9}a=\frac{13}{3}a=\frac{28}{3}</math>, and <math>a=\frac{28}{13}</math>.
 
Let <math>B=(0, 0)</math>, and <math>C=(12, 0)</math>. From there, we know that <math>A=(6, 8)</math>, so line <math>AB</math> is <math>y=\frac{4}{3}x</math>. Hence, <math>D=(a, \frac{4}{3}a)</math> for some <math>a</math>, and <math>BD=\frac{5}{3}a</math> so <math>AD=10-\frac{5}{3}a</math>. Now, notice that by symmetry, <math>E=(6+a, 8-\frac{4}{3}a)</math>, so <math>ED^2=6^2+(8-\frac{8}{3}a)^2</math>. Because <math>AD=ED</math>, we now have <math>(10-\frac{5}{3})^2=6^2+(8-\frac{8}{3}a)^2</math>, which simplifies to <math>\frac{64}{9}a^2-\frac{128}{3}a+100=\frac{25}{9}a^2-\frac{100}{3}a+100</math>, so <math>\frac{39}{9}a=\frac{13}{3}a=\frac{28}{3}</math>, and <math>a=\frac{28}{13}</math>.
 
It follows that <math>AD=10-\frac{5}{3}\times\frac{28}{13}=10-\frac{140}{39}=\frac{390-140}{39}=\frac{250}{39}</math>, and our answer is <math>250+39=\boxed{289}</math>.
 
It follows that <math>AD=10-\frac{5}{3}\times\frac{28}{13}=10-\frac{140}{39}=\frac{390-140}{39}=\frac{250}{39}</math>, and our answer is <math>250+39=\boxed{289}</math>.
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-Stormersyle
 
-Stormersyle
  
−
== Even Faster Law of Cosines(1 variable equation)==
+
== Solution 9 Even Faster Law of Cosines(1 variable equation)==
  
 
Doing law of cosines we know that <math>\cos A</math> is <math>\frac{7}{25}.</math>* Dropping the perpendicular from <math>D</math> to <math>AE</math> we get that <cmath>\frac{10-x}{2}=\frac{7x}{25}.</cmath>  
 
Doing law of cosines we know that <math>\cos A</math> is <math>\frac{7}{25}.</math>* Dropping the perpendicular from <math>D</math> to <math>AE</math> we get that <cmath>\frac{10-x}{2}=\frac{7x}{25}.</cmath>  

Revision as of 12:31, 3 March 2020

Problem 4

In $\triangle ABC, AB = AC = 10$ and $BC = 12$. Point $D$ lies strictly between $A$ and $B$ on $\overline{AB}$ and point $E$ lies strictly between $A$ and $C$ on $\overline{AC}$ so that $AD = DE = EC$. Then $AD$ can be expressed in the form $\dfrac{p}{q}$, where $p$ and $q$ are relatively prime positive integers. Find $p+q$.

Solution 1 (No Trig)

[asy] import cse5; unitsize(10mm); pathpen=black; dotfactor=3; pair B = (0,0), A = (6,8), C = (12,0), D = (2.154,2.872), E = (8.153, 5.128), F=(7.68,5.76), G=(7.077,6.564), H=(5.6,4.3), I=(4.5,6), J=(10,2.66); pair[] dotted = {A,B,C,D,E,F,G}; D(A--B); D(C--B); D(A--C); D(D--E); pathpen=dashed; D(B--F); D(D--G); dot(dotted); label("$A$",A,N); label("$B$",B,SW); label("$C$",C,SE); label("$D$",D,NW); label("$E$",E,NE); label("$F$",F,NE); label("$G$",G,NE); label("$x$",H,NW); label("$x$",I,NW); label("$x$",J,NE); [/asy]

We draw the altitude from $B$ to $\overline{AC}$ to get point $F$. We notice that the triangle's height from $A$ to $\overline{BC}$ is 8 because it is a $3-4-5$ Right Triangle. To find the length of $\overline{BF}$, we let $h$ represent $\overline{BF}$ and set up an equation by finding two ways to express the area. The equation is $(8)(12)=(10)(h)$, which leaves us with $h=9.6$. We then solve for the length $\overline{AF}$, which is done through pythagorean theorm and get $\overline{AF}$ = $2.8$. We can now see that $\triangle ABF$ is a $7-24-25$ Right Triangle. Thus, we set $\overline{AG}$ as $5-$$\tfrac{x}{2}$, and yield that $\overline{AD}$ $=$ $($ $5-$ $\tfrac{x}{2}$ $)$ $($ $\tfrac{25}{7}$ $)$. Now, we can see $x$ = $($ $5-$ $\tfrac{x}{2}$ $)$ $($ $\tfrac{25}{7}$ $)$. Solving this equation, we yield $39x=250$, or $x=$ $\tfrac{250}{39}$. Thus, our final answer is $250+39=\boxed{289}$. ~bluebacon008

Solution 2 (Easy Similar Triangles)

We start by adding a few points to the diagram. Call $F$ the midpoint of $AE$, and $G$ the midpoint of $BC$. We also have $\angle BAC = \theta$. Since triangle $ADE$ is isosceles, $\angle AED = \theta$, and $\angle ADF = \angle EDF = 90 - \theta$. Since $\angle DEA = \theta$, $\angle DEC = 180 - \theta$ and $\angle EDC = \angle ECD = \frac{\theta}{2}$. Since $FDC$ is a right triangle, $\angle FDC = 180 - 90 - \frac{\theta}{2} = \frac{180-m}{2}$.

Since $\angle BAG = \frac{\theta}{2}$ and $\angle ABG = \frac{180-m}{2}$, triangles $ABG$ and $CDF$ are similar by Angle-Angle similarity. Using similar triangle ratios, we have $\frac{AG}{BG} = \frac{CF}{DF}$. $AG = 8$ and $BG = 6$ because there are $2$ $6-8-10$ triangles in the problem. Call $AD = x$. Then $CE = x$ and $AE = 10-x$. Thus $CF = x + \frac{10-x}{2}$. Our ratio now becomes $\frac{8}{6} = \frac{x+ \frac{10-x}{2}{DF}$ (Error compiling LaTeX. Unknown error_msg). Solving for $DF$ gives us $DF = \frac{30+3x}{8}$. Since $DF$ is the height of the triangle $ADE$, $FE^2 + DF^2 = x^2$, or $DF = \sqrt{x^2 - (\frac{10-x}{2})^2}$. Solving the equation $\frac{30+3x}{8} = \sqrt{x^2 - (\frac{10-x}{2})^2}$ gives us $x = \frac{250}{39}$, so our answer is $250+39 = \boxed{289}$.

Solution 3 (Algebra w/ Law of Cosines)

As in the diagram, let $DE = x$. Consider point $G$ on the diagram shown above. Our goal is to be able to perform Pythagorean Theorem on $DG, GC$, and $DC$. Let $GE = 10-x$. Therefore, it is trivial to see that $GC^2 = \Big(x + \frac{10-x}{2}\Big)^2$ (leave everything squared so that it cancels nicely at the end). By Pythagorean Theorem on Triangle $DGE$, we know that $DG^2 = x^2 - \Big(\frac{10-x}{2}\Big)^2$. Finally, we apply Law of Cosines on Triangle $DBC$. We know that $\cos(\angle DBC) = \frac{3}{5}$. Therefore, we get that $DC^2 = (10-x)^2 + 12^2 - 2(12)(10-x)\frac{3}{5}$. We can now do our final calculation: \[DG^2 + GC^2 = DC^2 \implies x^2 - \Big(\frac{10-x}{2}\Big)^2 + \Big(x + \frac{10-x}{2}\Big)^2 = (10-x)^2 + 12^2 - 2(12)(10-x)\frac{3}{5}\] After some quick cleaning up, we get \[30x = \frac{72}{5} + 100 \implies x = \frac{250}{39}\] Therefore, our answer is $250+39=\boxed{289}$.

~awesome1st


Solution 4 (Coordinates)

Let $B = (0, 0)$, $C = (12, 0)$, and $A = (6, 8)$. Then, let $x$ be in the interval $0<x<2$ and parametrically define $D$ and $E$ as $(6-3x, 8-4x)$ and $(12-3x, 4x)$ respectively. Note that $AD = 5x$, so $DE = 5x$. This means that \begin{align*} \sqrt{36+(8x-8)^2} &= 5x\\ 36+(8x-8)^2 &= 25x^2\\ 64x^2-128x+100 &= 25x^2\\ 39x^2-128x+100 &= 0\\ x &= \dfrac{128\pm\sqrt{16384-15600}}{78}\\ x &= \dfrac{100}{78}, 2\\ \end{align*} However, since $2$ is extraneous by definition, $x=\dfrac{50}{39}\implies AD = \dfrac{250}{39}\implies\boxed{289}$ ~ mathwiz0803

Solution 5 (Law of Cosines)

As shown in the diagram, let $x$ denote $\overline{AD}$. Let us denote the foot of the altitude of $A$ to $\overline{BC}$ as $F$. Note that $\overline{AE}$ can be expressed as $10-x$ and $\triangle{ABF}$ is a $6-8-10$ triangle . Therefore, $\sin(\angle{BAF})=\frac{3}{5}$ and $\cos(\angle{BAF})=\frac{4}{5}$. Before we can proceed with the Law of Cosines, we must determine $\cos(\angle{BAC})=\cos(2\cdot \angle{BAF})=\cos^2(\angle{BAF})-\sin^2(\angle{BAF})=\frac{7}{25}$. Using LOC, we can write the following statement: \[(\overline{DE})^2=(\overline{AD})^2+\overline{AE}^2-2(\overline{AD})(\overline{AE})\cos(\angle{BAC})\implies\] \[x^2=x^2+(10-x)^2-2(x)(10-x)\left(\frac{7}{25}\right)\implies\] \[0=(10-x)^2-\frac{14x}{25}(10-x)\implies\] \[0=10-x-\frac{14x}{25}\implies\] \[10=\frac{39x}{25}\implies x=\frac{250}{39}\] Thus, the desired answer is $\boxed{289}$ ~ blitzkrieg21

Solution 6

In isosceles triangle, draw the altitude from $D$ onto $\overline{AD}$. Let the point of intersection be $X$. Clearly, $AE=10-AD$, and hence $AX=\frac{10-AD}{2}$.

Now, we recognise that the perpendicular from $A$ onto $\overline{AD}$ gives us two $6$-$8$-$10$ triangles. So, we calculate $\sin \angle ABC=\frac{4}{5}$ and $\cos \angle ABC=\frac{3}{5}$

$\angle BAC = 180-2\cdot\angle ABC$. And hence,

$\cos \angle BAC = \cos \angle (180-2\cdot\angle ABC) = -\cos (2\cdot\angle ABC) = \sin^2 \angle ABC - \cos^2 \angle ABC = 2\cos^2 \angle ABC - 1 = \frac{32}{25}-\frac{25}{25}=\frac{7}{25}$

Inspecting $\triangle ADX$ gives us $\cos \angle BAC = \frac{\frac{10-x}{2}}{x} = \frac{10-x}{2x}$ Solving the equation $\frac{10-x}{2x}=\frac{7}{25}$ gives $x= \frac{250}{39} \implies\boxed{289}$

~novus677

Solution 7 (Fastest via Law of Cosines)

We can have 2 Law of Cosines applied on $\angle A$ (one from $\triangle ADE$ and one from $\triangle ABC$),

$x^2=x^2+(10-x)^2-2(x)(10-x)\cdot \cos{A}$ and $12^2=10^2+10^2-2(10)(10)\cdot \cos{A}$

Solving for $\cos{A}$ in both equations, we get

$\cos{A} = \frac{(10-x)^2}{(2)(10-x)(x)}$ and $cos A = \frac{7}{25} \implies \frac{(10-x)^2}{(2)(10-x)(x)} = \frac{7}{25} \implies x = \frac{250}{39}$, so the answer is $\boxed {289}$

-RootThreeOverTwo

Solution 8 (Easiest way- Coordinates without bash)

Let $B=(0, 0)$, and $C=(12, 0)$. From there, we know that $A=(6, 8)$, so line $AB$ is $y=\frac{4}{3}x$. Hence, $D=(a, \frac{4}{3}a)$ for some $a$, and $BD=\frac{5}{3}a$ so $AD=10-\frac{5}{3}a$. Now, notice that by symmetry, $E=(6+a, 8-\frac{4}{3}a)$, so $ED^2=6^2+(8-\frac{8}{3}a)^2$. Because $AD=ED$, we now have $(10-\frac{5}{3})^2=6^2+(8-\frac{8}{3}a)^2$, which simplifies to $\frac{64}{9}a^2-\frac{128}{3}a+100=\frac{25}{9}a^2-\frac{100}{3}a+100$, so $\frac{39}{9}a=\frac{13}{3}a=\frac{28}{3}$, and $a=\frac{28}{13}$. It follows that $AD=10-\frac{5}{3}\times\frac{28}{13}=10-\frac{140}{39}=\frac{390-140}{39}=\frac{250}{39}$, and our answer is $250+39=\boxed{289}$.

-Stormersyle

Solution 9 Even Faster Law of Cosines(1 variable equation)

Doing law of cosines we know that $\cos A$ is $\frac{7}{25}.$* Dropping the perpendicular from $D$ to $AE$ we get that \[\frac{10-x}{2}=\frac{7x}{25}.\] Solving for $x$ we get $\frac{250}{39}$ so our answer is $289$.

-harsha12345

  • It is good to remember that doubling the smallest angle of a 3-4-5 triangle gives the larger (not right) angle in a 7-24-25 triangle.

See Also

2018 AIME I (Problems • Answer Key • Resources)
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Problem 3 		Followed by
Problem 5
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