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Difference between revisions of "2018 AIME I Problems/Problem 4"
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==Problem 4==
 
==Problem 4==
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In <math>\triangle ABC, AB = AC = 10</math> and <math>BC = 12</math>. Point <math>D</math> lies strictly between <math>A</math> and <math>B</math> on <math>\overline{AB}</math> and point <math>E</math> lies strictly between <math>A</math> and <math>C</math> on <math>\overline{AC}</math>) so that <math>AD = DE = EC</math>. Then <math>AD</math> can be expressed in the form <math>\dfrac{p}{q}</math>, where <math>p</math> and <math>q</math> are relatively prime positive integers. Find <math>p+q</math>.
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In <math>\triangle ABC, AB = AC = 10</math> and <math>BC = 12</math>. Point <math>D</math> lies strictly between <math>A</math> and <math>B</math> on <math>\overline{AB}</math> and point <math>E</math> lies strictly between <math>A</math> and <math>C</math> on <math>\overline{AC}</math> so that <math>AD = DE = EC</math>. Then <math>AD</math> can be expressed in the form <math>\dfrac{p}{q}</math>, where <math>p</math> and <math>q</math> are relatively prime positive integers. Find <math>p+q</math>.
  
 
==Solution 1 (No Trig)==
 
==Solution 1 (No Trig)==

Revision as of 18:48, 9 March 2018

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 $\overline{AB}$ = $2.8$. 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 $\overline{BD}$ $= 10-x$, so we have $\overline{AB}$ $=10=$ $10-x+$ $($ $5-$ $\tfrac{x}{2}$ $)$ $($ $\tfrac{25}{7}$ $)$. Solving this equation, we yield $39x=250$, or $x=$ $\tfrac{250}{39}$. Thus, our final answer is $250+39=\boxed{289}$. ~bluebacon008

Solution 2 (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 3 (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 4

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,

$\cos \angle BAC = \cos \angle (180-2\cdot\angle ABC) = -\cos (2\cdot\angle ABC) = \sin^2 \angle ABC - \cos^2 \angle ABC = \frac{16}{25}-\frac{9}{25}=\frac{7}{25}$

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

See Also

2018 AIME I (Problems • Answer Key • Resources)
Preceded by
Problem 3 		Followed by
Problem 5
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15
All AIME Problems and Solutions

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