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

(→‎Solution 1)
(Solution 2 (Law of Cosines))
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We draw the altitude from <math>B</math> to <math>\overline{AC}</math> to get point <math>F</math>. We notice that the triangle's height from <math>A</math> to <math>\overline{BC}</math> is 8 because it is a <math>3-4-5</math> Right Triangle. To find the length of <math>\overline{BF}</math>, we let <math>h</math> be the height and set up an equation by finding two ways to express the area. The equation is <math>(8)(12)=(10)(h)</math>, which leaves us with <math>h=9.6</math>. We then solve for the length <math>\overline{AF}</math>, which is done through pythagorean theorm and get <math>\overline{AB}</math> = <math>2.8</math>. We can now see that <math>\triangle ABF</math> is a <math>7-24-25</math> Right Triangle. Thus, we set <math>\overline{AG}</math> as <math>5-</math><math>\tfrac{x}{2}</math>, and yield that <math>\overline{AD}</math> <math>=</math> <math>(</math> <math>5-</math> <math>\tfrac{x}{2}</math> <math>)</math> <math>(</math> <math>\tfrac{25}{7}</math> <math>)</math>. Now, we can see <math>\overline{BD}</math> <math>= 10-x</math>, so we have <math>\overline{AB}</math> <math>=10=</math> <math>10-x+</math> <math>(</math> <math>5-</math> <math>\tfrac{x}{2}</math> <math>)</math> <math>(</math> <math>\tfrac{25}{7}</math> <math>)</math>. Solving this equation, we yield <math>39x=250</math>, or <math>x=</math> <math>\tfrac{250}{39}</math>. Thus, our final answer is <math>250+39=\boxed{289}</math>.
 
We draw the altitude from <math>B</math> to <math>\overline{AC}</math> to get point <math>F</math>. We notice that the triangle's height from <math>A</math> to <math>\overline{BC}</math> is 8 because it is a <math>3-4-5</math> Right Triangle. To find the length of <math>\overline{BF}</math>, we let <math>h</math> be the height and set up an equation by finding two ways to express the area. The equation is <math>(8)(12)=(10)(h)</math>, which leaves us with <math>h=9.6</math>. We then solve for the length <math>\overline{AF}</math>, which is done through pythagorean theorm and get <math>\overline{AB}</math> = <math>2.8</math>. We can now see that <math>\triangle ABF</math> is a <math>7-24-25</math> Right Triangle. Thus, we set <math>\overline{AG}</math> as <math>5-</math><math>\tfrac{x}{2}</math>, and yield that <math>\overline{AD}</math> <math>=</math> <math>(</math> <math>5-</math> <math>\tfrac{x}{2}</math> <math>)</math> <math>(</math> <math>\tfrac{25}{7}</math> <math>)</math>. Now, we can see <math>\overline{BD}</math> <math>= 10-x</math>, so we have <math>\overline{AB}</math> <math>=10=</math> <math>10-x+</math> <math>(</math> <math>5-</math> <math>\tfrac{x}{2}</math> <math>)</math> <math>(</math> <math>\tfrac{25}{7}</math> <math>)</math>. Solving this equation, we yield <math>39x=250</math>, or <math>x=</math> <math>\tfrac{250}{39}</math>. Thus, our final answer is <math>250+39=\boxed{289}</math>.
 
~bluebacon008
 
~bluebacon008
 +
 +
==Solution 2 (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:
 +
<cmath>(\overline{DE})^2=(\overline{AD})^2+\overline{AE}^2-2(\overline{AD})(\overline{AE})\cos(\angle{BAC})\implies</cmath>
 +
<cmath>x^2=x^2+(10-x)^2-2(x)(10-x)\left(\frac{7}{25}\right)\implies</cmath>
 +
<cmath>0=(10-x)^2-\frac{14x}{25}(10-x)\implies</cmath>
 +
<cmath>0=10-x-\frac{14x}{25}\implies</cmath>
 +
<cmath>10=\frac{39x}{25}\implies x=\frac{250}{39}</cmath>
 +
Thus, the desired answer is <math>\boxed{289}</math> ~ blitzkrieg21

Revision as of 18:52, 7 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

[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 $h$ be the height 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 (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

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