2018 AIME I Problems/Problem 4

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 $250+39=\boxed{289}$ .

Page

Toolbox

Search

2018 AIME I Problems/Problem 4

Problem 4

Solution 1