2018 AIME I Problems/Problem 4

1 Problem 4
2 Solution 1 (No Trig)
3 Solution 2 (Easy Similar Triangles)
4 Solution 3 (Algebra w/ Law of Cosines)
5 Solution 4 (Coordinates)
6 Solution 5 (Law of Cosines)
7 Solution 6
8 Solution 7 (Fastest via Law of Cosines)
9 Solution 8 (Easiest way- Coordinates without bash)
10 Solution 9 Even Faster Law of Cosines(1 variable equation)
11 Solution 10 (Law of Sines)
12 Video Solution
13 See Also

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 $BC$ . (Note that $DF$ and $AG$ are altitudes of their respective triangles). We also call $\angle BAC = \theta$ . 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 $CE = x$ , $AE = 10-x$ , and $EF = \frac{10-x}{2}$ . Thus $CF = x + \frac{10-x}{2}$ . Our ratio now becomes $\frac{8}{6} = \frac{x+ \frac{10-x}{2}}{DF}$ . Solving for $DF$ gives us $DF = \frac{30+3x}{8}$ . Since $DF$ is a 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.

Solution 10 (Law of Sines)

Let's label $\angle A = \theta$ and $\angle ECD = \alpha$ . Using isosceles triangle properties and the triangle angle sum equation, we get $180-(180-2\theta+\alpha) + \frac{180-\theta}{2} + \left(\frac{180-\theta}{2} - \alpha\right) = 180.$ Solving, we find $\theta = 2 \alpha$ .

Relabelling our triangle, we get $\angle ABC = 90 - \alpha$ . Dropping an altitude from $A$ to $BC$ and using the Pythagorean theorem, we find $[ABC] = 48$ . Using the sine area formula, we see $\frac12 \cdot 10 \cdot 12 \cdot \sin(90-\alpha) = 48$ . Plugging in our sine angle cofunction identity, $\sin(90-\alpha) = \cos(\alpha)$ , we get $\alpha = \cos{^{-1}}{\frac45}$ .

Now, using the Law of Sines on $\triangle ADE$ , we get $\frac{\sin{2\alpha}}{\frac{p}{q}} = \frac{\sin{(180-4\alpha)}}{10-\frac{p}{q}}.$ After applying numerous trigonometric and algebraic tricks, identities, and simplifications, such as $\sin{(180-4\alpha)}=\sin{4\alpha}$ and $\sin{\left(\cos{^{-1}}{\frac45}\right)} = \frac35$ , we find $\frac{p}{q} = \frac{10\sin{2\alpha}}{\sin{4\alpha}+\sin{2\alpha}} = \frac{250}{39}$ .

Therefore, our answer is $250 + 39 = \boxed{289}$ .

~Tiblis

Video Solution

https://www.youtube.com/watch?v=iE8paW_ICxw

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