Difference between revisions of "2018 AIME I Problems/Problem 4"
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==Problem 4== | ==Problem 4== | ||
− | 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> | + | 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== | + | ==Solution (Easiest Law of Cosines)== |
+ | We apply Law of Cosines on <math>\angle A</math> twice (one from <math>\triangle ADE</math> and one from <math>\triangle ABC</math>), | ||
+ | |||
+ | \begin{align*} | ||
+ | 12^2 &= 10^2 + 10^2 - 2(10)(10) \cdot \cos{A} \[5pt] | ||
+ | x^2&=x^2+(10-x)^2-2(x)(10-x)\cdot \cos{A} | ||
+ | \end{align*} | ||
+ | |||
+ | Solving for <math>\cos{A}</math> in both equations, we get | ||
+ | \begin{align*} | ||
+ | \cos{A} &= \frac{7}{25} \ | ||
+ | \cos{A} &= \frac{(10-x)^2}{(2)(10-x)(x)} = \frac{10-x}{2x} | ||
+ | \end{align*} | ||
+ | Setting the two equal, | ||
+ | \begin{align*} | ||
+ | \frac{10-x}{2x} &= \frac{7}{25} \[5pt] | ||
+ | 250-25x &= 14x \[5pt] | ||
+ | x &= \frac{250}{39}. | ||
+ | \end{align*} | ||
+ | Therefore, our answer is <math>\boxed{289}</math> | ||
+ | |||
+ | Note that this strategy of applying LOC on congruent or supplementary angles is very common. It's also how Stewart's Theorem is derived. | ||
+ | |||
+ | '''-RootThreeOverTwo, edits by epiconan''' | ||
+ | |||
+ | ==Solution 1 (No Trig)== | ||
<center> | <center> | ||
<asy> | <asy> | ||
+ | import olympiad; | ||
import cse5; | import cse5; | ||
unitsize(10mm); | unitsize(10mm); | ||
Line 10: | Line 36: | ||
dotfactor=3; | dotfactor=3; | ||
− | pair B = (0,0), A = (6,8), C = (12,0), D = ( | + | pair B = (0,0), A = (6,8), C = (12,0), D = intersectionpoints(circle(A,250/39),A--B)[0], E = intersectionpoints(circle(D,250/39),A--C)[0], F=intersectionpoints(circle(B,9.6),A--C)[0], G=A/2+E/2; |
pair[] dotted = {A,B,C,D,E,F,G}; | pair[] dotted = {A,B,C,D,E,F,G}; | ||
Line 29: | Line 55: | ||
label("$F$",F,NE); | label("$F$",F,NE); | ||
label("$G$",G,NE); | label("$G$",G,NE); | ||
− | label("$x$", | + | label("$x$",A--D,NW); |
− | label("$x$", | + | label("$x$",D--E,NW); |
− | label("$x$", | + | label("$x$",E--C,NE); |
+ | draw(rightanglemark(D,G,E)); | ||
+ | draw(rightanglemark(B,F,E)); | ||
</asy> | </asy> | ||
</center> | </center> | ||
+ | |||
+ | 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> represent <math>\overline{BF}</math> 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{AF}</math> = <math>2.8</math>. We can now see that <math>\triangle AFB</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>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 | ||
+ | |||
+ | Diagram edited by Afly | ||
+ | |||
+ | ==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>. (Note that <math>DF</math> and <math>AG</math> are altitudes of their respective triangles). We also call <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>, <math>AE = 10-x</math>, and <math>EF = \frac{10-x}{2}</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 a 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 = \frac{10-x}{2}</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> | ||
+ | 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} | ||
+ | </cmath> | ||
+ | After some quick cleaning up, we get | ||
+ | <cmath> | ||
+ | 30x = \frac{72}{5} + 100 \implies x = \frac{250}{39} | ||
+ | </cmath> | ||
+ | Therefore, our answer is <math>250+39=\boxed{289}</math>. | ||
+ | |||
+ | ~awesome1st | ||
+ | |||
+ | ==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 | ||
+ | <cmath>\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*}</cmath> | ||
+ | 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 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: | ||
+ | <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 | ||
+ | |||
+ | ==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>. | ||
+ | |||
+ | Now, we recognise that the perpendicular from <math>A</math> onto <math>\overline{AD}</math> gives us two <math>6</math>-<math>8</math>-<math>10</math> triangles. So, we calculate <math>\sin \angle ABC=\frac{4}{5}</math> and <math>\cos \angle ABC=\frac{3}{5}</math> | ||
+ | |||
+ | <math>\angle BAC = 180-2\cdot\angle ABC</math>. And hence, | ||
+ | |||
+ | <math>\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}</math> | ||
+ | |||
+ | Inspecting <math>\triangle ADX</math> gives us <math>\cos \angle BAC = \frac{\frac{10-x}{2}}{x} = \frac{10-x}{2x}</math> | ||
+ | Solving the equation <math>\frac{10-x}{2x}=\frac{7}{25}</math> gives <math>x= \frac{250}{39} \implies\boxed{289}</math> | ||
+ | |||
+ | ~novus677 | ||
+ | |||
+ | ==Solution 7 (Area into Similar Triangles)== | ||
+ | After calling <math>x=AD=DE=EC</math> and <math>10-x=AE=BD</math>, we see we have length ratios in terms of <math>x</math>, which motivates area ratios. We look at the area of triangle <math>ADC</math> in two ways in order to find <math>DG</math> (perpendicular from <math>D</math> to <math>AB</math>), and then use similar triangles to find <math>x</math>. | ||
+ | |||
+ | Using area ratios, <math>[ADC] = \frac{x}{10}\cdot[ABC] = \frac{x}{10} \cdot 48 = \frac{24x}{5}</math>. (To find the total area <math>[ABC] = 48</math>, drop the altitudes from <math>A</math> to <math>BC</math>, and call the foot of the altitude <math>F</math>. By the 6-8-10 triangle, the height <math>AF</math> is <math>8</math> and the area of <math>ABC</math> is <math>48</math>.) | ||
+ | |||
+ | The second way of finding the area of triangle <math>ACD</math> is <math>\frac{1}{2}bh</math>. The base is <math>AC=10</math>, and <math>DG</math> is the height. Therefore, | ||
+ | |||
+ | \begin{align*} | ||
+ | [ACD] = \frac{24x}{5} &= \frac{1}{2} \cdot 10 \cdot DG \[5pt] | ||
+ | \frac{24x}{25} &= DG | ||
+ | \end{align*} | ||
+ | |||
+ | Now we have <math>DG</math> in terms of <math>x</math>, we use the similar triangles <math>GCD</math> and <math>FAC</math> and set up the proportion | ||
+ | \begin{align*} | ||
+ | \frac{DG}{CF} &= \frac{GC}{FA} \[5pt] | ||
+ | \frac{\frac{24x}{25}}{6} &= \frac{5+\frac{x}{2}}{8} \[5pt] | ||
+ | x &= \frac{250}{39}. | ||
+ | \end{align*} | ||
+ | So, our answer is <math>\boxed{289}</math>. | ||
+ | '''-epiconan''' | ||
+ | |||
+ | ==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>. | ||
+ | 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>. | ||
+ | |||
+ | -Stormersyle | ||
+ | |||
+ | == Solution 9 one second accurate solve(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> | ||
+ | Solving for <math>x</math> we get <math>\frac{250}{39}</math> so our answer is <math>289</math>. | ||
+ | |||
+ | -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 <math>\angle A = \theta</math> and <math>\angle ECD = \alpha</math>. Using isosceles triangle properties and the triangle angle sum equation, we get <cmath>180-(180-2\theta+\alpha) + \frac{180-\theta}{2} + \left(\frac{180-\theta}{2} - \alpha\right) = 180.</cmath> Solving, we find <math>\theta = 2 \alpha</math>. | ||
+ | |||
+ | |||
+ | Relabelling our triangle, we get <math>\angle ABC = 90 - \alpha</math>. Dropping an altitude from <math>A</math> to <math>BC</math> and using the Pythagorean theorem, we find <math>[ABC] = 48</math>. Using the sine area formula, we see <math>\frac12 \cdot 10 \cdot 12 \cdot \sin(90-\alpha) = 48</math>. Plugging in our sine angle cofunction identity, <math>\sin(90-\alpha) = \cos(\alpha)</math>, we get <math>\alpha = \cos{^{-1}}{\frac45}</math>. | ||
+ | |||
+ | |||
+ | Now, using the Law of Sines on <math>\triangle ADE</math>, we get <cmath>\frac{\sin{2\alpha}}{\frac{p}{q}} = \frac{\sin{(180-4\alpha)}}{10-\frac{p}{q}}.</cmath> After applying numerous trigonometric and algebraic tricks, identities, and simplifications, such as <math>\sin{(180-4\alpha)}=\sin{4\alpha}</math> and <math>\sin{\left(\cos{^{-1}}{\frac45}\right)} = \frac35</math>, we find <math>\frac{p}{q} = \frac{10\sin{2\alpha}}{\sin{4\alpha}+\sin{2\alpha}} = \frac{250}{39}</math>. | ||
+ | |||
+ | |||
+ | |||
+ | Therefore, our answer is <math>250 + 39 = \boxed{289}</math>. | ||
+ | |||
+ | |||
+ | ~Tiblis | ||
+ | |||
+ | == Solution 11 (Trigonometry)== | ||
+ | We start by labelling a few angles (all of them in degrees). Let <math>\angle{BAC}=2\alpha = \angle{AED}, \angle{EDC}=\angle{ECD}=\alpha, \angle{DEC}=180-2\alpha, \angle{BDC}=3\alpha, \angle{DCB}=90-2\alpha, \angle{DBC}=90-\alpha</math>. Also let <math>AD=a</math>. By sine rule in <math>\triangle{ADE},</math> we get <math>\frac{a}{\sin{2\alpha}}=\frac{10-a}{\sin{4\alpha}} \implies \cos{2\alpha}=\frac{5}{a}-\frac{1}{2}</math> | ||
+ | Using sine rule in <math>\triangle{ABC}</math>, we get <math>\sin{\alpha}=\frac{3}{5}</math>. Hence we get <math>\cos{2\alpha}=1-2\sin^2{\alpha}=1-\frac{18}{25}=\frac{7}{25}</math>. Hence <math>\frac{5}{a}=\frac{1}{2}+\frac{7}{25}=\frac{39}{50} \implies a=\frac{250}{39}</math>. Therefore, our answer is <math>\boxed{289}</math> | ||
+ | |||
+ | Alternatively, use sine rule in <math>\triangle{BDC}</math>. (It’s easier) | ||
+ | |||
+ | ~Prabh1512 | ||
+ | |||
+ | == Solution 12 (Double Angle Identity)== | ||
+ | |||
+ | We let <math>AD=x</math>. Then, angle <math>A</math> is <math>2\sin^{-1}(\frac{3}{5})</math> and so is angle <math>AED</math>. We note that <math>AE=10-x</math>. We drop an altitude from <math>D</math> to <math>AE</math>, and we call the foot <math>F</math>. We note that <math>AF=\frac{10-x}{2}</math>. Using the double angle identity, we have <math>\sin(2\sin^{-1}(\frac{3}{5}))=2(\frac{3}{5})(\frac{4}{5})=\frac{24}{25}.</math> Therefore, <math>DG=\frac{24}{25}AD.</math> We now use the Pythagorean Theorem, which gives <math>(\frac{10-x}{2})^2+(\frac{24}{25}x)^2=x^2</math>. Rearranging and simplifying, this becomes <math>429x^2-12500x+62500=0</math>. Using the quadratic formula, this is <math>\frac{12500\pm\sqrt{12500^2-250000\cdot429}}{858}</math>. We take out a <math>10000</math> from the square root and make it a <math>100</math> outside of the square root to make it simpler. We end up with <math>\frac{12500\pm7000}{858}</math>. We note that this must be less than 10 to ensure that <math>10-x</math> is positive. Therefore, we take the minus, and we get <math>\frac{5500}{858}=\frac{250}{39} \implies \fbox{289}.</math> | ||
+ | |||
+ | ~john0512 | ||
+ | |||
+ | ==Video Solution== | ||
+ | |||
+ | https://www.youtube.com/watch?v=iE8paW_ICxw | ||
+ | |||
+ | |||
+ | https://youtu.be/dI6uZ67Ae2s ~yofro | ||
+ | |||
+ | ==See Also== | ||
+ | {{AIME box|year=2018|n=I|num-b=3|num-a=5}} | ||
+ | {{MAA Notice}} |
Latest revision as of 21:55, 14 December 2024
Contents
[hide]- 1 Problem 4
- 2 Solution (Easiest Law of Cosines)
- 3 Solution 1 (No Trig)
- 4 Solution 2 (Easy Similar Triangles)
- 5 Solution 3 (Algebra w/ Law of Cosines)
- 6 Solution 4 (Coordinates)
- 7 Solution 5 (Law of Cosines)
- 8 Solution 6
- 9 Solution 7 (Area into Similar Triangles)
- 10 Solution 8 (Easiest way- Coordinates without bash)
- 11 Solution 9 one second accurate solve(1 variable equation)
- 12 Solution 10 (Law of Sines)
- 13 Solution 11 (Trigonometry)
- 14 Solution 12 (Double Angle Identity)
- 15 Video Solution
- 16 See Also
Problem 4
In and
. Point
lies strictly between
and
on
and point
lies strictly between
and
on
so that
. Then
can be expressed in the form
, where
and
are relatively prime positive integers. Find
.
Solution (Easiest Law of Cosines)
We apply Law of Cosines on twice (one from
and one from
),
Solving for in both equations, we get
Note that this strategy of applying LOC on congruent or supplementary angles is very common. It's also how Stewart's Theorem is derived.
-RootThreeOverTwo, edits by epiconan
Solution 1 (No Trig)
We draw the altitude from to
to get point
. We notice that the triangle's height from
to
is 8 because it is a
Right Triangle. To find the length of
, we let
represent
and set up an equation by finding two ways to express the area. The equation is
, which leaves us with
. We then solve for the length
, which is done through pythagorean theorm and get
=
. We can now see that
is a
Right Triangle. Thus, we set
as
, and yield that
. Now, we can see
=
. Solving this equation, we yield
, or
. Thus, our final answer is
.
~bluebacon008
Diagram edited by Afly
Solution 2 (Easy Similar Triangles)
We start by adding a few points to the diagram. Call the midpoint of
, and
the midpoint of
. (Note that
and
are altitudes of their respective triangles). We also call
. Since triangle
is isosceles,
, and
. Since
,
and
. Since
is a right triangle,
.
Since and
, triangles
and
are similar by Angle-Angle similarity. Using similar triangle ratios, we have
.
and
because there are
triangles in the problem. Call
. Then
,
, and
. Thus
. Our ratio now becomes
. Solving for
gives us
. Since
is a height of the triangle
,
, or
. Solving the equation
gives us
, so our answer is
.
Solution 3 (Algebra w/ Law of Cosines)
As in the diagram, let . Consider point
on the diagram shown above. Our goal is to be able to perform Pythagorean Theorem on
, and
. Let
. Therefore, it is trivial to see that
(leave everything squared so that it cancels nicely at the end). By Pythagorean Theorem on Triangle
, we know that
. Finally, we apply Law of Cosines on Triangle
. We know that
. Therefore, we get that
. We can now do our final calculation:
After some quick cleaning up, we get
Therefore, our answer is
.
~awesome1st
Solution 4 (Coordinates)
Let ,
, and
. Then, let
be in the interval
and parametrically define
and
as
and
respectively. Note that
, so
. This means that
However, since
is extraneous by definition,
~ mathwiz0803
Solution 5 (Law of Cosines)
As shown in the diagram, let denote
. Let us denote the foot of the altitude of
to
as
. Note that
can be expressed as
and
is a
triangle . Therefore,
and
. Before we can proceed with the Law of Cosines, we must determine
. Using LOC, we can write the following statement:
Thus, the desired answer is
~ blitzkrieg21
Solution 6
In isosceles triangle, draw the altitude from onto
. Let the point of intersection be
. Clearly,
, and hence
.
Now, we recognise that the perpendicular from onto
gives us two
-
-
triangles. So, we calculate
and
. And hence,
Inspecting gives us
Solving the equation
gives
~novus677
Solution 7 (Area into Similar Triangles)
After calling and
, we see we have length ratios in terms of
, which motivates area ratios. We look at the area of triangle
in two ways in order to find
(perpendicular from
to
), and then use similar triangles to find
.
Using area ratios, . (To find the total area
, drop the altitudes from
to
, and call the foot of the altitude
. By the 6-8-10 triangle, the height
is
and the area of
is
.)
The second way of finding the area of triangle is
. The base is
, and
is the height. Therefore,
Now we have in terms of
, we use the similar triangles
and
and set up the proportion
.
-epiconan
Solution 8 (Easiest way- Coordinates without bash)
Let , and
. From there, we know that
, so line
is
. Hence,
for some
, and
so
. Now, notice that by symmetry,
, so
. Because
, we now have
, which simplifies to
, so
, and
.
It follows that
, and our answer is
.
-Stormersyle
Solution 9 one second accurate solve(1 variable equation)
Doing law of cosines we know that is
* Dropping the perpendicular from
to
we get that
Solving for
we get
so our answer is
.
-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 and
. Using isosceles triangle properties and the triangle angle sum equation, we get
Solving, we find
.
Relabelling our triangle, we get . Dropping an altitude from
to
and using the Pythagorean theorem, we find
. Using the sine area formula, we see
. Plugging in our sine angle cofunction identity,
, we get
.
Now, using the Law of Sines on , we get
After applying numerous trigonometric and algebraic tricks, identities, and simplifications, such as
and
, we find
.
Therefore, our answer is .
~Tiblis
Solution 11 (Trigonometry)
We start by labelling a few angles (all of them in degrees). Let . Also let
. By sine rule in
we get
Using sine rule in
, we get
. Hence we get
. Hence
. Therefore, our answer is
Alternatively, use sine rule in . (It’s easier)
~Prabh1512
Solution 12 (Double Angle Identity)
We let . Then, angle
is
and so is angle
. We note that
. We drop an altitude from
to
, and we call the foot
. We note that
. Using the double angle identity, we have
Therefore,
We now use the Pythagorean Theorem, which gives
. Rearranging and simplifying, this becomes
. Using the quadratic formula, this is
. We take out a
from the square root and make it a
outside of the square root to make it simpler. We end up with
. We note that this must be less than 10 to ensure that
is positive. Therefore, we take the minus, and we get
~john0512
Video Solution
https://www.youtube.com/watch?v=iE8paW_ICxw
https://youtu.be/dI6uZ67Ae2s ~yofro
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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