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 0== |
+ | By the Law of Cosines on <math>\triangle ABC</math>, we have: <cmath>\cos(A) = \frac{10^2+10^2-12^2}{2*10*10} = \frac{7}{25}</cmath> By the Law of Cosines on <math>\triangle ADE</math>, then <cmath>\frac{7}{25} = \frac{10-x}{2x} \iff x =\frac{250}{39}</cmath> So, our answer is <math>250+39=\boxed{289}</math>. | ||
+ | |||
+ | ==Solution 1 (No Trig)== | ||
<center> | <center> | ||
<asy> | <asy> | ||
+ | import olympiad; | ||
import cse5; | import cse5; | ||
unitsize(10mm); | unitsize(10mm); | ||
Line 10: | Line 14: | ||
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 33: | ||
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> | + | 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 | ~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 (Fastest via Law of Cosines)== | ||
+ | We can have 2 Law of Cosines applied on <math>\angle A</math> (one from <math>\triangle ADE</math> and one from <math>\triangle ABC</math>), | ||
+ | |||
+ | <math>x^2=x^2+(10-x)^2-2(x)(10-x)\cdot \cos{A}</math> and <math>12^2=10^2+10^2-2(10)(10)\cdot \cos{A}</math> | ||
+ | |||
+ | Solving for <math>\cos{A}</math> in both equations, we get | ||
+ | |||
+ | <math>\cos{A} = \frac{(10-x)^2}{(2)(10-x)(x)}</math> and <math>cos A = \frac{7}{25} \implies \frac{(10-x)^2}{(2)(10-x)(x)} = \frac{7}{25} \implies x = \frac{250}{39}</math>, so the answer is <math>\boxed {289}</math> | ||
+ | |||
+ | '''-RootThreeOverTwo''' | ||
+ | |||
+ | ==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 Even Faster Law of Cosines(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 20:07, 12 January 2024
Contents
- 1 Problem 4
- 2 Solution 0
- 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 (Fastest via Law of Cosines)
- 10 Solution 8 (Easiest way- Coordinates without bash)
- 11 Solution 9 Even Faster Law of Cosines(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 0
By the Law of Cosines on , we have:
By the Law of Cosines on
, then
So, our answer is
.
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 (Fastest via Law of Cosines)
We can have 2 Law of Cosines applied on (one from
and one from
),
and
Solving for in both equations, we get
and
, so the answer is
-RootThreeOverTwo
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 Even Faster Law of Cosines(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) | ||
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Followed by Problem 5 | |
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All AIME Problems and Solutions |
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