Difference between revisions of "2018 AIME I Problems/Problem 4"
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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>. | 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== | ||
+ | <math>\cos(A) = \frac{5^2+5^2-6^2}{2*5*5} = \frac{7}{25}</math>. Let <math>M</math> be midpoint of <math>AE</math>, then <math>\frac{7}{25} = \frac{10-x}{2x} \iff x =\frac{250}{39}</math>. | ||
==Solution 1 (No Trig)== | ==Solution 1 (No Trig)== | ||
<center> | <center> | ||
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* 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. | * 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. | ||
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+ | == 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>. | ||
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+ | 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>. | ||
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+ | 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>. | ||
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+ | Therefore, our answer is <math>250 + 39 = \boxed{289}</math>. | ||
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+ | ~Tiblis | ||
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+ | == 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> | ||
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+ | ==Video Solution== | ||
+ | |||
+ | https://www.youtube.com/watch?v=iE8paW_ICxw | ||
+ | |||
+ | |||
+ | https://youtu.be/dI6uZ67Ae2s ~yofro | ||
==See Also== | ==See Also== | ||
{{AIME box|year=2018|n=I|num-b=3|num-a=5}} | {{AIME box|year=2018|n=I|num-b=3|num-a=5}} | ||
{{MAA Notice}} | {{MAA Notice}} |
Revision as of 16:22, 3 October 2020
Contents
- 1 Problem 4
- 2 Solution
- 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 Video Solution
- 15 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
. Let be midpoint of , then .
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
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
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.