# Challenge 92: A Trisecting Median

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- Published on May 8, 2019
- Congratulations to Jaleb, Minh Cong Nguyen, Kevin Lu, Serengeti Ghasa, Arghyadeep Chatterjee, Nicolas Nauli, Midas Schonewille, and Mokou Fujiwara for successfully solving the last week's math challenge question! Jaleb was the first person to solve the question.

Welcome, everyone! LetsSolveMathProblems presents one weekly math challenge problem every Wednesday (continental U.S. time). To participate, please comment your proposed answer and explanation below (or discuss potential solutions with other viewers), keeping in mind that your comment should be left unedited to win the recognition prize. Up to the first ten people with correct solutions are recognized in the next challenge video. The solution to a challenge problem is posted as a separate video the following Wednesday.

For more Weekly Math Challenges:

usclip.net/p/PLpoKXj-PWCbaDXYHES37_zX4O-kCWxguM

LetsSolveMathProblems4 months ago^{+5}** Edit: Due to busy schedule, the solution to the last week's challenge will posted by this Tuesday at the absolute latest. I sincerely apologize for the delay.

* In that solution video, I will assume the knowledge of Euler's Totient Theorem and Chinese Remainder Theorem. If you are not familiar with these theorems, I recommend watching this video before this Sunday to acquire the necessary prerequisites: usclip.net/video/vv57_W5HMTg/video.html. The answer to the last week's challenge was 256.

Miyuki Umeki4 months agoIs okayyy, you did a very good job! Appreciate your effort!

Mokou Fujiwara4 months agoSolving challenging problems is healthy to your brain.

Smokie Bear 🔴🔵4 months ago^{+1}bruh AP Calc BC this year. When you aim for a perfect score but you realize you made *ONE* mistake on *ONE* part of *ONE* free response from looking at reddit after the test.

LetsSolveMathProblems4 months agoIf I recall correctly, I also missed one part of one free response question my year for AP Calculus BC, although I believe I missed one multiple choice question, as well. Getting a perfect score on any AP test requires some luck, even if you are very prepared. Don't stress yourself out too much because of it. 5 is an outstanding score. =)

JZ Animates4 months agoSub to Ruth almgill

Mr. Miscellaneous4 months agoHi!

I know this is irrelevant, but I was watching some of your AP Calculus BC FRQ solutions. I wanted to know if there is a specific way we need to organize or work/answer when solving the problem, or we can just right down our thinking however. Will we lose points for writing messily? Any response is appreciated!

Mr. Miscellaneous4 months agoNo problem, thank you for the info!

LetsSolveMathProblems4 months agoI do realize my response is late, and I offer my apologies. On the AP Calculus exam, as far as I am aware, you will not lose points for writing messily as long as all the required components are present in your solution. Certainly, a clearer organization would help prevent the graders from accidentally deducting points, but you should be more than fine as long as you put down your answers and, if appropriate, justifications. Best of luck on your score! =)

AMIT POKHRIYAL4 months agoAssuming triangle to be right angle it become easy...Let angle CAB be 90°.Then angle MAB is 30 and CAM is 60.After that we can find then every angle of triangle.Let AC=2y and AB=3y then BC=root(5)*y and MC=root(5)*y/2,DC=4*root(5)*y/13.Then we apply cosine law to find AM and AD using which comes out to be root(5)*y/2 and 6*root(5)*y/13 respectively.So AD/AM=12/13

fr dj4 months agoThe answer is 12/13 first let assume that lenght oof ab is 3x and ac is 2x we know that area of two triangle are aqual then we write ab time sin 2x over ac sinx is one the we found cos x is 3/4 by trig and we use cosx^2_1to find cosx the we write law cosine for adm and adc the am is 5/4 x then we use 2ac am Cosa /ac +bm then ad is 60/42 x then we divide

yüce yüce4 months ago^{+1}By using sine area formula, cos(alpha) = 3/4

after, apply cosine theorem for triangles ABM and AMC find some ratios about edges. so AD is angle bisector. rest is easy answer is 12/13

Mokou Fujiwara4 months agowhich angle is alpha?

larpoel Gaming4 months agoHelp me out in this problem The number of pairs of integers (x, y) satisfying the equation

xy(x + y + 1) = 5^2018 + 1 is:

(A) 0 (B) 2 (C) 1009 (D) 2018.

pascal delcombel4 months agoApply AlKashi on triangles ABM, ABG and ABC for angle B, simplify and get 12/13 ratio.

Nicola C4 months agoThe answer is 12/13.

We can label AB=3k and AC=2k, ∠BAM=θ and ∠MAC=2θ.

If we apply law of sines on triangles ABM and AMC we get

3k/sin(BMA)=BM/sin(θ)

and

2k/sin(AMC)=MC/(2 sin(θ)cos(θ)).

Considering that sin(BMA)=sin(AMC) because the angles are supplementary and MC=BM, dividing the first equation and the second one gives 3/2=2cos(θ), which yelds immediately cos(θ)=3/4.

Knowing this we can now use law of cosines on triangles ABM and AMC and set the equations equal (they are both equal to BM²)

We get, after subtracting both equations, 5k-9/2AM+1/2AM=0, therefore AM=5/4 k.

We can use this to get BM=√(79)/4 k.

Now if we use stewart's theorem on ABC with cevian AD we get, after a lot of work, AD=15/13 k.

Thus AD/AM=12/13 and we are done.

Nand Ram4 months agoI solved the question ,correct answer is 12/13 Let the Angle AMB=x and angle MAD=y . Now applying sine theorem in triangle AMB and AMC .After solution for cos theta, tanx .Again we apply sine theorem in triangles AMD and ADC .This gives us coty .Then we can calculate AD/AM =12/13

Mokou Fujiwara4 months agoSet up a coordinate system, so that

C at (1,0)

A at origin

Let angle BAC = θ

Coordinate of B: (3cosθ/2, 3sinθ/2)

Using midpoint formula:

Coordinate of M: (1/2+3cosθ/4, 3sinθ/4)

AM = [sqrt(13+12cosθ)]/4

Using section formula:

Coordinate of D: (9/13+6cosθ/13, 6sinθ/13)

AD=[3sqrt(13+12cosθ)]/13

See that? Both AD and AM have "sqrt(13+12cosθ)"!

AD/AM = (3/13) / (1/4)=12/13

By this method, It shows that even if we completely ignore the fact about trisecting median, we already have fixed value of AD/AM.

Wow, you may try to draw another triangles that do not have trisecting median properties!

Nicolas Nauli4 months agoThe answer is 12/13.

Let θ = angle BAM and x = AM

Using sine law, compare the smaller triangles with the whole triangle, and we can get 4 equations that relate x and θ with some other angles and sides which include the ratio relationship of 1.5 and 1, but in the end, they cancel out YAY and we can get 2 equations with x and θ being alone. Use the double angle formula for sine and the sines cancel out giving us cos θ as a fraction.

The two equations I got are:

2x = (sin3θ)/(sinθ)

4/3 x = (sin3θ)/(sin2θ)

And just by algebra , and a double angle formula for sine and other trig identities like sum of angles of sine and cosine etc., we get cos θ = 3/4 and x = 5/8.

Then we find MC by using cosine rule and get that MC = sqrt(79)/8. We can also get DC = sqrt(79)/8 x 1.6/2.6 = 0.684 , which is by the ratio given in the question.

I still need to find angle ACB so that I can use cosine rule to find AD. I just use sine rule and also the fact that I know cos θ and also BC because it is twice of MC, and find that it is around 33.9 degrees (but I keep the value in my calculator).

I now use the cosine rule and magically get a rational number! AD = 15/26

We now know x and AD, which gives us AD/x to be (15/26)/(5/8) = 12/13.

Bintang Alam Semesta W.A.M4 months ago^{+3}Solution:

let assume that the length of segment of AM is x, while the length of segment of AD is z.

Also assume that the length of segment of AC is q, and the length of segment of AB is 1.5q , while the length of segment of MD is t, and the length of segment of DC is 1.6t then the length of segment of MC is equals to the length of segment of BM which is 2.6k

Beside, the angle of

UbuntuLinux4 months ago^{+8}The 100th one should be a very very hard one

Mokou Fujiwara4 months agoMay not be nearly as hard as generating function one.

Aswini Banerjee4 months agoWithout any restriction i directly assumed

Michel Rodriguez4 months agoThe answer is 12/13. I ɡot the answers by usinɡ vectors. In trianɡle ABC , consider →AC= a̰ (consider →AC as vector a) in that case →AB=3a̰/2,

Let's consider trianɡle BAC,

→BC=→BA+→AC

→BC=−3 a̰/2 + a̰

→BC=− a̰/2

→CB= a̰/2

M is the moddle point of CB, so

→CM=→CB/2

→CM= a̰/4

→CM=→MB= a̰/4

DC/MD = 1.6

DC/MD=8/5

→CD=→CM*8/13

→CD= a̰ /4 *8/13

→CD=2 a̰/13

Let's consider trianɡle ACD,

→AD=→AC+→CD

→AD= a̰ + 2 a̰/13

→AD=15 a̰/13

Let's consider trianɡle ABM,

→AM=→AB+→BM

→AM= 3 a̰/2 + (− a̰/4 )

→AM= 3 a̰/2 − a̰/4

→AM= 5 a̰/4

→AD/→AM= (15 a̰/13)/( 5 a̰/4)

→AD/→AM= 12/13

So , AD/AM =12/13

Michel Rodriguez4 months ago@thefourseasons yeah! i've lost it. AB doesn't a parallel line for AC.so i made a mistake to take as AB =AC*3/2

thefourseasons4 months ago^{+1}Quite a valiant attempt, but I do not think your solution would work since having AB as a scalar product of AC i.e. AB = 3/2*AC would be assuming A,B,C collinear, which is the special/weird case. There is another case as well that would satisfy the conditions i.e. the thumbnail.

Serengeti Ghasa4 months agoThe answer is 12/13

Let AB is 1.5x and AC is x angle BAM is Y and AMB is Z

By sin formula AB/sinZ =BM/sinA and AC/sinZ=MC/sin2A

Working on above two results

We found that cosA=3/4 =>cos2A=1/8

Putting cosine formula for BM and MC and working on it we found AM=5x/8

It is easy to see that AC/AM=DC/DM=1.6 => AD is the angle bisector of CAM

=>AD=2.AM.AC.cosA/(AM+AC)=15x/26

=>AD/AM=(15x/26)/(5x/8)=12/13

With love and regards

Prabhat Ku Sahu

Hiren Bavaskar4 months agoLet's solve maths problem U gave us a spoiler through the title "trisecting median" which tells that AD bisects angle CAM and AM bisects angle DAB.. Let CM=MB=x.

CD=8x/13 DM=5x/13 AC=2 AB=3 (Scaling)

In triangle AMC , AD is angle bisector so DM/DC = AM/AC....(1)

In triangle ADB , AM is angle bisector so DM/MB= AD/ AB ...(2)

Do (2)/(1) and plug ratios

U get required answer as 12/13

KRISHNA KAMALAKANNAN4 months ago^{+3}The answer is 12/13

Assume that AB=3a, AC=2a, DC=8b, MD=5b. These are chosen so that AB/AC=1.5, DC/MD=1.6.

Notice that since MD+DC=MC, MC=8b+5b=13b. Therefore, BC=2*13b=26b.

Lastly, BD=BM+MD=13b+5b=18b.

Now we can use these lengths to plug into Stewart's and Apollonius's Theorems.

Therefore, (3a)^2 + (2a)^2 = 2(AM^2 + (13b)^2) and 8b*26b*18b + AD^2 * 26b = (3a)^2 * 8b + (2a)^2 * 18b.

Cleaning up, we have a^2 =(2/13)(AM^2 + 169b^2) and 3774b^2 + 26AD^2 = 144a^2

We can now substitute a^2 into the second equation to get 3774b^2 + 26AD^2 = (288/13)(AM^2 + 169b^2)

Distributing, we have 3774b^2+26AD^2 = (288/13)AM^2 + 3774b^2.

Subtracting 3774b^2 from both sides, we get 26AD^2 = (288/13)AM^2.

Dividing by 26AM^2 on both sides, we get AD^2/AM^2 = 144/169

Taking the positive sqaure root on both sides, we get AD/AM = 12/13.

I don't know why the angles weren't necessary, so I don't know if my answer is correct, but whatever :)

Bryan C4 months agoFirst, let AB = 3x, AC = 2x, AM = MC = y, MD = 5y/13 and DC =8y/13.

Using the fact that [AMB] = [AMC], we get that cos (theta) = 3/4. Now we use Stewart’s theorem in triangle AMB and the cosine law in triangle ABC (to get that 2y^2 = (79x^2)/8) to reach that AM^2 = (25x^2)/16 and AM = 5x/4. Using Stewart’s theorem once again for triangle AMC and using AM = 5x/4 and 2y^2 =(79x^2)/8, we reach that AD^2 = (225x^2)/169 and therefore AD = 15x/13. Now we simply do AD/AM which equals 12/13.

Notes: [ABC] = (1/2).a.b.sin(theta), theta being the angle between sides a and b.

sin(2.theta) = 2sin(theta)cos(theta)

cos(3.theta) = 4cos^3(theta) - 3cos(theta)

Luis Perez4 months ago^{+1}Let's play around D a bit. so CD/DM=8/5 adding 1 to both sides leads to CM/DM=BM/DM=13/5, from here DM/BM=5/13 adding 1 again leads to (1) BD/CM=18/13. Also again CD/DM=8/5 also means DM/CD=5/8 adding 1 leads to (2) CM/CD=13/8. Multiplaying (1) and (2) leaves BD/CD=18/8=9/4=AB^2/AC^2. This means AD is the simedian.

So

scaryowl0074 months ago^{+3}Applying law of sines, you get that cos theta is 3/4.

Then, applying law of cosines to whole triangle and substituting for cos, you get 144BM^2 = 79AB^2.

Applying law of cosines to the smaller triangles and subtracting the equations, you get 25AB^2=144AM^2.

Applying Stewarts theorem to triangle AMC, you get 20/117 AB^2+8/13AM^2=AD^2 +40/169MC^2.

by rearranging and substituting the cosine results to eliminate variables, you get a ratio of 12/13!!

JHawk244 months ago^{+5}Let AC = 2x, AB = 3x, BM = 13y, MD = 5y, DC = 8y, AM = z, and AD = i

By Stewarts theorem, (2x)^2*13y + (3x)^2*13y = (13y)^2(26y) + z^2(26y)

Simplifying we get x^2 = 26y^2 + (2/13)z^2

Also by Stewarts theorem, 20x^2*y + 8z^2*y = 13m^2y + 520y^2

Simplifying we get 20x^2 + 8z^2 = 13m^2 + 520y^2

Plugging in the first equation for x^2 we get,

520y^2 + 40/13z^2 + 104/13z^2 = 13m^2 + 520y^2

Simplifying we get 144/13z^2 = 13m^2

144/169 = m^2/z^2

m/z = 12/13

So therefore, AD/AM = 12/13

Idk how I didn't need to use the information about the angles though

JHawk244 months ago^{+1}imgur.com/6bQmmc1

JHawk244 months ago^{+1}So I don't edit this comment, when I said AD = i, I really meant to say AD = m. Sorry.

Thicc Kiwi4 months ago^{+1}CD=8y, DM=5y, therefore MB=CM=13y. Also call AC=2x, AB=3x, bam=theta, cam=2theta

From area = 1/2 ab sinc on triangles ACM and AMB, we get cos(theta)=3/4

law of cosine on the same triangles yields

169y^2=9x^2+AM^2-6x*AM*3/4

169y^2=4x^2+AM^2-4x*AM*1/8

Solving, AM is 5x/4, and plug in to find y^2 = 79/(16*169). Thus, angle bisector theorem tells us that AD bisects CAM.

Then use stewarts on ACM and plug in y^2 to solve for AD

8y*5y*13y+13y*AD^2=20x^2*y+25/2*x^2*y

AD=15x/13

Thus, AD/AM=12/13

Thicc Kiwi4 months agoactually for the last step, instead of stewarts you can just use angle bisector theorem on ADB. I'm not sure why I initially recognized AD was a bisector but didnt use it in my solution.

CoolBlue Bryan4 months agoAw man, this one is too complicated for me

Mokou Fujiwara4 months ago@Nicolas Nauli Actually, I found that for this particular problem, using coordinate geometry saves time.

Nicolas Nauli4 months ago@Mokou Fujiwara Well it is pretty tedious (well at least by hand), but yeah it is sort of like a hack with a price of a big amount of time.

Mokou Fujiwara4 months agoUse coordinate system to solve all geometric problems. Your life will be easier.