# Challenge 60: Cosine and Cyclic Quadrilateral

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**Published on Sep 27, 2018**- Congratulations to Walker Powell, attyfarbuckle, Kevin Tong, GreenMeansGO, CymoTec, Seth Harwood, Nicola C, Sean Skelly, SeongKyun Jung, and Bigg Barbarian for successfully solving the last week's math challenge question! Walker Powell was the first person to solve the question.

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Welcome, everyone! My channel hosts one weekly math challenge question per week (made by either myself, my family, or my friends), which will be posted every Wednesday. Please comment your proposed answer and explanation below! If you are among the first ten people with the correct answer, you will be recognized in the next math challenge video. The solution to this question and new question will be posted next Wednesday.

Haradhan Datta8 months agoIt's a very fine question.Answer:a+b=16.To solve the problem the following properties are used:1.Area of cyclic quad(1))rilateral=√((s_a)(s_b)(s_c)(s_d)) 2.Area of quadrilateral=2*Area of parallelogram formed by joining the midpoints of the quadrilateral.3.The diagonal of a parallelogram bisect it in two congruent triangles.4.The area of a triangle=(1/2)*ab sinC.5.The length of linesegment joining the midpoints of any two sides of a triangle=(1/2)*length of third side.6.Ptolemy's Theorem.

Let AB=a=k,BC=b=6k,CD=c=3k,DA=d=4k. So semipermier(s)=(a+b+c+d)/2=7k. So , Area (qdl.ABCD)=6√2k^2.(by(1)). Again, Area (qdl. ABCD)=2*(Area parallelogram EFGH, H being the midpoint of AB)=2*2*(Area of triangle EFG)=4*(1/2*EF*FG*sin(theta))=2* (1/2*BD)(1/2*AC)* sin(theta)=1/2*(AC*BD)*sin(theta)=1/2

*(AB.CD+AD.BC)*sin(theta)=1/2*(k. 3k+4k.6k)*sin(theta)=(27/2)k^2*sin(theta). Therefore, (27/2)k^2*sin(theta)=6√2k^2, or sin(theta)= 4√2/9 .So, cos^2(theta)=1-sin^2(theta)=1-(32/81)=49/81. So, cos(theta)=7/9=a/b.

a+b=16.

pekir8 months agoi got 17+33=50

Let AB=a, BC=6a, CD=3a, AD=4a

We know that cos(pi-x)=-cosx and using that and the cosine law on both of triangles ABD and BCD we can calculate that cosDAB=-7/11, cos BCD=7/11 and |DB|=a*sqrt(243/11).

Knowing that we can calculate |EF|, which equals |DB|/2=(a/2)*sqrt(243/11) (by midpoint theorem).

We do the same for triangles ABC and ADC and we get that cosABC=1/3, cosADC=-1/3 and |AC|=a*sqrt(33) => |GH|=(a/2)*sqrt(33).

Now we use law of sine to calculate sinEFC from the triangle EFC: |EF|/sinFCE=3a/sinEFC, where sinFCE=sinBCD=sqrt(1-cos^2BCD)=(6/11)*sqrt(2).

We isolate sinEFC=(3a/|EF|)*sinFCE=(4/11)*sqrt(22/3).

Then we do the same to calculate sinGFD: |GF|/sinGDE=2a/sinGFD, where sinGDF=sinADC=sqrt(1-cos^2ADC)=(2/3)*sqrt(2).

So we arrive at sinGFD=8/3*sqrt(2/33).

Now we use the fact that theta, angle GFD and angle EFC make up an angle of pi radians, so we get sin(pi-theta)=sin(GFD+EFC).

We can rewrite that as sin(theta)=sinGFD*cosEFC+sinEFC*cosGFD.

Now we just plug in the values, sinGFD=(8/3)*sqrt(2/33) => cosGFD=(13/3)*sqrt(1/33), sinEFC=(4/11)*sqrt(22/3) => cosEFC=sqrt(1/33).

and we arrive at sin(theta)=(20/33)*sqrt(2),

so we get that cos(theta)=17/33, so a=17, b=33, a+b=50.

It's my first time actively participating in those challanges and by that i mean typing my answer in comments and if someone catches a mistake please let me know.

Hung Hin Sun8 months ago^{+1}The mistake appears in this implication : sinEFC=(4/11)*sqrt(22/3) => cosEFC=sqrt(1/33) . The angle EFC is an obtuse angle and cosEFC = -sqrt(1/33). If you first find sinEFC and then use the identity (sinA)^2+(cosA)^2=1 to find cosEFC, then it should be careful to determine whether you take the positive or negative square root and also have to justify your decision. In this problem, cosine rule is suitable to be used since it can help us determine whether the angle is acute or obtuse when you find its value.

xamzx8 months agof*** i was so close, i got cos theta=-2/9

Albanovaphi78 months agoAnswer: a + b = 7 + 9 = 16.

To calculate: The cosine of the angle between the diagonals of a quadrilateral from two properties about its area.

Let's define AB = a, BC = b, CD = c and AD = d; theta angle between the diagonals AC and BD: = u; quadrilateral area: = S.

Every quadrilateral has an area S = (b ^ 2 + d ^ 2 -c ^ 2 -a ^ 2) tg (u). Furthermore, we know that, since the quadrilateral is inscriptable, it fulfills the Bramaguptha formula: S = sqrt ((p-a) * (p-b) * (p-c) * (p-d)). By squaring each equation and putting them together we have: tg ^ 2 (u) = [(pa) * (pb) * (pc) * (pd)] / [(b ^ 2 + d ^ 2 -c ^ 2 -a ^ 2) ^ 2], and using the relations between the sides of the quadrilateral it remains that tg ^ 2 (u) = sec ^ 2 (u) -1 = 32/49, sec (u) = 9/7, cos ( u) = 7/9. ▄

Respuesta: a+b=7+9=16.

Por calcular: El coseno del angulo entre las diagonales de un cuadrilátero a partir de dos propiedades acerca de su area.

Definamos AB=a, BC=b, CD=c y AD=d; ángulo teta entre las diagonales AC y BD := u; area del cuadrilatero := S.

Todo cuadrilatero tiene un area S = (b^2 +d^2 -c^2 -a^2)tg(u). Además, sabemos que, como el cuadrilátero es inscriptible, cumple la fórmula de Bramaguptha: S = sqrt((p-a)*(p-b)*(p-c)*(p-d)). Elevando al cuadrado cada ecuación y juntandolas nos queda: tg^2(u) = [(p-a)*(p-b)*(p-c)*(p-d)]/[(b^2 +d^2 -c^2 -a^2)^2], y usando las relaciones entre los lados del cuadrilátero nos queda que tg^2(u) = sec^2(u)-1=32/49, sec(u)=9/7, cos(u)=7/9. ▄

Jonathan Liu8 months agoforumgeom.fau.edu/FG2007volume7/FG200720.pdf easy win

Rajat Khandelwal8 months ago^{+1}I think ans is 16

Let sides of quadrilateral be 1,6,3,4

Now semi perimeter =7

Area of cyclic quadrilateral by herons formula =sqrt((7-1)(7-6)(7-3)(7-4))

=sqrt(72)

Now by potlemys theorem ac*bd=27

Now by similarity our required angle=angle between diagonals

Area of quadrilateral =1/2(ac)*(bd)sin(thita)

Putting valuea of following

We get sin(thita) =4sqrt(2)/9

Hence we get cos(thita) =7/9

Or a+b=16

Joseph Ho8 months ago^{+3}a+b=16

Let AB = x, BC = 6x, CD = 3x and AD = 4x. By Ptolemy theorem, AC × BD = AB×CD + BC×AD = 27x^2. By midpoint theorem, BD is parallel to EF and AC is parallel to FG. The angle between AC and BD is θ.

Semi perimeter of ABCD = 7x

Area of quadrilateral can be expressed into 2 ways:

AREA = sqrt((7x-x)(7x-3x)(7x-4x)(7x-6x))=6x^2×sqrt(2)

AREA = 1/2×AC×BD×sinθ = 27/2 x^2 sinθ.

After simplication, sinθ = 4/9 sqrt(2).

cos^2 θ = 1-16(2)/81 = 49/81. So cos θ = 7/9.

Aswini Banerjee8 months agoLet H is the midpoint of AB so EFGH is a Parallelogram. So EH=GF=x and GH=EF=y. Now as ABCD is cyclic

Minh Cong Nguyen8 months agoThe answer is 16

EF is parallel to BD and FG is parallel to AC, hence angle EFG is equal to the angle between the diagonals of the cyclic quadrilateral. Applying the formula tan(theta/2)=sqrt((s-y)*(s-t)/((s-x)*(s-z))) where x=1, y=6, z=3, t=4, s= 1/2*(x+y+z+t) = 7 we have tan(theta/2)= sqrt(1/8).

Then cos(theta) = (1- tan^2(theta/2)) / (1+tan^2(theta/2)) = (1-1/8)/ (1+1/8) = 7/9 = a/b

Therefore a+b = 16

CymoTec8 months agoThe answer does not exist. There is no cyclic quadrilateral which has the given ratio parameters.

Let M be the circumcenter of the cyclic quadrilateral. Connecting M to the points A, B, C, and D yields 4 isosceles triangles with the equal sides being the radius. To compute the length of the remaining side, there is the equation s = 2*R*sin(Gamma/2) with s being the side, R being the radius and Gamma being the angle at M. We can also rearrage it to get Gamma = 2*arcsin( s/(2R) ).

The ratios between the sides are given. We can multiply them with the length AB. AB on the other hand can be written in terms of its angle at M, which I'll call a. Plugging that into our second equation gives us the other angles relative to the angle a (b, c, and d are the angles at M for the triangles BCM, CDM, and ADM respectively):

a = a

b = 2*arcsin(6*sin(a/2))

c = 2*arcsin(3*sin(a/2))

d = 2*arcsin(4*sin(a/2))

Using the equation for b, we can constrain a: 2*arcsin(-1/6)

Juan Alberto Vargas Mesén8 months agoCymoTec By the way, there's a slight problem with my argument, which I realized only afterwards.

It still works, but sometimes you need to decrease rather than increase the circumference (precisely depending on whether the center is inside or outside the triangle). After thinking about this for a while, I can pretty confidently say that any arrangement of side lengths for an n-gon can be made cyclic, for as long as they satisfy triangle inequality

CymoTec8 months ago+Juan Alberto Vargas Mesén

Ahh, I see. That's pretty intuitive, yeah. Thanks for the explanation.

+attyfarbuckle

Of course! The arcsine is only defined until pi/2... That was the case I missed. Obviously, the quadrilateral doesn't have to contain the center.

Finally, plugging in your equation b = 2*(pi - arcsin(6*sin(a/2)) ) does yield a valid solution.

Thank you for your help. It did help greatly.

Juan Alberto Vargas Mesén8 months agoAn intuitive explanation as to why such quadrilateral must exist: start with 6, 3 & 4, which obviously form a triangle. There's a circumference that passes through those 3 points. In this case A & D coincide, so AD=0.

Leaving the lengths 6, 3 & 4 fixed, start making the circumference (continuously) bigger while at the same time maintaining points ABCD over said circumference, only stretching the "gap" between A and D. At some point you must reach a segment of length exactly 1 (since the limit of AD as the radius approaches infinity is 6+3+4=13).

attyfarbuckle8 months agoYour largest angle is a reflex angle.

So although sin(b/2) = 6*sin(a/2) is correct, b/2 > pi/2.

Which means

b/2 = pi - arcsin(6*sin(a/2))

Basically the centre of the circle is outside the quadrilateral.

Hope this helps.

Kevin Tong8 months agoAnswer is 7+9=16

WLOG, let AB=1, BC=6, CD=3, and DA=4. By Ptolemy's Theorem, we get

24+3=AC*BD=27

Furthermore, since triange CEF is similar to triangle BCD, and CE=CB/2, we have that EF=BD/2, and similarly, we get GF=AC/2. By drawing the midpoint, H, of side AB and making lines HE and HG, we get a similar relation that HE=GF=AC/2, and HG=EF=BD/2, so EFGH is a parallelogram. At this point, we could solve the problem using areas. Using Bretschneider's Formula, we get

[ABCD]=1/4 * sqrt(4*27^2-(36+16-1-9)^2)=1/4 * sqrt(1152)=6sqrt(2)

It's easy to see that [EFGH]=[ABCD]/2=3sqrt(2) due to similar triangles. Since EFGH was a parallelogram, we have [EFGH]=2[EFG], which we could find using

[EFG]=1/2 EF*FG*sin theta = 1/2 * 27/4 * sin theta = 3sqrt(2)/2

sin theta=4sqrt(2)/9

Since sin^2 theta + cos^2 theta=1, we have

cos theta = sqrt(1-(4sqrt(2)/9)^2)=sqrt(1-32/81)=sqrt(49/81)=7/9 => a+b=7+9=16

Hung Hin Sun8 months agoThe answer is 7+9 =16

WLOG we let AB=1, BC=6, CD=3 and DA=4. The required angle is the same as one of the angle between the diagonals, which is also equal to the sum of angle DBC and angle ACB. We find first DB=sqrt(243/11) and AC=sqrt(33) (by using the properties of cyclic quadrilateral and cosine rule). Consider triangle DBC, we can find cos DBC to be 5/sqrt(33) and when we consider triangle ACB, we can also find cosACB=17/(3sqrt(33)). By using the compound angle formula, we can easily find cosEFG=85/99-8/99=77/99 which can be simplified to 7/9. Hence the answer is 7+9=16

Richard Chen8 months ago^{+1}I got a+b=512

Set AB=1, BC=6, CD=3, DA=4. It is easy to get AC*BD=27, thus EF*FG=27/4. The area of the quadrilateral is sqrt(72), which equals to the sum of areaABD and area BCD. Then we get sinBAC=sinBCD=sqrt(72)/11. Then cosBCD=7/11.

When we know CE=3, CF=1.5, cosBCD=7/11, it is easy to calculate EF=sqrt(243/44). And EF*FG=27/4, we get FG=sqrt(33/4). Then we take in EG=2, it is easy to get cosEFG=a/b=215/297.

Thus, a+b=215+297=512

Jonathan Liu8 months agoRichard Chen aren't BAD and BCD supplementary? Not BAC and BCD.

Richard Chen8 months agoJonathan Liu BAC=180°-BCD，thus sinBAC=sinBCD

Jonathan Liu8 months agoRichard Chen How did you get sin BAC=sin BCD?

Richard Chen8 months agoThe properties of Cyclic Quadrilaterial I've used are:

The Area Equation: S=sqrt((p-AB)(p-BC)(p-CD)(p-DA)), when p=perimeter(ABCD)/2

The Equation: AB*CD+AD*BC=AC*BD

The Equation: EG=(AB+CD)/2, EF=BD/2, FG=AC/2

and the law of sine & the law of cosine:

S(ABD)=(AB*AD*sinBAD)/2, BD^2=AB^2+AD^2-2*AB*AD*cosBAD......