A Nice Geometry Problem | Best Math Olympiad Problems
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ABCD is inscribed in the circle of diameter BD.
The use of Pythagoras, Ptolemy and Heron theorems will give :
Phytagoras's theorem : BD = 15√2 , CD =3
Ptolemy's theorem : AB*CD + AD*BC = AC*BD ---> AC = 12√2
Heron's theorem : area of ADC = sqrt(p*(p-a)*(p-b)*(p-c)) with p = 0.5*(a+b+c) and a= DC, b=AC, c= AD --->
a= 3, b= 12√2, c = 15 , p = 9+ 6√2 ---> area of ADC = 18
There is simpler way. After getting the "3" dimension, the sine rule can be used. You need the sin of angle ADC. Use the sin (a+b) rule. Angle ADC = angle ADB + angle BDC. Sin(ADC) = COS(ADC) = 1/sqrt(2), sin(BDC) = 21/(15sqrt(2)), cos(BDC) = 3/(15sqrt(2)). Area ADC = (1/2)(15)(3)(1/sqrt(2))[(7/5)+(1/5)]/sqrt(2) = 18.
Triangle ABD is isosceles right triangle. Therefore angles ADB=ABD=45 and segment AD=15*sqrt(2)
Then I looked at right triangle BCD and also found segment CD=3
Now we all agreed that ABCD is a cyclic quadrilateral, so I called angle CBD=CAD=alfa.
Then I realized angle ACB=ADB=45. Since angle DCB=90, then angle DCA=45
On triangle ACD I applied law of sines using angles CAD=alfa and DCA=45. With this I could get the relation {sin(alfa)/3} = {sin(45)/15}. This resulted in sin(alfa)={sqrt(2)}/10
Since I forgot about Ptolemy's theorem, I looked at triangle ABC and applied law of sines one more time using angles CBA=(45+alfa) and ACB=45 to get the relation
{sin(45+alfa)/AC} = {sin(45)/15}. This resulted in AC=12*sqrt(2).
Now I could get the area of ADC=(1/2)*CD*AC*sin(45) ---> 18
Another simple problem for my favourite formula!
Having found DC, find AC with the cosine (supplement) rule, then the area of ACD with the semi-perimeter area formula.
This is a cyclic quadrilateral, as the hypotenuses of the two right-angle triangles are diameters of the same circumscribed circle.
Opposite angles of cyclic quadrilaterals are supplementary (180-).
The cosine of an angle is minus the cosine of it’s supplement, so we can use the cosine (supplement) rule to find AC.
Then find the area of triangle ACD with the semi-perimeter rule.
Remembering the cosine rule as:
In any triangle, the cosine of any angle is:
(The sum of the squares of the adjacent sides, less the square of the opposite side) divided by twice the product of the adjacent sides… and the cosine of the supplement of an angle is minus the cosine of the first angle. Let AC^2=x
In triangles ABC and ADC:
CosABC = -CosADC:
(15^2+21^2-x)/2*21*15=(x-3^2-15^2)/2*15*3
(666-x) = (x-234)*7
666 -x= 7x-1,638
2,304 = 8x
x = 288 = AC^2
AC = 16.9705
ADC=acos(15^2+3^2-16.97^2)/2*15*3=126.8698 degrees
Area ADC = 15*3*sinADC/2 = 18
Since AC is the bisector in triangle DCB, the solution becomes easier (since the angle subtending the diameter is 90 degrees and angle ABD and ACD are the same angle). In this case, the solution becomes simpler: (15x15+21x3)/2=144 gives the entire area. According to the angle bisector rule, shaded area = entire area/8 = 144/8=18. I recommend it to those who want a different and fast solution.
That's a really nice approach.
Excellent approach. However could you please explain this "According to the angle bisector rule, shaded area = entire area/8"? It would be of great benefit to all of us
@@rdesouza25
Of course, Mr. Souza,
if you see the bisector in the triangle DCB, it is obvious that the ratio of the areas of this triangle with base DB is 3/21, that is, when simplified, it is 1/7. If it is seen that the similar ratio is in the ABD triangle, the shaded area can be said to be 1/8 of the entire area. Enjoyable solutions👋
@@howardaltman7212
Thank you for your valuable comment, enjoyable solutions 👍
@@bestamitasar Thank you for the explanation. I understood it and I was able to calculate the (1/8) with the ratio of (1/7) on triangles ABD and CDB
Since it is a cyclic quadrilateral, area(∆ADC)/area(∆ABC) = (ADxDC)/(ABxBC)
Area of quadrilateral = area(∆ADC) + (∆ABC)
Also area of a cyclic quadrilateral can be found can be found using brahmagupta's formula
Find angle dbc in triangle dbc by first finding length of db and using sine rule. Angle dbc = angle dac (cyc. quad.). Then find angle acd using sine rule.Then that gives third angle adc. Then use area = 1/2 by |dc|.|ad| sine angle adc
A solution for the 2nd part: I have calculated that CD=3. Rotate clockwise ABCD round A with 90 degrees! We get a trapezoid DCC1B1. CC1=21+3=24 so area DCC1=3*24/2=36. DCB1 has the same area and DCA is the half ie. 18.
Find E on AC so DE is the height of red △. ABCD is cyclic because it has two oposites 90º angles. Then ∡DCA=∡ABD=45º (=∡CDE) and ∡DAC=∡DBC. ◺ABD is isosceles and has two 45º angles so DB =15√2. By Pythagoras on ◺BCD, DC=3. In ◺CDE by the 45º angles: EC=ED=3/√2. ◺ADE is similar to ◺BCD because ∡DAC=∡DBC, then AE/15=21/15√2 so AE=21/√2. AC=AE+EC=21/√2+3/√2=24/√2. Finally area of red △ is (AC)(ED)/2=(24/√2)(3/√2)/2=18.
Draw BD. As AB = DA = 15, ∆DAB is an isosceles right triangle.
Triangle ∆DAB:
DA² + AB² = BD²
BD² = 15² + 15²
BD² = 225 + 225
BD = √450 = 15√2
Triangle ∆BCD:
CD² + BC² = DB²
CD² + 21² = 15√2
CD² = 450 - 441 = 9
CD = √9 = 3
As ∆DAB is an isosceles right triangle, ∠ABD = ∠BDA = 45°, and cos(45°) = sin(45°) = 1/√2. Let ∠DBC = α and ∠CDB = β. As all side lengths for ∆BCD are known, we can calculate cos(α) = sin(β) = 21/15√2 = 7/5√2 and sin(α) = cos(β) = 3/15√2 = 1/5√2.
Triangle ∆ABC:
AC² = BC² + AB² - 2(BC)(AB)cos(α+45°)
AC² = 21² + 15² - 2(21)(15)cos(α+45°)
AC² = 441 + 225 - 630cos(α+45°)
AC² = 666 - 630cos(α+45°)
cos(α+45°) = cos(α)cos(45°) - sin(α)sin(45°)
cos(α+45°) = (7/5√2)(1/√2) - (1/5√2)(1/√2)
cos(α+45°) = 7/10 - 1/10 = 6/10 = 3/5
AC² = 666 - 630(3/5) = 666 - 378
AC = √288 = 12√2
Triangle ∆CDA:
A = (1/2)CD(DA)sin(β+45°)
A = (1/2)3(15)sin(β+45°)
A = (45/2)sin(β+45°)
sin(β+45°) = sin(β)cos(45°) + cos(β)sin(45°)
sin(β+45°) = (7/5√2)(1/√2) + (1/5√2)(1/√2)
sin(β+45°) = 7/10 + 1/10 = 8/10 = 4/5
A = (45/2)(4/5) = 180/10 = 18
A great problem which yields several solutions. I used the fact the quadrilateral is inscribable to facilitate a solution. The most important thing about these problems is not the nitty gritti, which most know about, but the thinking that lead to the analysis.
DB=15√2..DC=3(Pitagora)..carnot per il triangolo ADC,risulta l'angolo ADC=arccos(-3/5),perciò lamgolo ABC è 180-arccos(-3/5)...AH=15-63/5=12/5'com H la perpendicolare da C... quindi,in sintesi,A=(15*12/5)/2=18
Not the way I did it, but if you look at the dimensions (especially once you determine CD = 3), you can add a much smaller construction to aid in the solution than rhe rectangle shown. 15 is 20-5, and 21 is 25-4. If you extend AD and BC those distances, the figure becomes a 25-20-15 right triangle.
Even if it wasn't apparent that we're dealing with Pythagorean triples, the distances needed for AD and BC to be extended to intersect are two variables, and between Pythagoras and the necessary similarity of a triangle created by such extensions, you've got two equations.
Alternative method:
1. Find DC using Pythagoras theorem
2. Find AC using Ptolemy theorem
3. Find area of triangle using Heron's formula
You are an expert by copying my solution.
Be more creative
AN perpendikular BC , triangle ANB rotate 90 degrees , then B1=D , AN1CN square, AN=(21+3)/2=12, area ADC=3*12/2=18!
I did, eventually, work my way through your solution.
This is a clever solution, but it only works as we have two sides of length 15. My solution is more general.
You mentioned AN1CN is square. Where is point N1?
@@rdesouza25 After turning triangle ANB , point N1 it will be located to lines DC!
I also use Pythagoras first but then i finded the angle ADC by trigo and then area=1/2ab sintheta
I solved it in a few minutes like everyone else I think. However, I forgot one thing: before solving the exercises in this channel let's remember to rent a camper because to go from Florida to Maine we always have to pass through Oregon 😢😶🌫️🥶🤕
Sir since opposite angle add to 180 it is cyclic quadrilateral we can use Ptolemy theorem after finding 3 as DC to find AC
18 sq unit. It's a nice question & can easily be solved by using Pythagoras theorem & trigonometry 👌👌
That's OK!!!
How is CN height in triangle ADC??
In order for him to get AN=12, he needed AC=12*sqrt(2) and angle ACB=45. Then triangle ACN is isosceles right triangle with hypotenuse AC=12*sqrt(2)
I believe the height is AN1=12, since he rotated triangle ANB by 90 degrees. So the point N1 belongs to line that contains points C and D. By doing this, he got the square AN1CN. So the area ADC is CD*AN1*(1/2). That´s how {3*12/)2)}=18.
18
Well, I had to stare at it from time to time for the whole day, but a solution possibility finally came. I hope the answer is ... 18 (??!!?)
Now to watch the video.
Whoo hoo! Got it.
Although, my solution involved extending segments AD and BC to meet at, say, point P.
My solution also required solving for the length of segment CD (3).
Then, using similar triangles, I established that the new right triangle, DCP, was, in fact a 3-4-5 right triangle.
Once those dimensions were arrived at, it was a simple matter to find a height, h=(12/5), for the subject triangle, ADC, with base 15.
Thanks for a nice challenge.