I think there's a simpler solution. Let's say these two lines intersect at O. We know that BO = QO and that PO = AO. Given that AB = 16 and PQ = 14, we can derive that BO = QO = 1 and PO = AO = 15. Let's call the center of the two circles R and S. We can construct two similar triangles OBR and SAO (we know that they are similar because angle PSA = angle BOQ and we can apply AA similarity theorem). From this similarity we have BO/BR = AS/AO => BO/r = s/AO => rs = BO*AO = 15
It ain't that simple! Finding BO is a system of two equtions, equality of angles PSA and BOQ also has to be proven through sums of BOQ+POA and PSA+POA - that's extra steps too
Another solution, without using trig, and without squaring any variables: the format of the problem, which expects a constant solution but which does not state an angle of intersection, tells us that the value of rs is constant, and not dependent on then angle. So set the angle to a right angle. We know that AX = PX and BX = QX, and PX = PQ + QX. (This is true for all angles of intersection.) Therefore, for the case of the right angle, PX = 15 = S, and BX = 1 = R. Therefore RS = 15.
@verkuilb It's like solving a chess problem. Unlike normal chess, you can take advantage of the fact that there must be a known solution. It's not something you can use on real world problems.
You missed an easier solution, let's say the point where the lines meet is o, AO is 15 and BO is 1( bc AO=PO and BO=QO). now s/AO=BO/r because it's 2 triangle with same angles (AOS≈RBO). So now s/15=1/r--->sr=15. You can't get any simpler than this solution😅.
As a tradesman, I solve these problems graphically using FUSION (it used to be FUSION 360). I find it funny that I can get solutions in a limited range because of the limitations of calculation. When I saw rs= 14.9998, I figured that rs=15 was the right answer. Especially so when the answer switched to rs=15.0001 depending on where I was pushing the lines on the screen.
Using the fact that intersecting tangent points are equal in length between the point of tangency and the point of intersection, we can conclude that BO = QO = 1, where O is the point where the two lines intersect, and thus AO = PO = 15. Additionally, call R the center of the circle with radius r and S the center of the circle with radius S. The quadrilaterals RQOB and OPSA are similar, and corresponding parts of similar polygons are proportional. Therefore s / 1 = SA / QO = AO / RQ = 15 / r => rs = 15. Always fun that there are multiple ways to solve the same problem!
Alternate solution: Label the intersection of the two lines X. Let c1 be the centre of the small circle and c2 the centre of the large circle. AB = AX + XB = 16 and PQ = PX - XQ = 14. Note that BX = QX and AX = PX, so we can deduce that AX = 15 and BX = 1. Construct the right triangle C1 B X with angle b at X. r = tan(b) Construct the right triangle C2 A X with angle a at X. s = 15 tan(a) rs = 15 tan(a) tan(b). Observe that the angles a and b sum to pi/2, so tan(a) = 1/tan(b), thus rs = 15.
I am not good at geometry but i trust in algebra. From the intersection of lines to the circles centres the lines are bisectors and as such the angle between them will be 1/2 of the sum of angles that adf to 180 so 90 degrees. Drawing perpendicular from circle centres we get tan(alpha)=r/y tan(beta)=s/x Since 2 tangents from same point to a circle have the same length we can determine their sizes x+y=16 x-y=14 so y=1 and x=15 alpha+beta=90 so Cos(alpha+beta)=cos(alpha)cos(beta)-sin(alpha)sin(beta) 0=cos(alpha)cos(beta)-sin(alpha)sin(beta) => Tan(alpha)*tan(beta)=1 =>rs/xy=1 rs=15
They are two similar right-angled triangles. We can draw two small symmetric ones for the small circle with O, so QOB is equal to 16 - 14 (PQOB -PQ), and therefore OQ is 1. Thus, OP (OQP) will be 15. We have s and b1 as the legs of the larger triangle (b1=15), and a2 and r as the legs of the smaller triangle (a2 = 1). Since the triangles are similar, we have: r / 1 = 15 / s r * s = 15 r= cot(APO) s=15⋅tan(APO)
Another cool thing to consider so we can extremely simplify (16² - 14²)/4 is. Think of it as (16² - 14²)/2², then distribute to each → 16²/2² - 14²/2² → (16/2)² - (14/2)² {inverse power distribution rule} → 8² - 7² → (8 - 7)*(8 + 7) {difference between 2 squares rule} → 1*15 = 15
I understand completely how you derived the answer-but I’m left wondering whether “rs=15” has any significance, other than being the answer to this problem. Is there an area of 15 square units anywhere on the diagram which can be visually shown to be characteristic of the given information? Usually the question is seeking a length or area of something in the diagram. On this one, though, “rs=15” doesn’t seem to mean anything. It just “is”, abstractly…
@@hughcaldwell1034The fact that AX = 15 (length) and rs = 15 (area) have the same values is just coincidence, for only the case where AB - PQ = 2. AX will always be equal to (AB + PQ) / 2, but rs will not.
The significance of rs is that it is the quantity you can actually calculate. With the information given you can't calculate a value for r or for s but you can for the combination rs. I.e. this diagram isn't unique and there are many versions with different size circles, but they all obey rs=15 if the two labeled lengths have their given lengths.
Another solution ? X is intersection, n = |XA|, m = |XB| |PQ| = n-m |AB| = n+m -> n = 15, m = 1 now 2 * alpha = angle of PXA, 2 * beta = angle of BXQ and tan alpha = s / n tan beta = r / m but alpha + beta = 90 so tan beta = cotg alpha so s / n = m / r -> s * r = m * n = 15
As others have pointed out there is a much simpler solution than the one described in the video. For anyone near the beginning of their journey into geometry (or mathematics in general) I would offer the following guidance: When confronted by a problem like this one, where certain distances are given, try to solve the problem in the general case, and then plug in the known values at the end. So in this case we could let |PQ| = u and |AB| = v. Let the lines PQ and AB intersect at the point Z. Let BZ = QZ = x. So, we have PZ = u+x, and AZ = v-x. But PZ = AZ, so u+x = v-x, so x = (v-u)/2. As others have pointed out we can use similar triangles to show that x/r = s/(v-x). So, rs = x(v-x) = ((v-u)/2) * ((v+u)/2) = (v+u)(v-u)/4, or (v^2-u^2)/4 Plugging in the known values of u an v (14 and 16) gives rs = 15. There are several advantages to solving the general case rather than the particular case... 1) You gain an insight at to how the result depends on the given parameters. 2) During your working you can spot some errors very easily. For example, if at some point you have an expression such as "ux+v". You immediately know that you have done something wrong because you are adding together a term with dimensions L^2 ("uv", an area) and a term with dimension L ("v", a length). 3) You can test your answer by applying it to certain cases where the answer is obvious. For example, in the current puzzle, suppose r=0, then clearly rs should be zero and our formula (v^2-u^2)/4 correctly predicts that answer because, when r=0 it is clear than u=v. 4) When carrying numerical values through a series of steps in a proof it is easy to transcribe a digit wrongly, or do some piece of arithmetic wrongly. These problems are diminished if we are using letters rather than numerical values.
From an educational point of view: If this is a pure word problem, people will draw the radii r, s at the "right place" in a paper and will arrive at the solution (as only pythagoras theorem is involved). The mental blocker is the two radii drawn at two different random directions in the figure provided. Was it intentional to make it difficult? 🙂
1) Take intersection point as X. 2) XA=XP=x, XB=XQ=y 3) Given XA+XB=16 and XP-XQ=14 Translate them to x,y gives x+y=16, x-y=14 On solving, we get x=15, y=1 4) Now lets assume alpha be the angle subtanded by X with B and centre of circle with radius r, and beta be angle subtanded by X with P and centre of circle with radius s. 5) We know 2*alpha+2*beta=180, alpha+beta=90 6) tan(alpha)=r/1, tan(beta)=s/15 7) tan(alpha)*tan(beta)=(rs)/15 -> tan(alpha)*tan(90-alpha)=tan(alpha)*cot(alpha)=1=rs/15 -> Ans: rs=15
i believe theres simpler solution. lets draw lines from both centers to intersection of lines. lets call new-created angles α and β. so 2α+2β=180 => α=90-β. we also know that tg(90-β)=ctgβ. so r/1=15/s => rs=15
Let x be the distance between the centers of the two circles. Then, according to the Pythagorean Theorem, x² = 16² + (s - r)² x² = 14² + (s + r)² Since lefthandsides are equal, we can equate the righthandsides: 16² + (s - r)² = 14² + (s + r)² 16² + s² - 2sr + r² = 14² + s² + 2sr + r² 16² - 2sr = 14² + 2sr 16² = 14² + 4sr 16² - 14² = 4sr 8² - 7² = sr (8 - 7)(8 + 7) = sr sr = 1*15 = 15
Easiest solution(although probably not a proof) I see is rs must be the same regardless of the angle, so suppose it is a perpendicular ointersection, then r+s=16 and s-r=14, therefore r=1, s=15, rs=15.
I was just wondering if it was possible to work out what "r" and "s" are equal to individually ? (And other segments, but that shouldn't be too hard once you figure out the radii)
No, it’s not. Both r and s can increase/decrease as you change the angle of line intersection and the radii of the circles. But their product will always be 15, so long as the lengths between points of tangency remain 14 and 16 as stated.
@@verkuilb How do you know they will always be the same 15? Yes one increases while the other decrease when you change the angle but how do you know that the product specifically is constant?
@@ericpaul4575 What do you mean AB and PQ are fixed? Do you mean they are always the points where a line intersects the circle? Because I know that, but how does that make the product of the radius of each circle the same?
@@maxhagenauer24 No the lengths remain 16 and 14 respectively (AB and PQ). That means the ratio of the radii and other lengths would remain the same, hence the product is always 15.
How I solved it: 1) Let the point where PQ and AB cross be O 2) OQ = OB; OA = OP => OQ = 1; OA = 15 3) Let C1 be centre of small circle and C2 is the centre of bigger one 4) From triangles OBQ and APC2 we get that AP/BQ = s; (they have same angles) 5) From triangles BQC1 and OAP we get that r/15 = BQ/AP 6) rs = 15 Sorry for my English, I wish you understood)))
Wow! I did it a completely different way! I named the intersection between the lines Z, and said that PZ and AZ must be equal since they are tangents that meet at a point. Also BZ = QZ for the same reason. Therefore AB(16)=PQ - 2(ZB) This makes ZB=1=ZQ and therefore r is 1 because RB and RQ form a square with ZB and ZQ. Then since we can redraw the diagram with AB being perpendicular to PZ, PZ = s = 15 , so rs = 15x1 =15 This was a great problem, let me know if this method is incorrect but i happened to get the same answer when i clicked the solution! Thanks
You made an assumption that BZQ is a right angle. It does not have to be. However, I think your method works because r*s is the same regardless of the values of r and s (although I don’t know how to prove that)
If we call X the intersection of the two lines, it's very easy to see XB and XQ are of length 1 and XA and XP of length 15. If we call C the center of the small circle and D the center of the big circle, It's also easy to prove the right triangle CBX is similar to the the right triangle DAX, so the ratio of their legs must be equal and if s/15 = 1/r, then by cross product, rs = 15.
Incorrect. XB and XQ are only 1 (and XA and XP are 15) if the angle of intersection is a right angle. If you change the angle, you get different values for these lengths-but rs remains constant.
@@verkuilb No. XB=XQ and XA=XP no matter the angle (by virtue of a tangential circle having its center on the bisector of the angle). We are given AB = 16 and PQ = 14. Since AB = XA + XB and PQ = XP-XQ, XB and XQ are of length 1 and XA and XP of length 15. It is true r and s do change depending on the angle though.
@ekatvakushvaha1814 |PQ| = sqrt( (r+s)^2 - (r-s)^2 ) = sqrt(4rs), the formula is for two externally tangent circles, you can memorize it but it might be pretty intuitive to some others looking at it but it might not be for everyone.
@ekatvakushvaha1814 |PQ| = sqrt( (r+s)^2 - (r-s)^2 ) = sqrt(4rs), the formula is for two externally tangent circles, you can memorize it but it might be pretty intuitive to some others looking at it but it might not be for everyone.
you have WAY overcomplicated this. u literally need to use 3 properties which are 1) Similarity in triangle 2) Tangents from the same external point are equal 3) Angle sum property of quadrilaterals
I simply created a system of equations: s + r = 16 s - r = 14 and this would be correct regardless of the angle between the lines... and I get that s = 15 r = 1.
S+r does not necessarily equal 16. That’s only the case if the angle of intersection is a right angle. For other angles, other S and R will work. For example, (S,R) can be (5,3), or (30,0.5).
Irrelevant, as that only happens if the AB minus PQ is 2. If you change that difference, the arithmetic mean does not equal the product or rs. The fact that they’re the same here is pure coincidence.
I think there's a simpler solution. Let's say these two lines intersect at O. We know that BO = QO and that PO = AO. Given that AB = 16 and PQ = 14, we can derive that BO = QO = 1 and PO = AO = 15. Let's call the center of the two circles R and S. We can construct two similar triangles OBR and SAO (we know that they are similar because angle PSA = angle BOQ and we can apply AA similarity theorem). From this similarity we have BO/BR = AS/AO => BO/r = s/AO => rs = BO*AO = 15
This is how I approached the problem as well.
I think it might be even simpler, check my comment (which might be flawed ofc)
It ain't that simple! Finding BO is a system of two equtions, equality of angles PSA and BOQ also has to be proven through sums of BOQ+POA and PSA+POA - that's extra steps too
I grant you that you've devised a perfect alternative solution ( Congrats! ), but am not sure if it is simpler.
This is basically the way I solved it as well. It may be hard to follow because a few steps have been skipped over, such as showing that
Another solution, without using trig, and without squaring any variables: the format of the problem, which expects a constant solution but which does not state an angle of intersection, tells us that the value of rs is constant, and not dependent on then angle. So set the angle to a right angle. We know that AX = PX and BX = QX, and PX = PQ + QX. (This is true for all angles of intersection.) Therefore, for the case of the right angle, PX = 15 = S, and BX = 1 = R. Therefore RS = 15.
My solution! Yes cheating, assuming first there is a solution!
nice solution!
@@palpatinewasright Not cheating. Simply using the information provided, in a unique way.
@verkuilb It's like solving a chess problem. Unlike normal chess, you can take advantage of the fact that there must be a known solution. It's not something you can use on real world problems.
@@verkuilbbut tje format doesnt "expect" a constant solution sp are you sure that is accueate? Not sure sorry
Where is the music at the end?
There's no music at the end right??
You missed an easier solution, let's say the point where the lines meet is o, AO is 15 and BO is 1( bc AO=PO and BO=QO).
now s/AO=BO/r because it's 2 triangle with same angles (AOS≈RBO).
So now s/15=1/r--->sr=15.
You can't get any simpler than this solution😅.
As a tradesman, I solve these problems graphically using FUSION (it used to be FUSION 360). I find it funny that I can get solutions in a limited range because of the limitations of calculation. When I saw rs= 14.9998, I figured that rs=15 was the right answer. Especially so when the answer switched to rs=15.0001 depending on where I was pushing the lines on the screen.
Using the fact that intersecting tangent points are equal in length between the point of tangency and the point of intersection, we can conclude that BO = QO = 1, where O is the point where the two lines intersect, and thus AO = PO = 15. Additionally, call R the center of the circle with radius r and S the center of the circle with radius S. The quadrilaterals RQOB and OPSA are similar, and corresponding parts of similar polygons are proportional. Therefore
s / 1 = SA / QO = AO / RQ = 15 / r => rs = 15.
Always fun that there are multiple ways to solve the same problem!
Yes similar triangles much simpler than Pythagoras
Great! I would likely not come up with the solution on my own, but I'd love to see it on a test somewhere! Another cool math video, thanks!
perfect demonstration with the simpest terms. congrats.
Your animations to explain the steps are cool 👍
That was brilliant. Thanks.
Very elegant!!!
Damn! I knew the solution! I figure it out by myself! So satisfying!
Thank You!
Nice video! Even though I am new to your channel, the video was entertaining and fun. You deserved a like and subscribe from me! I wish you the best
Great Job
This was a good solution. I understood it all the way.
Asking for rs as the answer is a big hint that there are similar triangles lurking somewhere.
Alternate solution:
Label the intersection of the two lines X. Let c1 be the centre of the small circle and c2 the centre of the large circle.
AB = AX + XB = 16 and PQ = PX - XQ = 14.
Note that BX = QX and AX = PX, so we can deduce that AX = 15 and BX = 1.
Construct the right triangle C1 B X with angle b at X.
r = tan(b)
Construct the right triangle C2 A X with angle a at X.
s = 15 tan(a)
rs = 15 tan(a) tan(b).
Observe that the angles a and b sum to pi/2, so tan(a) = 1/tan(b), thus rs = 15.
Exactly what I did, except that a+b = pi/2.
@@RGP_Maths Damn, little brain slip there. Thanks. Fixed it.
I am not good at geometry but i trust in algebra. From the intersection of lines to the circles centres the lines are bisectors and as such the angle between them will be 1/2 of the sum of angles that adf to 180 so 90 degrees.
Drawing perpendicular from circle centres we get
tan(alpha)=r/y
tan(beta)=s/x
Since 2 tangents from same point to a circle have the same length we can determine their sizes
x+y=16
x-y=14 so y=1 and x=15
alpha+beta=90 so
Cos(alpha+beta)=cos(alpha)cos(beta)-sin(alpha)sin(beta)
0=cos(alpha)cos(beta)-sin(alpha)sin(beta)
=> Tan(alpha)*tan(beta)=1
=>rs/xy=1 rs=15
They are two similar right-angled triangles.
We can draw two small symmetric ones for the small circle with O, so QOB is equal to 16 - 14 (PQOB -PQ), and therefore OQ is 1. Thus, OP (OQP) will be 15.
We have s and b1 as the legs of the larger triangle (b1=15), and a2 and r as the legs of the smaller triangle (a2 = 1).
Since the triangles are similar, we have:
r / 1 = 15 / s
r * s = 15
r= cot(APO)
s=15⋅tan(APO)
Another cool thing to consider so we can extremely simplify (16² - 14²)/4 is.
Think of it as (16² - 14²)/2², then distribute to each → 16²/2² - 14²/2² → (16/2)² - (14/2)² {inverse power distribution rule} → 8² - 7² → (8 - 7)*(8 + 7) {difference between 2 squares rule} → 1*15 = 15
Great use of the Pythagorean Theorum
no data on the angle of the crossing lines . make it 90 degree like a regular cross . the answer is just there.
I understand completely how you derived the answer-but I’m left wondering whether “rs=15” has any significance, other than being the answer to this problem. Is there an area of 15 square units anywhere on the diagram which can be visually shown to be characteristic of the given information? Usually the question is seeking a length or area of something in the diagram. On this one, though, “rs=15” doesn’t seem to mean anything. It just “is”, abstractly…
Not sure if it really gives a better intuition, but if you label the intersection of the lines X, then |AX| * |AB| = 15 * 1 = 15.
@@hughcaldwell1034The fact that AX = 15 (length) and rs = 15 (area) have the same values is just coincidence, for only the case where AB - PQ = 2. AX will always be equal to (AB + PQ) / 2, but rs will not.
The relationship is:
15 = (16^2 - 14^2) / 4
The significance of rs is that it is the quantity you can actually calculate. With the information given you can't calculate a value for r or for s but you can for the combination rs. I.e. this diagram isn't unique and there are many versions with different size circles, but they all obey rs=15 if the two labeled lengths have their given lengths.
@ But rs will always be equal to AX * BX.
Another solution ?
X is intersection, n = |XA|, m = |XB|
|PQ| = n-m |AB| = n+m -> n = 15, m = 1
now 2 * alpha = angle of PXA, 2 * beta = angle of BXQ and
tan alpha = s / n tan beta = r / m but alpha + beta = 90 so tan beta = cotg alpha so s / n = m / r -> s * r = m * n = 15
As others have pointed out there is a much simpler solution than the one described in the video.
For anyone near the beginning of their journey into geometry (or mathematics in general) I would offer the following guidance:
When confronted by a problem like this one, where certain distances are given, try to solve the problem in the general case, and then plug in the known values at the end. So in this case we could let |PQ| = u and |AB| = v. Let the lines PQ and AB intersect at the point Z. Let BZ = QZ = x.
So, we have PZ = u+x, and AZ = v-x. But PZ = AZ, so u+x = v-x, so x = (v-u)/2.
As others have pointed out we can use similar triangles to show that x/r = s/(v-x).
So, rs = x(v-x) = ((v-u)/2) * ((v+u)/2) = (v+u)(v-u)/4, or (v^2-u^2)/4
Plugging in the known values of u an v (14 and 16) gives rs = 15.
There are several advantages to solving the general case rather than the particular case...
1) You gain an insight at to how the result depends on the given parameters.
2) During your working you can spot some errors very easily. For example, if at some point you have an expression such as "ux+v". You immediately know that you have done something wrong because you are adding together a term with dimensions L^2 ("uv", an area) and a term with dimension L ("v", a length).
3) You can test your answer by applying it to certain cases where the answer is obvious. For example, in the current puzzle, suppose r=0, then clearly rs should be zero and our formula (v^2-u^2)/4 correctly predicts that answer because, when r=0 it is clear than u=v.
4) When carrying numerical values through a series of steps in a proof it is easy to transcribe a digit wrongly, or do some piece of arithmetic wrongly. These problems are diminished if we are using letters rather than numerical values.
I solved this in my head!
Luckily the numbers were headable, lol
From an educational point of view: If this is a pure word problem, people will draw the radii r, s at the "right place" in a paper and will arrive at the solution (as only pythagoras theorem is involved). The mental blocker is the two radii drawn at two different random directions in the figure provided. Was it intentional to make it difficult? 🙂
two Pacmans were offended by each other on preview
I love geometry, always have.
I think after watching this, I might beat an extreme demon, thanks!
Bro, just use the DCT and TCT lengths formula...
Form two equations, square and then subtract them to get the value of rs
1) Take intersection point as X.
2) XA=XP=x, XB=XQ=y
3) Given XA+XB=16 and XP-XQ=14
Translate them to x,y gives
x+y=16, x-y=14
On solving, we get x=15, y=1
4) Now lets assume alpha be the angle subtanded by X with B and centre of circle with radius r,
and beta be angle subtanded by X with P and centre of circle with radius s.
5) We know 2*alpha+2*beta=180, alpha+beta=90
6) tan(alpha)=r/1, tan(beta)=s/15
7) tan(alpha)*tan(beta)=(rs)/15
-> tan(alpha)*tan(90-alpha)=tan(alpha)*cot(alpha)=1=rs/15
-> Ans: rs=15
Am I the only one who misses the old "pause the video to give this problem a try and when you're ready..."?
i believe theres simpler solution. lets draw lines from both centers to intersection of lines.
lets call new-created angles α and β. so 2α+2β=180 => α=90-β.
we also know that tg(90-β)=ctgβ. so r/1=15/s => rs=15
I solved it the same way nice.
Much easier solution can be found by kongruent triangles and the relationships 1/s = r/(16-1), so rs =15×1=15
can you calculate the angles of the triangle and the lenght of A and P to the center, where the lines connect
Let x be the distance between the centers of the two circles. Then, according to the Pythagorean Theorem,
x² = 16² + (s - r)²
x² = 14² + (s + r)²
Since lefthandsides are equal, we can equate the righthandsides:
16² + (s - r)² = 14² + (s + r)²
16² + s² - 2sr + r² = 14² + s² + 2sr + r²
16² - 2sr = 14² + 2sr
16² = 14² + 4sr
16² - 14² = 4sr
8² - 7² = sr
(8 - 7)(8 + 7) = sr
sr = 1*15 = 15
Aneone else think the smaller circle looked mad while bigger one looked a little upset?
Easiest solution(although probably not a proof) I see is rs must be the same regardless of the angle, so suppose it is a perpendicular ointersection, then r+s=16 and s-r=14, therefore r=1, s=15, rs=15.
r+s can't be 16 because BA is 16 and clearly there is a gap between circles, and s-r can't be 14
@@MautozTech Incorrect if the angle of intersection of the two lines is 90 degrees, then s=15, r=1, and their sum is indeed 16.
We can have two mathematically similar kites as well
(s+r)²+14²=(s-r)²+16²
(s+r)²-(s-r)²=16²-14²
4sr=(16-14)(16+14)
sr=15
I was just wondering if it was possible to work out what "r" and "s" are equal to individually ?
(And other segments, but that shouldn't be too hard once you figure out the radii)
No, it’s not. Both r and s can increase/decrease as you change the angle of line intersection and the radii of the circles. But their product will always be 15, so long as the lengths between points of tangency remain 14 and 16 as stated.
@@verkuilb How do you know they will always be the same 15? Yes one increases while the other decrease when you change the angle but how do you know that the product specifically is constant?
@@maxhagenauer24 because AB and PQ are fixed.
@@ericpaul4575 What do you mean AB and PQ are fixed? Do you mean they are always the points where a line intersects the circle? Because I know that, but how does that make the product of the radius of each circle the same?
@@maxhagenauer24 No the lengths remain 16 and 14 respectively (AB and PQ). That means the ratio of the radii and other lengths would remain the same, hence the product is always 15.
How I solved it:
1) Let the point where PQ and AB cross be O
2) OQ = OB; OA = OP => OQ = 1; OA = 15
3) Let C1 be centre of small circle and C2 is the centre of bigger one
4) From triangles OBQ and APC2 we get that AP/BQ = s; (they have same angles)
5) From triangles BQC1 and OAP we get that r/15 = BQ/AP
6) rs = 15
Sorry for my English, I wish you understood)))
Aren't tangent drawn from external pont equal?
15, I solved using trigonometry
Cool!
Wow! I did it a completely different way! I named the intersection between the lines Z, and said that PZ and AZ must be equal since they are tangents that meet at a point. Also BZ = QZ for the same reason. Therefore AB(16)=PQ - 2(ZB)
This makes ZB=1=ZQ and therefore r is 1 because RB and RQ form a square with ZB and ZQ. Then since we can redraw the diagram with AB being perpendicular to PZ, PZ = s = 15 , so rs = 15x1 =15
This was a great problem, let me know if this method is incorrect but i happened to get the same answer when i clicked the solution! Thanks
You made an assumption that BZQ is a right angle. It does not have to be. However, I think your method works because r*s is the same regardless of the values of r and s (although I don’t know how to prove that)
rs=15 u^2
15
This is the most pro-Elon Musk channel I've seen. I mean, it does thank him after every video 😂
hey we get the formula a-b/2 = r1•r2
If we call X the intersection of the two lines, it's very easy to see XB and XQ are of length 1 and XA and XP of length 15. If we call C the center of the small circle and D the center of the big circle, It's also easy to prove the right triangle CBX is similar to the the right triangle DAX, so the ratio of their legs must be equal and if s/15 = 1/r, then by cross product, rs = 15.
Incorrect. XB and XQ are only 1 (and XA and XP are 15) if the angle of intersection is a right angle. If you change the angle, you get different values for these lengths-but rs remains constant.
@@verkuilb No. XB=XQ and XA=XP no matter the angle (by virtue of a tangential circle having its center on the bisector of the angle). We are given AB = 16 and PQ = 14. Since AB = XA + XB and PQ = XP-XQ, XB and XQ are of length 1 and XA and XP of length 15. It is true r and s do change depending on the angle though.
Slight mis-speak at 5:49 to see who is paying attention.
∣AB∣^2 − ∣PQ∣^2 = 4*r*s
16^2 - 14^2 = 4*r*s
r*s = 15
How did you get to this conclusion?
@ekatvakushvaha1814 |PQ| = sqrt( (r+s)^2 - (r-s)^2 ) = sqrt(4rs), the formula is for two externally tangent circles, you can memorize it but it might be pretty intuitive to some others looking at it but it might not be for everyone.
@@maxhagenauer24 oh ok thanks
@ekatvakushvaha1814 Yes your welcome no problem, it's just a formula but you can might ve able to see it intuitively, just hard to describe.
@ekatvakushvaha1814 |PQ| = sqrt( (r+s)^2 - (r-s)^2 ) = sqrt(4rs), the formula is for two externally tangent circles, you can memorize it but it might be pretty intuitive to some others looking at it but it might not be for everyone.
Geometry
Hey!
Again I will show you an illegal shortcut: set the angle to be 90 degrees and you can see s has to be (16+14)/2 and r has to be (16-14)/2 so rs=15.
🇮🇳🇮🇳🇮🇳🇮🇳🇮🇳🇮🇳🇮🇳🇮🇳🇮🇳
you have WAY overcomplicated this. u literally need to use 3 properties which are
1) Similarity in triangle
2) Tangents from the same external point are equal
3) Angle sum property of quadrilaterals
@2:08 why are we allowed to assume that s>r?
It doesn’t matter if it’s negative or not I’m pretty sure as it’s instantly squared.
@ agreed, just something that should be pointed out. Can write |s-r| to cover all cases
@@Happy_Abe that’s true, it would be nicer that way.
Can one find individual values for r and s? My intuitive answer:
r = 3
s = 5
No. They will change as the angle of intersection changes. For example, with a right angle, R=1 and S=15.
I simply created a system of equations:
s + r = 16
s - r = 14
and this would be correct regardless of the angle between the lines...
and I get that
s = 15
r = 1.
S+r does not necessarily equal 16. That’s only the case if the angle of intersection is a right angle. For other angles, other S and R will work. For example, (S,R) can be (5,3), or (30,0.5).
i sent more complex problem to you
Arithmetic mean in geometry (14+16):2=15.
Irrelevant, as that only happens if the AB minus PQ is 2. If you change that difference, the arithmetic mean does not equal the product or rs. The fact that they’re the same here is pure coincidence.
I have a much simpler solution with far less steps...
Hello sir,today is My birthday (31.1.2025)🎉
Happy belated birthday 🥳🤍
Fifteen square units, not fifteen units squared.