2 nice inscribed circle problems

0:00 intro
0:45 review
2:16 problem 1
5:06 problem 2

Problem 1
Problem 2

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[randomguy]

I just got a straight edge and compass yesterday, it's amazing how much more intuitive these problems are with these tools, even if you don't fully construct the solution.

[danmerget]

Whoa! Someone wrote a math puzzle involving a right triangle with integral sides, and for once it ISN'T a 3-4-5 triangle. I didn't know that was possible! 😊

[angeluomo]

On the second problem, I worked with similar triangles. Comparing the two legs of two similar right triangles, the following ratio is valid: (7-r)/r = r/(3-r). This becomes 21-10r+r^2 = r^2 and simplifies to 21 = 10r, or r= 2.1. Done.

[wayneyadams]

Ilove this type of problem. Inscribed circles shapes inscribed in circles, various overlapping cirgles and other figures are all very interesting.

[paparmar]

Problem 1 can be solved without invoking Pythagoras by using the characteristic of right triangles wherein the radius of the inscribed circle is [product of legs]/perimeter. From the information given and derived @ 3:56: 12 = (a+12) (56)/[2*(a + 56)]; hence a = 21.

[merterenacar]

fun fact: Hey, this is Presh Talwalkar

[big_numbers]

Haven't watched the video yet but I did these by using the fact that the 2 legs of a kite you can make with by connecting the center of the circle with the tangent points are the same. Got a=21 for the first one and r=2.1 for the second one.

[sergeyg]

If you recall the double angle tangent formula, the answer to prob 1 pops right out: tan(½α)=12/44; and so the other leg is 33/56*(12+44)=33 and a=33-12=21.

[geetashashikanth]

Question:
Two circles intersect at points A and B. PQ is a common tangent to the circles, touching them at P and Q respectively. Prove that the angles APB and AQB are supplementary (their sum is 180 degrees).


I had a doubt plz reply 😢

[jaimeyomayuza]

Gracias por la dedicación y trabajo, excelente página!!!
From. Bogotá D.C. COLOMBIA

[GillesF]

About the first exercise: because << The center of the circle inscribed in any triangle is the point of intersection of the bisectors >>, we can get the right answer this way:

• consider triangle (of the picture) is ABC (with A being the top angle and C the 90° angle)

• consider O as the center of the circle

then:

angle ABO = arctan(12/44) = 15.25°

angle ABC = 2·ABO = 2·15.25° = 30.5°

angle BAC = 180° - 90° - 30.5° = 59.5°

angle BAO = BAC/2 = 59.5°/2 = 29.75°

=>

tan(29.75°) = 12/x

x·tan(29.75°) = 12

x = 12/tan(29.75°)

x ≈ 20.99

■ final result: x = 21 units of length

🙂 AND THANK YOU FOR THE VIDEO!!!

[walterfletcher]

9:20 - I'm going to have to take off points for units appearing out of nowhere. Centimeters are as good a unit as any but that was not stated in the original problem.

[IvanToshkov]

There's a better way for solving equation at the end of the first problem:

1. We start with (a + 12)^2 + 56^2 = (a + 44)^2
2. Subtract (a + 12)^2 from both sides => 56^2 = (a + 44)^2 - (a + 12)^2
3. Use the formula A^2 - B^2 = (A - B)(A + B) => 56^2 = 32(2a + 56)
4. 56 factored to primes is 2^3*7, so 56^2 = 2^6*7^2. 32 is 2^5. => 2^6 * 7^2 = 2^5 * (2a + 56)
5. Extract 2 from the parenthesis => 2^6 * 7^2 = 2^6 * (a + 28)
6. Divide both sides by 2^6 => a + 28 = 49
7. a = 21

[wayneyadams]

1:47 It is not a supposition that OB and OC are the same length since they are both radii, it is a FACT!!!!

[BobOBob]

As far as I was able to observe, the setup of Problem 2 did not say anything about units.
The introduction of centimeters at 9:20 does not seem to be called for.

[logic_lane-zu]

As a student of 9th I can confidently say that " *I CAN'T SOLVE THIS PROBLEM* "

[marcgriselhubert]

Let's use an orthonormal center O, the lower left corner of the trapezoïd and horizotal first axis. Be A the upper right corner of
the trapezoïd, we have A(3; 2.R) with R the radius of the circle. Be B the lower right corner of the trapezoïd, we heve B(7; 0).
It's easy to have the equation of (AB): R.x +2.y -7.R =0. The center P of the circle is P(R; R) and the distance from P to (AB) is:
abs((R^2) +2.R -7.R)/sqrt((R^2) +4) = (R.(5 - R)) /sqrt((R^2) + 4) as R<5. This distance is equal to R, then we have:
5 - R = sqrt((R^2) + 4). We square: (R^2) -10.R + 25 = (R^2) + 4, and then 10.R = 25 4 = 21, so finally R = 21/10.

[devondevon]

a = 21
This is the easiest one of the two.
The base = 56 ( 12 + 44 ) tangent circle theorem
The height = 12 + a tangent circle theorem
The hypotenuse = 44 + a (given)
Pythagorean Theorem:
(44 + a)^2 = (12 + a)^2 + 56^2
a^2 + 88a + 1936 = a^2 + 24a + 144 + 3136
88a - 24a = 3280 - 1936
64a = 1344
a = 1344/64
a =21

[tiborfarkas]

to the 2nd prob. let's generalize. 3 - > x, 7-> y.
based on the similar triangles, the solution is r = xy/(x+y) = 1/(1/x+1/y),
so the diameter of the circle is the HARMONIC MEAN of the two parallel sides of the trapezoid.

[gregoryknapen]

For the second problem, you have similar triangles so r/(7-r) = (3-r)/r. Solving for r gives the same answer as in the video.