This downside appeared on the 2022 JEE Superior Paper 1 Arithmetic part as Query 8. Due to Rahul and Battu for the suggestion! The JEE Superior is an especially tough examination for admission into India’s prestigious IIT colleges. This is among the simpler questions of the paper. I give credit score to the Unacademy Atoms video concerning the paper which helped me perceive the best way to resolve the issue.
Let ABC be the triangle with AB = 1, AC = 3, and ∠BAC = π/2. If a circle of radius r > 0 touches the edges AB, AC and in addition touches internally the circumcircle of the triangle ABC, then the worth of r is ___.
As typical, watch the video for an answer.
JEE 2022 Triangle Circumcircle And Inscribed Circle
Or hold studying. . .
“All can be properly in case you use your thoughts in your selections, and thoughts solely your selections.” Since 2007, I’ve devoted my life to sharing the enjoyment of sport principle and arithmetic. MindYourDecisions now has over 1,000 free articles with no adverts due to neighborhood help! Assist out and get early entry to posts with a pledge on Patreon.
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Since BAC = π/2, angle A is a proper angle and BC is its hypotenuse equal to √(32 + 12) = √10. Then angle A is 90 levels, so it is going to subtend a 180 diploma arc of the triangle’s circumcircle. This implies BC is a diameter, the circumcircle’s radius is half its size (0.5√2), and the circumcircle’s heart M is the midpoint of the hypotenuse–which is midway horizontally between A and C (distance 0.5) and midway vertically between A and C (distance 1.5).
The circle D is constructed to the touch sides AB, AC and is internally tangent at T to the circumcircle. As a result of M and D are tangent circles, their facilities and the tangent level are collinear. Since MT is a radius of the circumcircle (radius 0.5√10), and DT is that of the smaller circle (radius r), MD is the distinction in lengths of the radii, or 0.5√10 – r.
Assemble tangent radii to AB and AC from circle D. Every radius is r. Since angle A is a proper angle, there’s a sq. fashioned between A and D with facet lengths equal to r. Thus we will calculate the horizontal and vertical distances between D and M as |r – 0.5| and |r – 1.5|. We then have a proper triangle MED with legs ME = |r – 1.5| and ED = |r – 0.5| and hypotenuse 0.5√10 – r. We thus have:
Since r > 0 was given, the one resolution is r = 4 – √10 &approx 0.84.
References
2022 JEE Superior paper 1 reply sheet (see Q.8) https://jeeadv.ac.in/paperwork/jeeadv-2022-paper1.pdf
Unacademy Atoms video options to paper 1 (Q8 is round 59:00) https://youtu.be/prgX_Wu_O6A?t=3544
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