South Korea’s CSAT test is a standardized college admissions test that is known to be difficult. This was an interesting question.

The quadrilateral ABCD is inscribed in a circle, with AB = 5, AC = 3√5, and AD = 7. Also angle BAC = angle CAD. What is the radius of the circle?

As usual, watch the video for a solution.

Fun Problem From Korea’s CSAT Test

Or keep reading. . .

"All will be well if you use your mind for your decisions, and mind only your decisions." It costs thousands of dollars to run a website and your support matters. If you like the posts and videos, please consider a monthly pledge on Patreon. You may also consider a one-time donation to support my work. . .

. . . . M I N D . Y O U R . D E C I S I O N S . P U Z Z L E . . . . Answer To Fun Problem From Korea’s CSAT Test

(Pretty much all posts are transcribed quickly after I make the videos for them–please let me know if there are any typos/errors and I will correct them, thanks).

Let BAC = CAD = α. By the inscribed angle theorem, these inscribed angles will subtend arcs BC and CD with double the measure 2α. Chords that subtend arcs of equal measure will have the same length (to see why, construct triangles from the center of the circle. The central angles will have the same measure as the arcs, and all radii length are equal. So the two triangles will be congruent, and therefore the subtending chords will have equal length.)

Let the equal chords BC and CD have a length x.

Apply the Al-Kashi’s law of cosines to triangles ABC and ACD to get:

x2 = 52 + (3√5)2 – 2(5)(3√5) cos α x2 = 72 + (3√5)2 – 2(7)(3√5) cos α

Subtract the second equation from the first:

0 = 52 + 72 – 4(3√5) cos α 0 = 24 – 12√5 cos α 2/√5 = cos α

Substituting into the first equation, we get:

x2 = 52 + (3√5)2 – 2(5)(3√5)(2/√5) x2 = 25 + 45 – 60 x = √10

We can solve for sin α as:

(sin α)2 + (cos α)2 = 1 (sin α)2 + (2/√5)2 = 1 (sin α)2 + 4/5 = 1 sin α = 1/√5

(we can ignore the negative square root since we know the sine is positive for this acute angle α)

Finally we can apply the Al-Tusi’s law of sines to triangle ABC to calculate the circumradius R of the triangle:

BC/(sin α) = 2R √10/(1/√5) = 2R √50 = 2R (5√2)/2 = R

Therefore the circle’s radius is (5√2)/2 ≈ 3.536. What a nice problem!

Special thanks this month to:

Daniel Lewis Lee Redden Kyle

Thanks to all supporters on Patreon and YouTube!

References

Cornerstones of Math video https://www.youtube.com/watch?v=0WbDrYnrMbk

Images https://commons.wikimedia.org/wiki/File:Jamshid_al_Koshiy.jpg Leonid Kaydalov, CC BY 4.0 , via Wikimedia Commons https://commons.wikimedia.org/wiki/File:Stamps_of_Azerbaijan,_2009-861.jpg

PRESH TALWALKAR

I run the MindYourDecisions channel on YouTube, which has over 1 million subscribers and 200 million views. I am also the author of The Joy of Game Theory: An Introduction to Strategic Thinking, and several other books which are available on Amazon.

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