Russian Math Olympiad Problem | A Very Nice Geometry Challenge | 2 Different Methods
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- Опубліковано 1 кві 2024
- Russian Math Olympiad Problem | A Very Nice Geometry Challenge | 2 Different Methods
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Another solution without using trigonometry:
Locate center of semicircle, O.
Construct AC and BO, intersecting at E.
BO bisects angle AOC, and AO=BO=OC, so triangles AOB and BOC are congruent.
AC is perpendicular to BO, so angle AEO and angle ACD are both right angles.
Angle EAO = Angle CAD, so triangle AEO is similar to triangle ACD.
Since AD = 2AO, triangle ACD has sides twice as long as those of triangle AEO.
OE = half of CD = 7/2
BE = R - 7/2
Right triangles ABE and AOE share side AE.
By the Pythagorean theorem, AE = 15^2 - (R-7/2)^2
By the Pythagorean theorem, AE = R^2 - (7/2)^2
15^2 - (R-7/2)^2 = R^2 - (7/2)^2
Simplifying gives us 2R^2 -7R -225 = 0
Solving the quadratic equation gives us R = 25/2 or -9
R must be positive, so R = 25/2
Great solution ........ 👍
I'm a couple of weeks late, but... here goes:
1. Complete the circle. Let O represent its center. Draw radius BO, extending to the other side of the completed circle to create diameter BE (passing through O, of course.)
2. Draw AC, intersecting BE at F. By 'triangle inscribed in semicircle is a right triangle', establish that △ACD is a right triangle.
3. AD is a diameter of the circle/semicircle, therefore AD = 2r. Also, CD = 7, so, by Pythagoras:
AC² = (2r)² - 7². This can be reduced to:
AC = 2 • √(r² - (7/2)²).
4. Since AB = BC = 15, △ABC is isosceles, with base AC. Given this fact, we still need to establish that BO (actually, BOE) is the perpendicular bisector of AC.
We can do that by drawing one additional radius, OC. This will establish the congruency of △ABO and△CBO (SSS). Once we have that we can easily get that AF = CF along with the perpendicularity of BO and AC.
5. The last thing we need is the relationship between the chords AC (AFC) and BE (BFE), intersecting at F.
That relationship is:
AF x FC = BF x FE
Substituting (from #3) and assigning BF = a, we get:
√(r² - (7/2)²) • √(r² - (7/2)²) = (a) • (2r - a)
Simplifying this (I'll show a few intermediate steps) to get an equation in 2 variables, we first get:
r² - 49/4 = 2ar - a², and then,
r² - 2ar + a² = 49/4, and,
(r - a)² = (7/2)², or,
a = r - 7/2
6. We can get a second such equation (in a and r) by applying Pythagoras to △AFB:
a² + √(r² - (7/2)²)² = 15², simplifying to,
a² + r² - 49/4 = 225, or,
a² + r² = 225 + 49/4 = 949/4, i.e.,
a² + r² = 949/4
7. Substituting #5 (a = r - 7/2), we get:
(r - 7/2)² + r² = 949/4, or,
2r² - 7r + 49/4 = 949/4, then,
2r² - 7r - 900/4 = 0
8. Employing the quadratic formula yields:
r = (7 ± 43)/4, or,
r = 25/2
??
Now to watch the video to see if I got it right.
Cheers!
Got it!
Forgot about the cyclic quadrilateral theorem...
Oh, well. That makes my solution a 3rd method.
Thanks for the challenge!
R sin((PI - 2X)/2) = 7/2; and R sin(x/2) = 15/2, from the construction above. Then using trig identities, one gets 2R^2 - 7R - 225 = 0.
Posto α=ABC. Teorema del coseno 15^2+15^2-2*15*15*cosα=4r^2+7^2-2*7*2rcos(180-α)=4r^2-7^2(quest'ultima equazione perché ACD è rettangolo)..dalle 2 equazioni risulta r=12,5..cosα=-7/25..α=106,26
A shortcut in Math’s booster excellent second solution.
Ptolemy's theorem:
For a cyclic quadrilateral (that is, a quadrilateral inscribed in a circle), the product of the diagonals equals the sum of the products of the opposite sides.
AC BD = AB CD + BC AD => AC BD = 15•7 + 15• d ( d = diameter )
=> AC BD = 15(d+7)
=> AC²⋅BD²=15 ² (d+7)² => (d²-7² )(d²-15² )=15² (d+7)²
=> (d+7)(d-7)⋅(d²-15² )=15 ²(d+7)²
=> (d-7)⋅(d²-15² )=15² (d+7)
=> d³ -15² d - 7d²+7⋅15²=15²⋅d+7⋅15²
=> d³ - 7d² - 2•15² d =0
=> d² - 7d - 450 = 0 cause d>0
=> d = 25 or d = -18 ( is rejected )
So R=d/2 => R=25
Good morning from Greece .
I used the law of cosines method, but I realized that
Drop a perpendicular from O to AB and label the intersection E.
Hello. I find interesting geometry problem:
Line l touches the circumcircle of triangle ABC at point A. Points D and E
are such that CD and BE are perpendicular to l, and angles DAC and EAB are right angles. Prove that
BD and CE intersect at the height of triangle ABC from vertex A.
Please, can you solve it?)
You have: 2 arc sin 15/2r + arc sin 7/2r = π/2. Just solve it using trigonemtric indetities.
Хотел написать примерно то же самое 👍
This is very clever and excellent! To fill in and make it easier to follow: DB has been constructed.
excellent!
@@jimlocke9320
Thanks for filling in those details.
Yes, very clever solution.
The radius of the semi circle is r = 22.33
Excellent step-by-step explanation....🙂
15/2 = R sin (a/2)
7/2 = R sin (b/2)
2a + b = pi
b = pi - 2a
7/2 = R sin ((pi -2a)/2) = R cos a
7/2 = R (1 - 2 sin ^ 2 (a/2))
7/2 = R (1 - 2 (15/(2R)^2)
Équation du 2nd degré en R, etc.
awesome
Those answers that aren't possible, do they get meaning in the complex or another domain?