2
votes
Accepted
Interesting property about triangles I don't know about.
The answer is on the comments.
According to Calvin Lin it's a "well known" olympiad problem.
A more elegant proof that mine involves 60 degrees rotations. IE, BCC' and BA'A are the same ...
- 171
2
votes
Accepted
How to show that given two acute angles, the sine ratio of the greater angle is greater than the sine ratio of lesser angle?
Let us consider $\triangle{ABC}$ such that $\angle{ABC}=\psi$ and $\angle{ACB}=\theta$.
Let $D$ be a point on $BC$ such that $AD\perp BC$.
We use the following claim. A proof of the claim is written ...
- 135k
1
vote
Find the area of BGHF
The following is your diagram with the horizontal line $JK$ through $G$, plus the vertical lines $GL$ and $HM$, added to it:
Define $s = \sqrt{40}$ to be the square's side lengths. Next, since $\...
- 46.5k
1
vote
Surface area of sphere coming out as $\pi^2 r^2$
Comment only.
Not clear. You wanted to extend to 3d from the following 2d situation? May be Pappu's thm would help if location of CG is known.
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1
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Simple bisection geometry
Note that
$$\angle DIK = \angle DAF = \frac 12 \angle CAB = \angle CAD = \angle IKD,$$
hence $DK=DI$. Therefore the projection of $D$ onto $IK$ is the midpoint of $IK$. Note that $F$ is the projection ...
- 10.8k
1
vote
Area of a right angled triangle given the dimensions of an inscribed rectangle
We can show that the configuration forces two of the rectangle vertices to coincide, as shown in the figure below, where the two rectangles are $APQR$ and $STUV$.
Let $\overline{TB}=x$. Then $\...
- 7,192
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How to show that given two acute angles, the sine ratio of the greater angle is greater than the sine ratio of lesser angle?
Consider a circle of radius $1$ centered at the point $O$. Fix any point $A$ on the circle. Fix a point $B$ on the circle such that $\angle BOA=\frac{\pi}2-\psi$. Let $BB'$ be the altitude of the ...
- 85.7k
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