MIRB14

10 Questions
23 Answers

+1

1685

4

+115

Geometry proof

Let PA and PC be tangents from P to a circle. Let B and D be points on the circle such that B, D, and P are collinear. Prove that AB*CD=BC*DA.
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MIRB14 Mar 4, 2018

+1

1497

1

+115

In the figure, $PA$ is tangent to semicircle $SAR$, $PB$ is tangent to semicircle $RBT$, and $SRT$ is a straight line. If arc $AS$ is $58^\c

In the figure, PA is tangent to semicircle SAR, PB is tangent to semicircle RBT, and SRT is a straight line. If arc AS is 58 degrees and arc BT is 37 degrees, then find angle APB, in degrees.
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MIRB14 Feb 19, 2018

+3

3173

2

+115

In triangle $ABC$, $AB = 5$, $AC = 6$, and $BC = 7$. Circles are drawn with centers $A$, $B$, and $C$, so that any two circles are externall

In triangle ABC, AB=5, AC=6, and BC=7. Circles are drawn with centers A, B, and C, so that any two circles are externally tangent. Find the sum of the areas of the circles.
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MIRB14 Feb 7, 2018

+2

2281

1

+115

In regular octagon $ABCDEFGH$, $M$ and $N$ are midpoints of $\overline{BC}$ and $\overline{FG}$ respectively. Compute $[ABMO]/[EDCMO]$. ($[A

In regular octagon ABCDEFGH, M and N are midpoints of line BC and line FG respectively. Compute [ABMO]/[EDCMO]. ([ABCD] denotes the area of polygon ABCD.)

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MIRB14 Jan 29, 2018

+1

3139

4

+115

In square $ABCD$, $E$ is the midpoint of $\overline{BC}$, and $F$ is the midpoint of $\overline{CD}$. Let $G$ be the intersection of $\overl

In square ABCD, E is the midpoint of line BC, and F is the midpoint of line CD. Let G be the intersection of line AE and line BF. Prove that DG = AB.

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MIRB14 Jan 20, 2018

+1

3250

2

+115

The smaller square in the diagram has side length 1. Equilateral triangles are constructed on its four sides, as shown. What is the area of

The smaller square in the diagram has side length 1. Equilateral triangles are constructed on its four sides, as shown. What is the area of the larger square?
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MIRB14 Jan 17, 2018

+2

1613

4

+115

A circle is drawn that intersects all three sides of triangle PQR as shown below. Prove that if AB = CD = EF, then the center of the circle

A circle is drawn that intersects all three sides of triangle PQR as shown below. Prove that if AB = CD = EF, then the center of the circle is the incenter of triangle PQR.
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MIRB14 Dec 17, 2017

+2

2413

1

+115

Angle bisectors TX and UY of triangle TUV meet at point I. Find all possible values of angle V if angle TIU = 109 degrees. As your answer,

Angle bisectors TX and UY of triangle TUV meet at point I. Find all possible values of angle V if angle TIU = 109 degrees. As your answer, enter the number of degrees in angle V. If you find more than one possibility, list the possible values in increasing read more ..

MIRB14 Dec 13, 2017

+3

3896

3

+115

A square $DEFG$ varies inside equilateral triangle $ABC,$ so that $E$ always lies on side $\overline{AB},$ $F$ always lies on side $\overlin

A square varies inside equilateral triangle so that always lies on side always lies on side and always lies on side The point starts on side and ends read more ..

MIRB14 Dec 8, 2017

+2

2198

1

+115

In the staircase-shaped region below, all angles that look like right angles are right angles, and each of the eight congruent sides marked

In the staircase-shaped region below, all angles that look like right angles are right angles, and each of the eight congruent sides marked with a tick mark have length 1 foot. If the region has area 53 square feet, what is the number of feet in the perimeter read more ..

MIRB14 Dec 4, 2017

+115

+2

Thanks, but is there a way to not use trig?

MIRB14Mar 4, 2018

+115

+1

Nevermind I solved it...

MIRB14Feb 19, 2018

+115

+1

triangle BCD is similar to triangle KJL because of SSS similarity.

triangle FGH is not similar to triangle SRQ.

triangle CDE is not similar to triangle MLC.

triangle LMN is not similar to triangle FEG.

MIRB14Feb 19, 2018